We note that \(y=\frac { 3 }{ 4 } \sqrt { { x }^{ 2 }-16 } \Rightarrow \frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 9 } =1\). So, the given curve is a portion of the hyperbola \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 9 } =1\) between the lines y = 1 and y = 6 and it lies above the x-axis.

The region to be revolved is sketched.
Since revolution is made about y-axis, we write the equation of the portion of the hyperbola as \(x=\frac { 4 }{ 3 } \sqrt { 9+{ y }^{ 2 } } .\) 
So, the volume of the solid generated is given by 
\(V=\pi \int _{ 1 }^{ 6 }{ { x }^{ 2 }dy=\pi \int _{ 1 }^{ 6 }{ { \left( \frac { 4 }{ 3 } \sqrt { 9+{ y }^{ 2 } } \right) }^{ 2 } } } dy=\pi \left( \frac { 16 }{ 9 } \right) \int _{ 1 }^{ 16 }{ \left( 9+{ y }^{ 2 } \right) dy } \)
\(=\pi \left( \frac { 16 }{ 9 } \right) { \left( 9y+\frac { { y }^{ 3 } }{ 3 } \right) }_{ 1 }^{ 6 }=\pi \left( \frac { 16 }{ 9 } \right) \left[ (54+72)-(9+\frac { 1 }{ 3 } ) \right] =\frac { 5600 }{ 27 } \pi \)

The ellipse is symmetric about both the axes. The major axis lies along x-axis. The region to be revolved is sketched
Hence, the required volume is given by
\(V=\pi \int _{ -a }^{ a }{ { y }^{ 2 }dx } =\pi \int _{ -a }^{ a }{ \frac { { b }^{ 2 } }{ { a }^{ 2 } } \left( { a }^{ 2 }-{ x }^{ 2 } \right) dx } \)
\(=\frac { 2\pi { b }^{ 2 } }{ { a }^{ 2 } } \int _{ 0 }^{ a }{ ({ a }^{ 2 }-{ x }^{ 2 })dx } \) since the integrand is an even function
\(=\frac { { 2\pi b }^{ 2 } }{ { a }^{ 2 } } { \left( { a }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=\frac { 2\pi { b }^{ 2 } }{ 3 } ={ \left( { a }^{ 2 }-\frac { { a }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=\frac { 2\pi { b }^{ 2 } }{ 3 } \left( \frac { { 2a }^{ 3 } }{ 3 } \right) =\frac { 4\pi { ab }^{ 2 } }{ 3 } \)

Consider the triangular region in the first quadrant which is bounded by the line y = \(\frac{r}{h}\)x. x-axis, the lines x = 0 and x = h. revolving the region about the x-axis, we get a cone of base radius r and height h.

Hence, the volume of the cone is given by
\(\\ \\ V=\pi \int _{ 0 }^{ h }{ { y }^{ 2 }dx=\pi \int _{ 0 }^{ h }{ { \left( \frac { r }{ h } x \right) }^{ 2 } } dx=\pi { \left( \frac { r }{ h } \right) }^{ 2 } } { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ h }=\frac { \pi { r }^{ 2 }h }{ 3 } \)

We recall that the equation of the tangent to the circle x² + y² = a² at (x₁, y₁) is xx₁+yy₁ = a². So, the equation of the tangent to the circle x² + y² + = 4 at (1, \(\sqrt 3\)) is x+y\(\sqrt 3\) = 4; that is y = -\(\frac{1}{\sqrt 3}\)( x - 4). The tangent meets the x-axis at the point (4,0). The slope of the tangent is-\(\frac{1}{\sqrt 3}\). So the slope of the normal is \(\sqrt 3\) and hence equation of the normal is y-\(\sqrt 3\) =\(\sqrt 3\)(x-1); that is y = \(\sqrt 3\) x and it passes through the origin. The area to be found is shaded in the adjoining figure. It can be found by two methods.

Equation of AB is \(\frac { y-1 }{ 2-1 } =\frac { x+1 }{ 3+1 } or\quad y=\frac { 1 }{ 4 } (x+5)\)
Equation of BC is \(\frac { y-5 }{ 2-5 } =\frac { x-0 }{ 3-0 } or\quad y=-x+5\)
Equation of AC is \(\frac { y-1 }{ 5-1 } =\frac { x+1 }{ 0+1 } or\quad y=4x+5\)
\(\therefore\) Area of \(\Delta\)ABC = Area DACO+ Area of OCBE − Area of DABE
\(=\int _{ -1 }^{ 0 }{ (4x+5)dx+\int _{ 0 }^{ 3 }{ (-x+5)dx-\frac { 1 }{ 4 } \int _{ -1 }^{ 3 }{ (x+5)dx } } } \)
\(\\ \\ \\ ={ \left[ \frac { { 4x }^{ 2 } }{ 2 } +5x \right] }_{ -1 }^{ 0 }+{ \left[ -\frac { { x }^{ 2 } }{ 2 } +5x \right] }_{ 0 }^{ 3 }-\frac { 1 }{ 4 } { \left[ \frac { { x }^{ 2 } }{ 2 } +5x \right] }_{ -1 }^{ 3 }\)
\(=0-(+2-5)+\left( -\frac { 9 }{ 2 } +15 \right) -0-\frac { 1 }{ 4 } \left[ \frac { 9 }{ 2 } +15 \right] +\frac { 1 }{ 4 } \left[ \frac { 1 }{ 2 } -5 \right] =\frac { 15 }{ 2 } \)

First, we get the points of intersection of the parabolas. For this, we solve y² x = 4 and x² y = 4 simultaneously Eliminating y between them, we get x⁴ = 64x and so x = 0 and x = 4. Then the points of intersection are (0, 0) and (4, 4). The required region is sketched.

Viewing in the direction of y -axis, the equation of the upper boundary is y = 2\(\sqrt x\) for 0\(\le x \le\) 4 and the equation of the lower boundary is \(y =\frac {x^2}{4}\)for \(0 \leq x \leq 4\). So, the required area \(\Delta\) is
\(A=\int_{0}^{4}\left(y_{U}-Y_{L}\right) d x=\int_{0}^{4}\left(2 \sqrt{x}-\frac{x^{2}}{4}\right) d x=\left[2\left(\frac{2 x^{3 / 2}}{3}\right)-\frac{x^{3}}{12}\right]_{0}^{4}=\left[2\left(\frac{2 \times 8}{3}\right)-\frac{64}{12}\right]-0=\frac{16}{3}\)

The given curve is \(y=\begin{cases} cosx,0\le x\le \frac { \pi }{ 2 } \\ -cosx,\frac { \pi }{ 2 } \le x\le \pi \end{cases}\)
It lies above the x − axis. The required area is sketched. So, the required area is given by

\(A=\int _{ 0 }^{ \pi }{ ydx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ cosxdx } +\int _{ \frac { \pi }{ 2 } }^{ \pi }{ (-cosx)dx } ={ [sin\quad x] }_{ 0 }^{ \frac { \pi }{ 2 } }-{ [sin\quad x] }_{ \frac { \pi }{ 2 } }^{ \pi } } \)
= [1-0]-[0-1] = 2

The equation of the parabola is ( y + 2)² = −(x−9). The parabola crosses the y- axis at (0, −5) and (0, 1). The vertex is at (9, −2) and the axis of the parabola is y = −2. The required area is sketched
Viewing in the positive direction of x-axis, and making horizontal strips, the required area A is given by

\(A=\int _{ -5 }^{ 1 }{ xdy } =\int _{ -5 }^{ 1 }{ (5-4y-{ y }^{ 2 })dy={ \left[ 5y-2{ y }^{ 2 }-\frac { { y }^{ 3 } }{ 3 } \right] }_{ -5 }^{ 1 }=\frac { 8 }{ 3 } -\left( -\frac { 100 }{ 3 } \right) =36 } \)

Given \(\int _{ 0 }^{ \infty }{ { e }^{ -a{ x }^{ 2 } }{ x }^{ 3 }dx=32,\alpha >0} \)

\(\Rightarrow \int _{ 0 }^{ \infty }{ { e }^{ { -\alpha x }^{ 2 } } } .{ x }^{ 2 }.dx=32\)
\(put\ t= { x }^{ 2 }\)
\(\Rightarrow dt=2x\ dx\)
\(\\ \Rightarrow \frac { dt }{ 2 } = xdx\)
\( \int _{ 0 }^{ \infty }{ { e }^{ -\alpha t }\frac{dt}{2}} \frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ { e }^{ -\alpha t }.tdt } \)
\( \frac { 1 }{ 2 } \times\frac{1} { \alpha}^{ 2 } \) ...... (1)
Given \(\int^x_0e^{-ax^2}x^3 dx = 32\)
By (1),
\(\Rightarrow \frac { 1! }{ { \alpha }^{ 2 } } =32\times 2\left[ \because \int _{ 0 }^{ \infty }{ { e }^{ -ax }{ x }^{ n }dx=\frac { n! }{ { a }^{ n+1 } } } \right] \)
\(\frac{1}{2}\times \frac{1}{\alpha^2}= 32 \Rightarrow \frac {1}{\alpha ^2}= 64\\ \alpha ^2 = \frac{1}{64}\)
\(\Rightarrow \alpha =\frac { 1 }{ 8 } \)

\(=\int _{ 0 }^{ \pi /2 }{ { e }^{ -tan\quad x }{ sec }^{ 6 }xdx\ \left[ \because sec\ x=\frac { 1 }{ cos\quad x } \right] } \)
\(Put\quad t=tan\quad x\Rightarrow dt={ sec }^{ 2 }xdx\)
\(=\int _{ 0 }^{ \pi /2 }{ { e }^{ -tanx }{ sec }^{ 4x }{ sec }^{ 2x }dx } \)
\(=\int _{ 0 }^{ \pi /2 }{ { e }^{ -tanx }{ (1+{ tan }^{ 2 }) }^{ 2 } } { sec }^{ 2 }xdx\)
\(put\quad t=tanx\Rightarrow dt={ sec }^{ 2 }xdx\)

\(=\int _{ 0 }^{ \infty }{ { e }^{ -t }{ (1+{ t }^{ 2 }) }^{ 2 } } dt\)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -t }(1+{ t }^{ 4 }+{ 2t }^{ 2 })dt } \)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -t }(1)dt+\int _{ 0 }^{ \infty }{ { e }^{ -t }{ t }^{ 4 }dt+2 } } \int _{ 0 }^{ \infty }{ { e }^{ -t }{ t }^{ 2 }dt } \)
\(=0!+4!+2(2!)\left[ \because \int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n }dx=n! } \right] \)
\(=1+4\times 3\times 2\times 1+2\times 2=1+24+4\)
\(\therefore I=29\)

Applying integration by parts, we get
\(\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n } } dx={ \left[ { x }^{ n }(-{ e }^{ -x }) \right] }_{ 0 }^{ \infty }-\int _{ 0 }^{ \infty }{ (-{ e }^{ -x }) } ({ nx }^{ n-1 })dx=n\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n-1 }dx } \)
\(Let\quad { I }_{ n }=\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n }dx.\ Then,\ { I }_{ n }={ nI }_{ n-1 } } \)
So, we get I_n= n(n−1) I_n−2.
Proceeding in this way, we get ultimately,
I_n = n(n-1)(n-2)...(2)(1)I₀.
But, \({ I }_{ 0 }=\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ 0 }dx } ={ \left( -{ e }^{ -x } \right) }_{ 0 }^{ \infty }=0+1=1.\) So, we get n = n(n-1)(n-2)(2)(1) = n!
Hence, we get
\(\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n }dx=n! } \), where n is a nonnegative integer.

Put x = 2a cos²\(\theta\).
Then, dx = -4a cos \(\theta\) sin \(\theta\)d\(\theta\)
when x = 0, 2a cos²\(\theta\) = 0 and so \(\theta\) = \(\frac{\pi}{2}\)
When x = 2a, 2a cos²\(\theta\) = 2a and so \(\theta\) = 0
Hence, we get
\(I=\int _{ 0 }^{ 2a }{ { x }^{ 2 }\sqrt { 2ax-{ x }^{ 2 } } dx } \)
\(\int _{ \frac { \pi }{ 2 } }^{ 0 }{ { 4a }^{ 2 }{ cos }^{ 2 }\theta \sqrt { { 4a }^{ 2 }{ cos }^{ 2 }\theta -4{ a }^{ 2 }{ cos }^{ 4 }\theta } } (-4a\ cos\theta sin\ \theta )d\theta \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { 4a }^{ 2 }{ cos }^{ 2 }\theta\ 2a\ cos\ \theta sin\ \theta (4a\ cos\ \theta sin\theta )d\theta } \)
\(=32{ a }^{ 4 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 4 }\theta { sin }^{ 2 }\theta d\theta } \)
\(=32{ a }^{ 4 }\times \frac { 1 }{ 6 } \times \frac { 3 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 2 } =\pi { a }^{ 4 }\) 

\(put\sqrt { t } =z\Rightarrow \frac { 1 }{ 2\sqrt { t } } dt=dz\)
\(\Rightarrow \frac { 1 }{ 2z } dt=dz\Rightarrow dt=2zdz\)

\(\therefore I=\int _{ 0 }^{ sin\quad x }{ 2z{ sin }^{ -1 }zdz } +\int _{ 0 }^{ cosx }{ 2z{ cos }^{ -1 }zdz } ...(1)\)
= I₁+ I₂ 
\(=2\int _{ 0 }^{ x }{ \theta sin\theta cos\theta d\theta +2\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta cos\theta (-sin\theta )d\theta } } \)
\(=\int _{ 0 }^{ x }{ \theta 2sin\theta cos\theta d\theta -\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta 1sin\theta cos\theta d\theta } } \)
\(=\int _{ 0 }^{ x }{ \theta sin2\theta d\theta -\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta sin2\theta d\theta } } \)
\([\because sin2\theta =sin\theta cos\theta ]\)
\(=\int _{ 0 }^{ x }{ \theta sin2\theta d\theta \int _{ x }^{ \frac { \pi }{ 2 } }{ \theta sin2\theta d\theta } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \theta sin2\theta d\theta } } \)
\(\left[ \because \int _{ a }^{ c }{ f(x)dx+\int _{ c }^{ b }{ f(x)dx=\int _{ a }^{ b }{ f(x)dx } } } \right] \)
\(={ \left[ -\frac { \theta cos2\theta }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }+\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { cos2\theta }{ 2 } d\theta } \)

\(={ \left[ -\frac { \theta cos2\theta }{ 2 } +\frac { sin2\theta }{ 4 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(=\frac { -\frac { \pi }{ 2 } cos\pi }{ 2 } +\frac { sin\pi }{ 4 } -(0+0)\)
\(=\frac { -\frac { \pi }{ 2 } (-1) }{ 2 } +\frac { 0 }{ 4 } =\frac { \pi }{ 4 } \)

\(=\int _{ 0 }^{ \frac { 3 }{ 5 } }{ -(5x-3)dx+\int _{ \frac { 3 }{ 5 } }^{ 1 }{ -(5x-3)dx } } \)
\(={ { \left[ 3x-\frac { { 5x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ \frac { 3 }{ 5 } }+\left[ \frac { { 5x }^{ 2 } }{ 2 } -3x \right] }_{ 3 }^{ 1 }\)
\(=\frac { 9 }{ 5 } -\frac { 5 }{ 2 } \left( \frac { 9 }{ 25 } \right) -0+\frac { 5 }{ 2 } -3-\frac { 5 }{ 2 } \left( \frac { 9 }{ 25 } \right) +\frac { 9 }{ 5 } \)
\(=\frac { 9 }{ 5 } -\frac { 9 }{ 10 } +\frac { 5 }{ 2 } -3-\frac { 9 }{ 10 } +\frac { 9 }{ 5 } \)
\(=\frac { 18-9+25-30-9+18 }{ 10 } =\frac { 13 }{ 10 } \)

Let \(I=\int _{ 0 }^{ 2\pi }{ { sin }^{ 4 }x{ cos }^{ 3 }xdx } \quad ...(1)\)
By the property \(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=\int _{ 0 }^{ a }{ f(a-x)dx } } \right] \)
we get 
\(I=\int _{ 0 }^{ 2\pi }{ { \left[ sin(2\pi -x) \right] }^{ 4 } } { \left[ cos(2\pi -x) \right] }^{ 3 }dx\)
\(=\int _{ 0 }^{ 2\pi }{ { [-sin\quad x] }^{ 4 }{ [-cos\quad x] }^{ 3 }dx } \)
\(=-\int _{ 0 }^{ 2\pi }{ { sin }^{ 4 }x{ cos }^{ 3 }xdx\quad ..(2) } \)
\(\therefore (1)+(2)\Rightarrow \)
\(2I=\int _{ 0 }^{ 2\pi }{ { sin }^{ 4 }x{ cos }^{ 3 }xdx } \)
\(-\int _{ 0 }^{ 2\pi }{ { sin }^{ 4 }x{ cos }^{ 3 }xdx=0 } \)
\(\Rightarrow I=0\)

\(=\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { x }^{ 5 }dx } +\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx\quad dx+ } \int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { tan }^{ 3 }xdx+ } \int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ dx } \)
\(=0+0+0+{ [x] }_{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }\)
\(=\frac { \pi }{ 2 } -\left( -\frac { \pi }{ 2 } \right) \)
\(=\frac { \pi }{ 2 } +\frac { \pi }{ 2 } \)
\(=\pi \)
\(\\ \\ \\ \\ \\ \because \int { { x }^{ 5 } } dx\) is an odd function \(\int { { x } } \) cos xdx is an odd function \(\int { { tan }^{ 3 } } \)xdx is an odd function

Put I = \(\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { dx }{ a^{ 2 }sin^{ 2 }x+b^{ 2 }cos^{ 2 }x } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sec^{ 2 }xdx }{ a^{ 2 }tan^{ 2 }x+b^{ 2 } } \)
Put u = tan x 
Then du = sec² x dx
When x = 0 , we have u = tan 0 = 0 
When x = \(\frac{\pi}{4}\), we have u = tan\(\frac{\pi}{4}\) = 1
\(\therefore I=\int _{ 0 }^{ 1 } \frac { du }{ a^{ 2 }u^{ 2 }+b^{ 2 } } =\frac { 1 }{ a^{ 2 } } \int _{ 0 }^{ 1 } \frac { du }{ u^{ 2 }+\left( \frac { b }{ a } \right) ^{ 2 } } =\frac { 1 }{ a^{ 2 } } { \left[ \frac { a }{ b } tan^{ -1 }(\frac { au }{ b } ) \right] }_{ 0 }^{ 1 }=\frac { 1 }{ ab } tan^{ -1 }\left( \frac { a }{ b } \right) \)
We derive some more properties of definite integrals

\(\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ sin^{ 4 }x+cos^{ 4 }x } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ (sin^{ 2 }x+cos^{ 2 }x)-2sin^{ 2 }xcos^{ 2 }x } \)
\(\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ (1-\frac { 1 }{ 2 } (2sinxcosx)^{ 2 } } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { 2sin2xdx }{ 2-sin^{ 2 }2x } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { 2sin2xdx }{ 1+cos^{ 2 }2x } \)
Put u = cos 2x, Then, du = −2sin 2x dx
When x = 0 , we have u = cos 0 = 1. When x = \(\frac{\pi}{4}\) we have u = cos\(\frac{\pi}{4}\) = 0
∴ I =  \(\int^{0}_{1} \frac {-du}{1+u^2}\) = \(\int^{0}_{1} \frac {du}{1+u^2}\) = [tan^-1 u\(]^1_0\) = \(\frac{\pi}{4}\)

We know that the greatest integer function [x] is the largest integer less than or equal to x. In other words, it is defined by [x] = n, if n \(\le\) x \(\le\) (n +1) , where n is an integer.
So, we get \(\begin{cases} \begin{matrix} 0 & if & 0\le x\le 1 \end{matrix} \\ \begin{matrix} 1 & if & 1\le x<\sqrt { 2 } \end{matrix} \\ \begin{matrix} 2 & if & \sqrt { 2 } \le x\le 1.5 \end{matrix} \end{cases}\)
We note that the above function is not continuous on [0,1.5]
But, it is continuous in each of the sub-intervals [0, 1] , [1, \(\sqrt2\)] and [\(\sqrt2\), 1.5]; that is, it is piece-wise continuous on [0, 1.5].

\(\int _{ 0 }^{ 1.5 }{ \left[ { x }^{ 2 } \right] dx=\int _{ 0 }^{ 1 }{ [{ x }^{ 2 }] } dx+\int _{ 1 }^{ \sqrt { 2 } }{ [{ x }^{ 2 }]dx+ } } \int _{ \sqrt { 2 } }^{ 1.5 }{ \left[ { x }^{ 2 } \right] } dx=\int _{ 1 }^{ \sqrt { 2 } }{ 1dx } +\int _{ \sqrt { 2 } }^{ 1.5 }{ 2dx } \)
\(={ 0+(x) }_{ 1 }^{ \sqrt { 2 } }+{ (2x) }_{ \sqrt { 2 } }^{ 1.5 }=(\sqrt { 2 } -1)+(3-2\sqrt { 2 } )=2-\sqrt { 2 } \)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\sqrt { tan\ x } +\sqrt { cot\ x } )dx } \) Then we get
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \sqrt { \frac { sinx }{ cosx } } +\sqrt { \frac { cosx }{ sinx } } \right) dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx+cosx }{ \sqrt { sinxcosx } } dx } =\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx+cosx }{ \sqrt { 2sinxcosx } } dx } \)
\(=\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{2 } }{ \frac { (sinx+cosx)dx }{ \sqrt { 1-sinx-cosx{ ) }^{ 2 } } } } \)
Put u = sin x − cos x 
Then, du = (cos x + sin x)dx .
When x = 0, u = −1
When x =\(\frac{\pi}{2}\), u = 1
\(\therefore I=\sqrt { 2 } \int _{ -1 }^{ 1 }{ \frac { du }{ \sqrt { 1-{ u }^{ 2 } } } =\sqrt { 2 } { [{ sin }^{ -1 }u] }_{ -1 }^{ 1 }=\sqrt { 2 } \left[ { sin }^{ -1 }(1)-{ sin }^{ -1 }(-1)) \right] } =\pi \sqrt { 2 } \)

Let I = \(\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \frac { { sin }^{ -1 }x }{ { (1-{ x }^{ 2 }) }^{ \frac { 3 }{ 2 } } } dx } \)
Put u = sin^-1 x. Then, x = sin u and so, du =\(\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx\)
When x = 0, u = 0 
When \(x=\frac { 1 }{ \sqrt { 2 } } ,u=\frac { \pi }{ 4 } .\)
\(\therefore I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { u }{ { cos }^{ 2 }u } } du=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ u{ sec }^{ 2 }udu={ [utanu] }_{ 0 }^{ \frac { \pi }{ 4 } } } -\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tan\quad udu={ [utanu] }_{ 0 }^{ \frac { \pi }{ 4 } }+{ \left[ logcosu \right] }_{ 0 }^{ \frac { \pi }{ 4 } } } \)
\(=\frac { \pi }{ 4 } +log\frac { 1 }{ \sqrt { 2 } } =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } log2\)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { cos\theta }{ (1+sin\theta )(2+sin\theta ) } } d\theta \)
Put u = 1 + sin\(\theta\)
Then, du = cos\(\theta\) d\(\theta\)
When \(\theta\) = 0, u = 1 
When \(\theta =\frac{\pi}{2}, u=2\)
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { du }{ u(1+u) } } =\int _{ 1 }^{ 2 }{ \frac { (1+u)-u }{ u(1+u) } du } =\int _{ 1 }^{ 2 }{ \left( \frac { 1 }{ u } -\frac { 1 }{ 1+u } \right) du=[logu-log(1+u)]_{ 1 }^{ 2 } } \)
\(=(log2-log3)-(log1-log2)=2log2-log3=log\frac { 4 }{ 3 } .\)

Here a = 1, b = 2,f(x) = 4x²-1
\(\therefore f(a+(b-a)\frac { r }{ n } )=f(1+1(\frac { r }{ n } ))\)
\(=f\left( 1+\frac { r }{ n } \right) \)
\(=4{ \left( 1+\frac { r }{ n } \right) }^{ 2 }-1=4\left( 1+\frac { { r }^{ 2 } }{ { n }^{ 2 } } +\frac { 2r }{ n } \right) -1\)
\(=4+\frac { { 4r }^{ 2 } }{ { n }^{ 2 } } +\frac { 8r }{ n } -1=3+\frac { { 4r }^{ 2 } }{ { n }^{ 2 } } +\frac { 8r }{ n } \)
\(\int _{ a }^{ b }{ f(x)dx= } \underset { n\rightarrow \infty }{ lim } \frac { b-a }{ n } \sum _{ r=1 }^{ n }{ f } \left( a+(b-a)\frac { r }{ n } \right) \)
\(\therefore \int _{ 1 }^{ 2 }{ ({ 4x }^{ 2 }-1)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \left( 3+\frac { { 4r }^{ 2 } }{ { n }^{ 2 } } +\frac { 8r }{ n } \right) } \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ 3+\frac { 1 }{ n } } \sum _{ r=1 }^{ n }{ \frac { { 4r }^{ 2 } }{ { n }^{ 2 } } } +\frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { 8r }{ n } } \right] \)
\(=\left[ \underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } .3n+\frac { 1 }{ n } \frac { 4 }{ { n }^{ 2 } } ({ 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ n }^{ 2 })+\frac { 1 }{ n } .\frac { 8 }{ n } (1+2+3....+n) \right] \)\(=\underset { n\rightarrow \infty }{ lim } \left[ 3+\frac { 4 }{ { n }^{ 3 } } \frac { n(n+1)(2+1) }{ 6 } +\frac { 8 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } \right] \)
\(\left[ \because \sum { r } =\frac { n(n+1) }{ 2 } \sum { { r }^{ 2 }=\frac { n(n+1)(2n+1) }{ 6 } } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ 3+\frac { 4 }{ { n }^{ 3 } } \frac { { n }^{ 3 }(1+\frac { 1 }{ n } )(2+\frac { 1 }{ n } ) }{ 6 } +\frac { 8 }{ { n }^{ 2 } } \frac { { n }^{ 2 }(1+\frac { 1 }{ n } ) }{ 2 } \right] \)
\(=[3+\frac { 2 }{ 3 } (1+0)(2+0)+4(1+0)]\)
\([\because when\quad n\rightarrow \infty ,1/n\rightarrow 0]\)
\(=3+\frac { 4 }{ 3 } +4=\frac { 9+4+12 }{ 3 } =\frac { 25 }{ 3 } \)
\(\therefore \int _{ 1 }^{ 2 }{ (4{ x }^{ 2 }-1)dx=\frac { 25 }{ 3 } } \)

Here a = 0, b = 1,f (x) = 5x + 4
\(\therefore f(a+(b-a)\frac { r }{ n } )=f\left( 0+1(\frac { r }{ n } ) \right) =f(\frac { r }{ n } )=5(\frac { r }{ n } )+4\)
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { b-a }{ n } \sum _{ r=1 }^{ n }{ f(a+(b-a)\frac { r }{ n } } \)
\(\therefore \int _{ 0 }^{ 1 }{ (5x+4)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \left( \frac { 5r }{ n } +4 \right) } \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { 5r }{ n } +\frac { 1 }{ n } \sum _{ r=1 }^{ n }{ 4 } } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 1 }{ n } .\frac { 5 }{ n } .(1+2+3+...+n)+\frac { 1 }{ n } .4n \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 5 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } +4 \right] \)
\([\because \sum { r=\frac { n(n+1) }{ 2 } ;\sum { { r }^{ 2 }=\frac { n(n+1)(2n+1) }{ 6 } } } ]\)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 5 }{ { n }^{ 2 } } { n }^{ 2 }\frac { (1+\frac { 1 }{ n } ) }{ 2 } +4\)
\(=\frac { 5 }{ 2 } (1+0)+4=\frac { 5 }{ 2 } +4\)
\([wehen\quad n\rightarrow \infty ,1/n\quad \rightarrow 0]\)
\(=\frac { 5+8 }{ 2 } =\frac { 13 }{ 2 } \)
\(\therefore \int _{ 0 }^{ 1 }{ (5x+4)dx } =\frac { 13 }{ 2 } \)

We use the formula
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { b-a }{ n } \sum _{ r=1 }^{ n }{ f } \left( a+(b-a)\frac { r }{ n } \right) \)
Here f(x) = 2x²+3, a = 1 and b = 4
So, we get
\(f\left( a+(b-a)\frac { r }{ n } \right) =f\left( 1+(4-1)\frac { r }{ n } \right) =f\left( 1+\frac { 3r }{ n } \right) =2{ \left( 1+\frac { 3r }{ n } \right) }^{ 2 }+3=5+\frac { { 18r }^{ 2 } }{ { n }^{ 2 } } +\frac { 12r }{ n } \)Hence, we get
\(\int _{ 1 }^{ 4 }{ ({ 2x }^{ 2 }+3) } dx=\underset { n\rightarrow \infty }{ lim } \frac { 3 }{ n } \sum _{ r=1 }^{ n }{ \left( 5+\frac { { 18r }^{ 2 } }{ { n }^{ 2 } } +\frac { 12r }{ n } \right) } =\underset { n\rightarrow \infty }{ lim } \left[ \frac { 15 }{ n } \sum _{ r=1 }^{ n }{ 1+\frac { 54 }{ { n }^{ 3 } } \sum _{ r=1 }^{ n }{ { r }^{ 2 } } +\frac { 36 }{ { n }^{ 2 } } \sum _{ r=1 }^{ n }{ r } } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 15 }{ n } n+\frac { 54 }{ { n }^{ 3 } } ({ 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ n }^{ 2 })+\frac { 36 }{ { n }^{ 2 } } (1+2+..+n) \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ 15+\frac { 54 }{ { n }^{ 3 } } \frac { n(n+1)(2n+1) }{ 6 } +\frac { 36 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ 15+9\left( 1+\frac { 1 }{ n } \right) \left( 2+\frac { 1 }{ n } \right) +18\left( 1+\frac { 1 }{ n } \right) \right] \)
= 15+9(1+ 0)(2 + 0) +18(1+ 0) = 51.

Here a = 1, b = 1.5, n = 5 and f(x) = 2-x
So, the width of each subinterval is
\(h=\Delta x=\frac { b-a }{ n } =\frac { 1.5-1 }{ 5 } =\frac { 0.5 }{ 5 } =0.1\)
The partition of the interval is given by
x₀ = 1 
x₁ = 1.1
x₂ = 1.2
x₃ = 1.3
x₄ = 1.4
x₅ = 1.5
Mid-point rule for Riemann sum with equal widh \(\Delta x\) is
\(S=\left[ f\left( \frac { { x }_{ 0 }+{ x }_{ 1 } }{ 2 } \right) +f\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \right) +f\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } \right) +f\left( \frac { { x }_{ 3 }+{ x }_{ 4 } }{ 2 } \right) +f\left( \frac { { x }_{ 4 }+{ x }_{ 5 } }{ 2 } \right) \right] \Delta x\)
\(=\left[ f\left( \frac { 1+1.1 }{ 2 } \right) +f\left( \frac { 1.1+1.2 }{ 2 } \right) +f\left( \frac { 1.2+1.3 }{ 2 } \right) +f\left( \frac { 1.3+1.4 }{ 2 } \right) +f\left( \frac { 1.4+1.5 }{ 2 } \right) \right] 0.1\)
= [f(1.05)+f(1.15)+f(1.25)+f(1.35)+f(1.45)](0.1)
= 0.95+0.85+0.75+0.65+0.55) \(\times\)(0.1)
= (2-1.05)+(2-1.15)+(2-1.25)+(2-1.35)+(2-1.45)\(\times\)(0.1)
= 3.75\(\times\)0.1 = 0.375

Here a = 1, b = 1.5, n = 5, f(x) = x²
So, the width of each subinterval is
\(h=\Delta x=\frac { b-a }{ n } =\frac { 1.5-1 }{ 5 } =\frac { 0.5 }{ 5 } =0.1\)
The right end rule for Riemann sum with equal width \(\Delta\)x
x₁ = 1.1
x₂ = 1.2
x₃ = 1.3
x₄ = 1.4
x₅ = 1.5
The right - end rule for Riemann sum with equal width Ax is
S = [fx₁)+f(x₂)+f(x₃)+......+f(x_n)]\(\Delta\)x
S = [f(1.1)+f(1.2)+f(1.3)+f(1.4)+f(1.5)](0.1)
= (1.21+1.44+1.69+1.96+2.25) \(\times\)(0.1)
= (8.55)\(\times\)(0.1)
= 0.855

To find \(\int _{ 1 }^{ 1.5 }{ xdx } \) in  {1.1, 1.2, 1.3, 1.4, 1.5}.
Here a = 1,
b = 1.5, n = 5, f(x) = x
\(\therefore h=\Delta x=\frac { b-a }{ n } =\frac { 1.5-1 }{ 5 } =\frac { 0.5 }{ 5 } =0.1\)
The partition of the interval is given by 
x₀ = 1
x₁ = x₀ + h = 1 + 0.1 = 1.1
x₂ = x₁ + h = 1.1 + 0.1 = 1.2
x₃ = x₂ + h = 1.2 + 0.1 = 1.3
x₄ = x₃ + h = 1.3 + 0.1 = 1.4
x₅ = x₄ + h = 1.4 + 0.1 = 1.5
The left end rule for Riemann sum with equal width \(\Delta\)x is
\(\int _{ a }^{ b }{ xdx } =[f({ x }_{ 0 })+f({ x }_{ 1 })+...+f({ x }_{ (n-1 })]\Delta x\)
S = [f(1) + f(1.1) + f(1.2) + f(1.3) + f(1.4)](0.1)
= (1 + 1.1 + 1.2 + 1.3 + 1.4) (0.1)
= 6 \(\times\) 0.1
S = 0.6

The parabola x = y² +1 is y² x = −1. It is symmetrical about x-axis and has the vertex at (1, 0) and focus at \(\left( \frac { 5 }{ 4 } ,0 \right) \). The region for revolution is shaded. Hence, the required volume is given by
\(V=\pi \int _{ -1 }^{ 1 }{ { x }^{ 2 }dy } \)
\(=\pi \int _{ -1 }^{ 1 }{ { ({ y }^{ 2 }+1) }^{ 2 }dy } \)
\(=2\pi \int _{ 0 }^{ 1 }{ \left( { y }^{ 4 }+{ 1y }^{ 2 }+1 \right) dy } \), since the integrand is an even function
\(=2\pi { \left( \frac { { y }^{ 5 } }{ 5 } +2\frac { { y }^{ 3 } }{ 3 } +y \right) }_{ 0 }^{ 1 }=2\pi \left( \frac { 1 }{ 5 } +\frac { 2 }{ 3 } +1 \right) \pi \)

The region to be revolved about the x-axis is sketched as in

Hence, the required volume is given by
\(V=\pi \int _{ 0 }^{ 1 }{ { y }^{ 2 }dx=\pi \int _{ 0 }^{ 1 }{ { ({ x }^{ 2 }+4x+5) }^{ 2 } } } dx\)
\(=\pi \int _{ 0 }^{ 1 }{ ({ x }^{ 4 }+{ 16 }x^{ 2 }+25+{ 8x }^{ 3 }+{ 40x+10x }^{ 2 }) } dx\)  
\(=\pi { \left( \frac { { x }^{ 5 } }{ 5 } +8\frac { { x }^{ 4 } }{ 4 } +26\frac { { x }^{ 3 } }{ 3 } +40\frac { { x }^{ 2 } }{ 2 } +25x \right) }_{ 0 }^{ 1 }\)
\(=\pi \left( \frac { 1 }{ 5 } +2+\frac { 26 }{ 3 } +20+25 \right) =\frac { 838 }{ 15 } \pi \)

By revolving the upper semicircular region enclosed between the circle x² + y² = a² and the x-axis, we get a sphere of radius a.

The boundaries of the region are y = \(\\ \\ \\ \\ \\ \\ \\ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \) x-axis, the lines x = −a and x = a. Hence, the volume of the sphere is given by
\(v=\pi \int _{ -a }^{ a }{ { y }^{ 2 }dx=\pi } \int _{ -a }^{ a }{ \left( { a }^{ 2 }-{ x }^{ 2 } \right) } dx\) 
\(=2\pi \int _{ 0 }^{ a }{ ({ a }^{ 2 }-{ x }^{ 2 })dx } \) since the integrand (a²-x²) is an even function
\(=2\pi { \left( { a }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=2\pi \left( { a }^{ 3 }-\frac { { a }^{ 3 } }{ 3 } \right) =\frac { 4 }{ 3 } { \pi a }^{ 3 }\)

Using the formula \(n={ e }^{ { log }_{ e }n },\) we get
\(I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { n }^{ x } } } dx=\int _{ 0 }^{ \infty }{ { n }^{ -x }{ x }^{ n }dx=\int _{ 0 }^{ \infty }{ { n }^{ -x }{ x }^{ n } } dx=\int _{ 0 }^{ \infty }{ { ({ e }^{ log\quad n }) }^{ -x }{ x }^{ n }dx=\int _{ 0 }^{ \infty }{ { e }^{ -xlog\quad n }{ x }^{ n }dx } } } \)
Using the substitution u = x log n, we get dx \(\frac { du }{ log\ n } \)
When x = 0,
we get u = 0
When x = \(\infty\), we get u = \(\infty\)
\(\therefore I=\int _{ 0 }^{ \infty }{ { e }^{ -u }{ \left( \frac { u }{ log\quad n } \right) }^{ n }\frac { du }{ log\quad n } } \)
\(=\frac { 1 }{ { (log\quad n) }^{ n+1 } } \int _{ 0 }^{ \infty }{ { e }^{ -u }{ u }^{ (n+1)-1 }du=\frac { Γ(n+1) }{ { (log\quad n) }^{ n+1 } } =\frac { n! }{ { (log) }^{ n+1 } } } \)

\(I=\int _{ 0 }^{ 1 }{ { x }^{ 2 } } { (1-x) }^{ 3 }dx\)
We know \(\int _{ 0 }^{ 1 }{ { x }^{ m }{ (1-x) }^{ n }dx=\frac { m!\times n! }{ (m+n+1)! } } \)
\(\therefore I=\frac { 2!\times 3! }{ (2+3+1)! } =\frac { 2\times 3\times 2\times 1 }{ 6\times 5\times 4\times 3\times 2\times 1 } \)
\(I=\frac { 1 }{ 60 } \)

\(Let\quad t=\frac { x }{ 4 } \Rightarrow dt=\frac { dx }{ 4 } \Rightarrow dx=4dt\)
Here n = 7, which is odd
\(\therefore I=4\int _{ 0 }^{ \pi /2 }{ { sin }^{ 7 }tdt } \)

\(=4\times \frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1\)
\(I=\frac { 64 }{ 35 } \)

\(Let\quad t=3x\\\Rightarrow dt=3dx\\\Rightarrow \frac { dt }{ 3 } =dx\)
Here n = 6, which is even
\(\therefore I=\frac { 1 }{ 3 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }t\quad dt=\frac { 1 }{ 3 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } \)

\(=\frac { 8 }{ 45 } \)

\(Let\quad t=2x\Rightarrow dt=2dx\Rightarrow \frac { dt }{ 2 } =dx\)
\(\therefore I=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 6 }t\quad dt=\frac { 1 }{ 2 } { I }_{ 6 } } \)

\([\int _{ 0 }^{ \pi /2 }{ { sin }^{ n }xdx=\frac { n-1 }{ n } { I }_{ n-2 }, } n\ge 2]\)
\(\\ =\frac { 1 }{ 2 } \times \frac { 5 }{ 6 } \times \frac { 3 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 2 } =\frac { 5\pi }{ 64 } \)

Put x = sin \(\theta\). Then, dx = cos \(\theta\) d \(\theta\)
when x = 0, sin \(\theta\) = 0 and so \(\theta\) = 0. When x = 1, sin \(\theta\) = 1 and so \(\theta =\frac{\pi}{2}\)
Hence, we get
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }\theta { (1-{ sin }^{ 2 }\theta ) }^{ 5 }cos\theta d\theta } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }\theta { cos }^{ 11 }\theta d\theta =\frac { 10 }{ 16 } \times \frac { 8 }{ 14 } \times \frac { 6 }{ 12 } \times \frac { 4 }{ 10 } \times \frac { 2 }{ 8 } \times \frac { 1 }{ 6 } =\frac { 1 }{ 336 } } \)
By applying the reduction formula III iteratively, we get the following results (stated without proof):
\(\int _{ 0 }^{ 1 }{ { x }^{ m }{ (1-x) }^{ n }dx } =\frac { m!\times n! }{ (m+n+1)! } \), where m and n are positive integers

Dividing the numerator and denominator by cos²x we get,
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \frac { 1 }{ { cos }^{ 2 } } dx }{ \frac { 5 }{ { cos }^{ 2 }x } +\frac { 4{ sin }^{ 2 }x }{ { cos }^{ 2 }x } } } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ 5{ sec }^{ 2 }x+4{ tan }^{ 2 }x } dx } \)
\(\\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }xdx }{ 5(1+{ tan }^{ 2 }x)+4{ tan }^{ 2 }x } } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ 5+5{ tan }^{ 2 }x+4{ tan }^{ 2 }x } } dx\)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ 5+9{ tan }^{ 2 }x } dx\quad put\quad u=tan\quad x } \)

\(\Rightarrow du={ sec }^{ 2 }x,dx\)
\(\therefore I=\int _{ 0 }^{ \infty }{ \frac { du }{ 5+9{ u }^{ 2 } } =\frac { 1 }{ 9 } \int _{ 0 }^{ \infty }{ \frac { du }{ \frac { 5 }{ 9 } +{ u }^{ 2 } } =\frac { 1 }{ 9 } \int _{ 0 }^{ \infty }{ \frac { du }{ { \left( \frac { \sqrt { 5 } }{ 3 } \right) }^{ 2 }+{ u }^{ 2 } } } } } \)
\(=\frac { 1 }{ 9 } \frac { \pi }{ 2\left( \frac { \sqrt { 5 } }{ 3 } \right) } \left[ \because \int _{ 0 }^{ \infty }{ \frac { dx }{ { a }^{ 2 }+{ x }^{ 2 } } } =\frac { \pi }{ 2a } \right] \)
\(=\frac { \pi }{ 2.3\sqrt { 5 } } =\frac { \pi }{ 6\sqrt { 5 } } \times \frac { \sqrt { 5 } }{ \sqrt { 5 } } =\frac { \pi }{ 6\sqrt 5 } \)

\(Let\ u={ x }^{ 2 }\ v=cos 2x\ dx\)
\(u'=2x\quad { v }_{ 1 }=\frac { sin2x }{ 2 } \)
\(u"=2\quad { v }_{ 2 }=-\frac { cos2x }{ 4 } \)
\(\\ { v }_{ 3 }=-\frac { sinx }{ 8 } \)
Bernoulli's' formula:
\(\int { uvdx } =u{ v }_{ 1 }-u'{ v }_{ 2 }+u''{ v }_{ 3 }\)
\(\therefore \int _{ 0 }^{ \pi /2 }{ { x }^{ 2 }cos2x\quad dx={ \left[ \frac { { x }^{ 2 }sin2x }{ 2 } +\frac { 2xcos2x }{ 4 } -\frac { 2sin2x }{ 8 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } } } \)
\(=\left[ \frac { { \pi }^{ 2 } }{ 8 } sin\pi +\frac { 2\pi }{ 8 } cos\pi -\frac { sin\pi }{ 8 } \right] -[0+0-0]\)
\(=\frac {- 2\pi }{ 8 } (-1)=\frac { \pi }{ 4 }\quad [\because sin\pi =0,cos\pi =-1]\)

\(put\ t={ sin }^{ -1 }x\Rightarrow dt=\frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
\(\therefore \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { e }^{ at }t\quad dt=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ t } { e }^{ at }dt } \)
\( u=t\ \ v={ e }^{ at }dt\)
\(u'=1\quad { v }_{ 1 }= e ^t\)
\(u'' = 0 \ \ { v }_{ 2 }=e^t\)
\(\int { uvdx } ={ uv }_{ 1 }-u'{ v }_{ 2 }\)
\(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ t{ e }^{ at }dt } ={ \left[ t\frac { { e }^{ at } }{ a } -1\frac { { e }^{ at } }{ { a }^{ 2 } } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(dt = \frac{1}{\sqrt 1-x^2}dx\)
I = \(\int ^\frac{\pi}{4}_0 e^t t \ dt\)
\([t e^t-t]^{\frac{\pi}{4}}_0\)
\(=\frac { { e }^{ \frac { \pi }{ 4 } } }{ { a }^{ 2 } } \left( \frac { a\pi }{ 4 } -1 \right) -\frac { { e }^{ 0 } }{ { a }^{ 2 } } (-1)\)
\(=1+ e ^ \frac { \pi }{ 4 } (\frac { \pi }{ 4 }-1) \)

Let \(\int _{ 0 }^{ 1 }{ \frac { sin(3{ tan }^{ -1 }x){ tan }^{ -1 }x }{ 1+{ x }^{ 2 } } } dx\)
\(put\quad { tan }^{ -1 }x=t\Rightarrow \frac { 1 }{ 1+{ x }^{ 2 } } dx=dt\)
\(\Rightarrow \int _{ 0 }^{ \frac { \pi }{ 4 } }{ t\quad sin(3t)dt } =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tsin3t } dt\)

\(Let\ u=t\ \ v=sin3tdt\)
\(u'=1\ \ { v }_{ 1 }=\frac { -cos3t }{ 3 } \)
\(u'' = 0 \quad { v }_{ 2 }=\frac { -sin3t }{ 9 } \)
Bernoulli's formula:
\(\int { uvdx } =u{ v }_{ 1 }-u'{ v }_{ 2 }\)
\(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tsin3tdt={ \left[ t\left( \frac { -cos3t }{ 3 } \right) -1\left( \frac { -sin3t }{ 9 } \right) \right] }_{ 0 }^{ \frac { \pi }{ 4 } } } \)
\(=\frac { -1 }{ 9 } { [3tcos3t-sin3t] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=\frac { -1 }{ 9 } \left[ \left( 3\frac { \pi }{ 4 } cos\frac { 3\pi }{ 4 } -sin\frac { 3\pi }{ 4 } \right) -(0-0) \right] \)
\(=\frac { -1 }{ 9 } \left[ \frac { 3\pi }{ 4 } \left( -\frac { 1 }{ \sqrt { 2 } } \right) -\frac { 1 }{ \sqrt { 2 } } \right] \)
\(=\frac { -1 }{ 9 } \left[ \frac { -3\pi -4 }{ 4\sqrt { 2 } } \right] =\frac { 3\pi }{ 36\sqrt { 2 } } +\frac { 4 }{ 36\sqrt { 2 } } \)
\(=\frac { \pi }{ 12\sqrt { 2 } } +\frac { 1 }{ 9\sqrt { 9 } } =\frac { 1 }{ \sqrt { 9 } } \left[ \frac { \pi }{ 12 } +\frac { 1 }{ 9 } \right] \)

Let \(u={ x }^{ 3 }\quad dv={ e }^{ -2x }\)
\(u_1=3{ x }^{ 2 }{ v }_{ 1 }=\frac { { e }^{ -2x } }{ -2 } \)
\(u_2=6x\quad { v }_{ 2 }=\frac { { e }^{ -2x } }{ 4 } \)
\(u_3=6\quad { v }_{ 3 }=\frac { { e }^{ -2x } }{ -8 } \)
\(u_4 = 0 \quad \ { v }_{ 4 }=\frac { { e }^{ -2x } }{ 16 } \)
Bernoulli's' formula:
\(\int { uvdx } ={ uv }_{ 1 }-u'{ v }_{ 2 }+u"{ v }_{ 3 }-u"'{ v }_{ 4 }+...\)
\(\therefore \int _{ 0 }^{ 1 }{ { x }^{ 3 }{ e }^{ -2x }dx={ \left[ { x }^{ 3 }\left( \frac { { e }^{ -2x } }{ -2 } \right) -3{ x }^{ 2 }\left( \frac { { e }^{ -2x } }{ 4 } \right) +6x\left( \frac { { e }^{ -2x } }{ -8 } \right) -6\left( \frac { { e }^{ -2x } }{ 16 } \right) \right] }_{ 0 }^{ 1 } } \)
\(={ \left[ { e }^{ -2x }\left( \frac { -{ x }^{ 3 } }{ 2 } -\frac { -3{ x }^{ 2 } }{ 4 } -\frac { 3x }{ 4 } -\frac { 3 }{ 8 } \right) \right] }_{ 0 }^{ 1 }\)
\(=\left[ { e }^{ -2 }\left( -\frac { 1 }{ 2 } -\frac { 3 }{ 4 } -\frac { 3 }{ 4 } -\frac { 3 }{ 8 } \right) -{ e }^{ -0 }\left( \frac { -3 }{ 8 } \right) \right] \)
\(={ e }^{ -2 }\left( \frac { -4-12-3 }{ 8 } \right) +\frac { 3 }{ 8 } \)
\(={ e }^{ -2 }\left( \frac { -19 }{ 8 } \right) +\frac { 3 }{ 8 } =\frac { 3 }{ 8 } -\frac { 19 }{ 8 } { e }^{ -2 }\)

Taking u = 1−x² and v = e^{−\(\lambda\)x}, and applying the Bernoulli’s formula, we get
\(I=\int _{ -1 }^{ 1 }{ { e }^{ -\lambda x } } (1-{ x }^{ 2 })dx={ \left[ (1-{ x }^{ 2 })\left( \frac { { e }^{ -\lambda x } }{ -\lambda } \right) -(-2x)\left( \frac { { e }^{ -\lambda x } }{ { \lambda }^{ 2 } } \right) +(-2)\left( \frac { { e }^{ -\lambda x } }{ -{ \lambda }^{ 3 } } \right) \right] }_{ -1 }^{ 1 }\)
\(=2\left( \frac { { e }^{ -\lambda } }{ { \lambda }^{ 2 } } \right) +2\left( \frac { { e }^{ -\lambda } }{ { \lambda }^{ 3 } } \right) +2\left( \frac { { e }^{ \lambda } }{ { \lambda }^{ 2 } } \right) -2\left( \frac { { e }^{ \lambda } }{ { \lambda }^{ 3 } } \right) \)
\(=\frac { 2 }{ { \lambda }^{ 2 } } ({ e }^{ \lambda }+{ e }^{ -\lambda })-\frac { 2 }{ { \lambda }^{ 3 } } ({ e }^{ \lambda }-{ e }^{ -\lambda })\)

Taking u = x² and v = sin nx, and applying the Bernoulli’s formula, we get
\(I=\int _{ 0 }^{ 2\pi }{ { x }^{ 2 } } sin\quad nx\quad dx={ \left[ \left( { x }^{ 2 } \right) \left( -\frac { cos\quad nx }{ n } \right) -(2x)\left( -\frac { sin\quad nx }{ { n }^{ 2 } } \right) +(2)\left( \frac { cos\quad nx }{ { n }^{ 3 } } \right) \right] }_{ 0 }^{ 2x }\)
\(=\left[ (4{ \pi }^{ 2 })\left( -\frac { 1 }{ n } \right) -0+(2)\left( \frac { 1 }{ { n }^{ 3 } } \right) \right] -\left[ 0-0+(2)\left( \frac { 1 }{ { n }^{ 3 } } \right) \right] \) since cos 2n\(\pi\) = 1 and sin 2n \(\pi\) = 0
\(=-\frac { 4{ \pi }^{ 2 } }{ n } +\frac { 2 }{ { n }^{ 3 } } -\frac { 2 }{ { n }^{ 3 } } =\frac { { 4\pi }^{ 2 } }{ n } \)

Taking u = 1 + x − 2x³ and v = e^-2x, and applying the Bernoulli’s formula, we get
I = \(\\ \int _{ 0 }^{ 1 }{ { e }^{ -2x }(1+x-{ 2x }^{ 3 })dx } \)
\(={ \left[ (1+x-{ 2x }^{ 3 })\left( \frac { { e }^{ -2x } }{ -2 } \right) -(1-6{ x }^{ 2 })\left( \frac { { e }^{ -2x } }{ -4 } \right) +(-12x)\left( \frac { { e }^{ -2x } }{ -8 } \right) -(-12)\left( \frac { { e }^{ -2x } }{ 16 } \right) \right] }_{ 0 }^{ 1 }\)
\(={ \left[ \frac { { e }^{ -2x } }{ 16 } (16{ x }^{ 3 }+24{ x }^{ 2 }+16x) \right] }_{ 0 }^{ 1 }\)
\(\\ =\frac { 7 }{ 2{ e }^{ 2 } } \)

Taking u = x² and v = cos nx, and applying the Bernoulli’s formula, we get
\(I=\int _{ 0 }^{ \pi }{ { x }^{ 2 }cos\quad nxdx } ={ \left[ ({ x }^{ 2 })\left( \frac { sin\quad nx }{ n } \right) -(2x)\left( -\frac { cos\ nx }{ { n }^{ 2 } } \right) +(2)\left( -\frac { sin\quad nx }{ { n }^{ 3 } } \right) \right] }_{ 0 }^{ \pi }\)
\(=\frac { 2\pi { (-1) }^{ n } }{ { n }^{ 2 } } \), since cos n\(\pi\) = (-1)ⁿ and sin n\(\pi\) = 0

Let \(I=\int _{ 0 }^{ \pi }{ xlog\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) \left( \frac { 3-cos\quad x }{ 3+cos\quad x } \right) dx } \quad ...(1)\)
\(I=\int _{ 0 }^{ 2\pi }{ (2\pi -x)log\left( \frac { 3+cos(2\pi -x) }{ 3-cos(2\pi -x) } \right) } dx\)
\(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=\int _{ 0 }^{ a }{ (a-x)dx } } \right] \)
\(=\int _{ 0 }^{ 2\pi }{ 2\pi log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx } \)
\(-\int _{ 0 }^{ 2\pi }{ xlog\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx } \)
\(\left[ \because cos(2\pi -x)=cos\quad x \right] \)
\(I=2\pi \int _{ 0 }^{ 2\pi }{ log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx-I } [using(1)]\)
\(\\ 2I=2\pi \int _{ 0 }^{ 2\pi }{ log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx } \quad \quad ...(2)\)
\(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=2\int _{ 0 }^{ \frac { a }{ 2 } }{ f(x)dx\quad if(a-x)=f(x) } } \right] \)
I = 2 pi integral limit 0 to p_i
\(\left[ log3+\frac { x }{ 3-x } \right] dx\)
\(\Rightarrow I=2\pi \int _{ 0 }^{ \pi }{ log\left( \frac { 3+cos(\pi -x) }{ 3-cos(\pi -x) } \right) dx } \)
\(\Rightarrow I=2\pi \int _{ 0 }^{ \pi }{ log\left( \frac { 3-cos\quad x }{ 3+cos\quad x } \right) dx...(3) } \)
\([\because cos(\pi -x)=cosx]\)
Adding (2) and (3) we get,
\(2I=\int _{ 0 }^{ \pi }{ log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) \left( \frac { 3-cos\quad x }{ 3+cos\quad x } \right) dx } \)

\(\Rightarrow 2\pi \int _{ 0 }^{ 2\pi }{ 0dx=0 } \)
\(\Rightarrow 2I=0\Rightarrow I=0\)
\(\therefore \int _{ 0 }^{ 2\pi }{ xlog\left( \frac { 3+cos\ x }{ 3-cos\ x } \right) dx=0 } \)

Let \(f(x)={ sin }^{ 2 }x\)
\(f(-x)={ (sin(-x)) }^{ 2 }={ sin }^{ 2 }x=f(x)\)
\(\therefore \int _{ -\frac { \pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } =2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } \)
\(cos\ 2x=1-2{ sin }^{ 2 }x\)
\(2{ sin }^{ 2 }x=1-cos2\)
\({ sin }^{ 2 }x=\frac { 1-cos\quad 2x }{ 2 } \)
\(=2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 1-cos2x }{ 2 } \right) dx } \)
\(=\frac { 2 }{ 2 } { \left[ x-\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=\frac { \pi }{ 4 } -\frac { sin\frac { \pi }{ 4 } }{ 2 } -0+\frac { sin0 }{ 2 } \)
\(\\ =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } =\frac { \pi -2 }{ 4 } \)

Let \(f(x)=xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\(f(-x)=-x\quad cos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\(=-x\quad cos\left( \frac { { e }^{ \frac { 1 }{ x } }-1 }{ { e }^{ \frac { 1 }{ x } }+1 } \right) \)
\(=-x\quad cos\left( \frac { 1-{ e }^{ x } }{ 1+{ e }^{ x } } \right) \)
\(=-x\quad cos\left( -\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \right) \)
\(=-xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\([\because cos(-\theta )=cos\theta ]\)
= -f(x)
\(\therefore\) f(x) is an odd function
\(\therefore \int _{ -5 }^{ 5 }{ xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) } dx=0\)

I = \(\int _{ 0 }^{ 1 }{ \frac { 1-{ x }^{ 2 } }{ { (1+{ x }^{ 2 }) }^{ 2 } } } dx\)
[Dividing the numerator and denominator by x]
\(I=\int _{ 0 }^{ 1 }{ \frac { \frac { 1 }{ { x }^{ 2 } } -\frac { { x }^{ 2 } }{ { x }^{ 2 } } }{ 0\left( \frac { 1 }{ x } +\frac { { x }^{ 2 } }{ { x }^{ 2 } } \right) } } dx=\int _{ 0 }^{ 1 }{ \frac { { x }^{ \frac { 1 }{ 2 } -1 } }{ { (\frac { 1 }{ x } +x) }^{ 2 } } } dx\)
\(put\ x+\frac { 1 }{ x } =t\Rightarrow (1-\frac { 1 }{ { x }^{ 2 } } )dx=dt\)
\(\Rightarrow \left( \frac { 1 }{ { x }^{ 2 } } -1 \right) dx=-dt\)

\(\therefore I=\int _{ \infty }^{ 2 }{ -\frac { dt }{ { t }^{ 2 } } } =\int _{ 2 }^{ \infty }{ \frac { dt }{ { t }^{ 2 } } } \)
\(=\int _{ 2 }^{ \infty }{ { t }^{ -2 }dt } ={ \left[ \frac { { t }^{ -1 } }{ -1 } \right] }_{ 2 }^{ \infty }={ \left[ -\frac { 1 }{ t } \right] }_{ 2 }^{ \infty }\)
\(=-\frac { 1 }{ \infty } +\frac { 1 }{ 2 } =0+\frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 2 } \therefore \int _{ 0 }^{ 1 }{ \frac { 1-{ x }^{ 2 } }{ { (1+{ x }^{ 2 }) }^{ 2 } } dx=\frac { 1 }{ 2 } } \)

\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ \frac { 1 }{ 2 } }\theta (1-{ cos }^{ 2 }) } sin\ \theta\ d\theta \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( { cos }^{ \frac { 1 }{ 2 } }\theta -{ cos }^{ \frac { 5 }{ 2 } }\theta \right) } sin\theta\ d\theta \)
\(Put\ cos\theta =t\)
\(\\ \Rightarrow -sin\theta\ d\theta =dt\)
\(\Rightarrow sin\theta\ d\theta =-dt\)

\(=-\int _{ 1 }^{ 0 }{ ({ t }^{ \frac { 1 }{ 2 } }-{ t }^{ \frac { 5 }{ 2 } }) } dt=\int _{ 0 }^{ 1 }{ ({ t }^{ \frac { 1 }{ 2 } }-{ t }^{ \frac { 5 }{ 2 } }) } dt\)
\(={ \left[ \frac { { t }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -\frac { { t }^{ \frac { 7 }{ 2 } } }{ \frac { 7 }{ 2 } } \right] }_{ 0 }^{ 1 }=\left[ \frac { 2 }{ 3 } (1)-\frac { 2 }{ 7 } (1)-0 \right] \)
\(=\frac { 2 }{ 3 } -\frac { 2 }{ 7 } =\frac { 14-6 }{ 21 } =\frac { 8 }{ 21 } \)

\(Let\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { sinx }{ 1+cosx } \right) dx } \)
\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { 1+sinx }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(\left[ \because cosx=2{ cos }^{ 2 }x-1\Rightarrow 1+cos2x=2{ cos }^{ 2 }x \right] \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { 1 }{ { cos }^{ 2 }\frac { x }{ 2 } } +\frac { sinx }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( { sec }^{ 2 }\frac { x }{ 2 } +\frac { 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(\\ [\because sin2x=2sinx\quad cosx]\)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( { sec }^{ 2 }\frac { x }{ 2 } 2tan\frac { x }{ 2 } \right) dx } \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }(f'(x)+f(x)dx } \)
Where \(f(x)=2tan\frac { x }{ 2 } \)
\(=\frac { 1 }{ 2 } .{ e }^{ x }.f(x)={ \left[ \left( \frac { 1 }{ 2 } { e }^{ x }.2tan\frac { x }{ 2 } \right) \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(={ \left[ { e }^{ x }tan\frac { x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(={ e }^{ \frac { \pi }{ 2 } }tan\frac { \pi }{ 4 } -{ e }^{ 0 }tan0={ e }^{ \frac { \pi }{ 2 } }(1)\)
\(\therefore I={ e }^{ \frac { \pi }{ 2 } }\)

I \(=\int _{ 0 }^{ 1 }{ \sqrt { \frac { (1-x)(1-x) }{ (1+x)(1-x) } } } dx\)
\(=\int _{ 0 }^{ 1 }{ \sqrt { \frac { (1-x) 2}{ (1-x)^2 } } } dx\)
\(=\int _{ 0 }^{ 1 }{ \frac { 1-x }{ \sqrt { 1-{ x }^{ 2 } } } } dx\)
Put x = sin u, dx = cosu.du, 0u = sin^-1(x)
For x = 0, u = 0 and x = 1, u = \(\frac{\pi}{2}\)
\(=\int _{ 0 }^{ 1 }{ \frac { 1-sin\ u }{ \sqrt { 1-{ sin }^{ 2 }u } } } \times cos u.du\)
\(=\int _{ 0 }^{ 1 } \frac { 1-sin\ u }{cos u} \times cos u.du\)
= \( [u+cos u]^\frac{\pi }{2}_0\)= (\(\frac{\pi}{2}\)+0)-(0+1) = \(\frac{\pi}{2}\) -1
\(=\frac { \pi }{ 2 } -0-1=\frac { \pi }{ 2 } -1\)

We write \(\int _{ 0 }^{ 2a }{ f(x)dx\int _{ 0 }^{ a }{ f(x)dx } } +\int _{ 0 }^{ 2a }{ f(x)dx } \) ....(1)
Consider \(\int _{ 0 }^{ 2a }{ f(x)dx } \)
Substituting x = a + u, we have dx = du ; when x = a, u = 0 and when x = 2a,u = a.
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx=\int _{ 0 }^{ a }{ f(a+u)du } =\int _{ 0 }^{ a }{ d(u)du } } \), since f(x) = f(a+x)
\(\\ \\ \\ =\int _{ 0 }^{ a }{ f(x)dx } \) ..(2)
Substituting (2) in (1), we get
\(\int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx } \)

\(\int _{ 0 }^{ 1 }{ \frac { 2x+7 }{ { 5x }^{ 2 }+9 } dx } =\int _{ 0 }^{ 1 }{ \frac { 2x }{ { 5x }^{ 2 }+9 } } +7\int _{ 0 }^{ 1 }{ \frac { dx }{ (5{ x }^{ 2 })+{ 3 }^{ 2 } } =\frac { 1 }{ 5 } } log{ [{ 5x }^{ 2 }+9] }_{ 0 }^{ 1 }+\frac { 7 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }{ \left( \frac { 3 }{ \sqrt { 5 } } \right) }^{ 2 } } } \)
\(=\frac { 1 }{ 5 } [log14-log9]+\frac { 7 }{ 5 } \times \frac { \sqrt { 5 } }{ 3 } { \left[ { tan }^{ -1 }\frac { x }{ \left[ \frac { 3 }{ \sqrt { 5 } } \right] } \right] }_{ 0 }^{ 1 }=\frac { 1 }{ 5 } log\frac { 14 }{ 9 } +\frac { 7 }{ 3\sqrt { 5 } } { tan }^{ -1 }\frac { \sqrt { 5 } }{ 3 } \)

\(\int _{ 0 }^{ 3 }{ (3{ x }^{ 2 }-4x+5) } dx=\int _{ 0 }^{ 3 }{ 3{ x }^{ 2 } } dx-\int _{ 0 }^{ 3 }{ 4x } dx+\int _{ 0 }^{ 3 }{ 5 } dx\)
\(=3\int _{ 0 }^{ 3 }{ { x }^{ 2 } } dx-4\int _{ 0 }^{ 3 }{ c } dx+5\int _{ 0 }^{ 3 }{ dx } \)
\(=3{ \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }-4{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 3 }+5{ \left[ x \right] }_{ 0 }^{ 3 }\)
= (27 − 0) − 2(9 − 0) + 5(3− 0)
= 27 −18 +15 = 24.

The region is sketched. It lies above the x − axis. Hence, the required area is given by 

\(A=\int _{ -1 }^{ 3 }{ ydx } =\int _{ -1 }^{ 3 }{ \left( \frac { 30-6x }{ 5 } \right) dx={ \left( \frac { 30x-3{ x }^{ 2 } }{ 5 } \right) }_{ -1 }^{ 3 } } \)
\(=\left( \frac { 90-27 }{ 5 } \right) -\left( \frac { -30-3 }{ 5 } \right) =\frac { 96 }{ 5 } \)

\( \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx}=\frac { n! }{ { a }^{ n+1 } } \)
\(n=5,\quad a=3 \)
\(=\frac { 5! }{ { 3 }^{ 6 } } \)

Take 2a = 2\(\pi\) and f(x) = g(cosx)
Then, f (2a−x) = f(2\(\pi\)-x) = g(cos(2\(\pi\)-x)) = g(cos x) = f(x)
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx=2 } \int _{ 0 }^{ a }{ f(x)dx } \)
\(\therefore \int _{ 0 }^{ 2\pi }{ g(cosx)dx=2\int _{ 0 }^{ \pi }{ g(cosx)dx } } \)

We know that \(\int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x) } dx\quad if(2a-x)=f(x)\)
Take 2a = \(\pi\) and f(x) = g(sinx)
Then, f(2a-x) = g(sin(\(\pi\)-x)) = g(sinx) = f(x).
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx } \)
\(\int _{ 0 }^{ \pi }{ g(sinx)dx } =2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ g(sinx)dx } \)