\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 4 }x{ cos }^{ 6 }xdx=\frac { (6-1) }{ (6+4) } .\frac { (6-3) }{ (6+4-2) } .\frac { (6-5) }{ (6+4-4) } .\frac { (4-1) }{ (4) } .\frac { (4-3) }{ (4-2) } .\frac { \pi }{ 2 } } \)
\(=\frac { (5) }{ (10) } \frac { (3) }{ (8) } \frac { (1) }{ (6) } \frac { (3) }{ (4) } \frac { (1) }{ (2) } \frac { \pi }{ 2 } =\frac { 3\pi }{ 512 } \)
Also, \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 4 }x{ cos }^{ 6 }xdx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 6 }x{ cos }^{ 4 }xdx } =\frac { (3) }{ (10) } \frac { (1) }{ (8) } \frac { (5) }{ (6) } \frac { (3) }{ (4) } \frac { (1) }{ (2) } \frac { \pi }{ 2 } =\frac { 3\pi }{ 512 } \)

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:
\(\int { uvdx } ={ uv }_{ 1 }-u'{ v }_{ 2 }+u"{ v }_{ 3 }-u"'{ v }_{ 4 }+...\)
\(\therefore \int _{ 0 }^{ 1 }{ { x }^{ 3 }{ e }^{ -2x }dx={ \left[ { x }^{ 3 }\left( \frac { { e }^{ -2x } }{ -2 } \right) -3{ x }^{ 2 }\left( \frac { { e }^{ -2x } }{ 4 } \right) +6x\left( \frac { { e }^{ -2x } }{ -8 } \right) -6\left( \frac { { e }^{ -2x } }{ 16 } \right) \right] }_{ 0 }^{ 1 } } \)
\(={ \left[ { e }^{ -2x }\left( \frac { -{ x }^{ 3 } }{ 2 } -\frac { -3{ x }^{ 2 } }{ 4 } -\frac { 3x }{ 4 } -\frac { 3 }{ 8 } \right) \right] }_{ 0 }^{ 1 }\)
\(=\left[ { e }^{ -2 }\left( -\frac { 1 }{ 2 } -\frac { 3 }{ 4 } -\frac { 3 }{ 4 } -\frac { 3 }{ 8 } \right) -{ e }^{ -0 }\left( \frac { -3 }{ 8 } \right) \right] \)
\(={ e }^{ -2 }\left( \frac { -4-12-3 }{ 8 } \right) +\frac { 3 }{ 8 } \)
\(={ e }^{ -2 }\left( \frac { -19 }{ 8 } \right) +\frac { 3 }{ 8 } =\frac { 3 }{ 8 } -\frac { 19 }{ 8 } { e }^{ -2 }\)

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
\(\therefore \int _{ -5 }^{ 5 }{ xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) } dx=0\)

Let us put I = \(\int ^{3}_{2} \frac{\sqrt {x}}{\sqrt {5-x}+\sqrt {x}}\) dx --- (1)
Applying the formula \(\int ^{a}_{b}\) f(x) dx =\(\int ^{a}_{b}\) f(a+b -x) dx, we get
I =  \(\int ^{3}_{2} \frac{\sqrt {(2+3-x)}}{\sqrt {5-(2+3-x)}+\sqrt { {(2+3-x)}}}\) dx = \(\int ^{3}_{2} \frac{\sqrt {(5-x)}}{\sqrt {x}+\sqrt { {(5-x}}}\) dx     -- (2)
Adding (1) and (2), we get 
2I =  \(\int ^{3}_{2} \frac{\sqrt {x} + \sqrt {5-x} }{\sqrt {x}+\sqrt { {5-x}}}\) dx =\(\int ^{3}_{2} \) dx = \([x]^{3}_{2}\) = 3 - 2 = 1
Hence, we get I = \(\frac{1}{2}\)

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

Given equation of line is 3x - 2y + 6 = 0
2y = 3x + 6 \(\Rightarrow\) y = \(\frac{3x+6}{2}\)

x	0	-2
t	3	0

\(\therefore Area=\int _{ -3 }^{ -2 }{ -ydx+ydx } +\int _{ -2 }^{ 1 }{ ydx } \)
[\(\because\) the Area is below the x - axis]

\(=\frac { -1 }{ 2 } \int _{ -3 }^{ -2 }{ (3x+6)dx+\frac { 1 }{ 2 } \int _{ -2 }^{ 1 }{ (3x+6)dx } } \)
\(={ \left[ \frac { { 3x }^{ 2 } }{ 2 } +6x \right] }_{ -3 }^{ -2 }+\frac { 1 }{ 2 } { \left[ \frac { { 3x }^{ 2 } }{ 2 } +6x \right] }_{ -2 }^{ 1 }\)
\(=-\frac { 1 }{ 2 } \left[ \left( \frac { 12 }{ 2 } -12 \right) -\left( \frac { 27 }{ 2 } -18 \right) \right] +\frac { 1 }{ 2 } \left[ \left( \frac { 3 }{ 2 } +6 \right) -\left( \frac { 12 }{ 2 } -12 \right) \right] \)
\(=-\frac { 1 }{ 2 } \left[ (-6)-\left( \frac { 27-36 }{ 2 } \right) \right] +\frac { 1 }{ 2 } \left[ \left( \frac { 3+12 }{ 2 } \right) -(-6) \right] \)
\(=-\frac { 1 }{ 2 } \left[ -6+\frac { 9 }{ 2 } \right] +\frac { 1 }{ 2 } \left[ \frac { 15 }{ 2 } +6 \right] \)
\(\\ =-\frac { 1 }{ 2 } \left[ \frac { -3 }{ 2 } \right] +\frac { 1 }{ 2 } \left[ \frac { 27 }{ 2 } \right] =\frac { 3 }{ 4 } +\frac { 27 }{ 4 } =\frac { 30 }{ 4 } =\frac { 15 }{ 2 } \)
\(\therefore\)  A = 7.5 sq.units

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

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

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

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ n }x } =\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2\)
\(Let\quad { I }_{ 10 }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ 10 }xdx=\frac { 9 }{ 10 } { I }_{ 8 } } \)
\(=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times { I }_{ 6 }=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times { I }_{ 4 }\)
\(\\ =\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times \frac { 3 }{ 4 } { I }_{ 2 }\)
\(=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times \frac { 3 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 2 } \)
\(=\frac { 315 }{ 1280 } \times \frac { \pi }{ 2 } =\frac { 63\pi }{ 256(2) } =\frac { 63\pi }{ 512 } \)

Let f (x) = x cos x
Then f (−x) = (−x) cos(−x) = −x cos x = − f (x).
So f (x) = x cos x is an odd function.
Hence, applying the property, for odd function f(x), \(\int _{ -a }^{ a }{ f(x)dx=0 } \)
\(\therefore\) we get \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)

\(\int _{ 0 }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ [2x] } dx+\int _{ \frac { 1 }{ 2 } }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ 0dx+ } \int _{ \frac { 1 }{ 2 } }^{ 1 }{ 1 dx} = 0+[x]^1_{\frac{1}{2}} = 1 -\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

Volume of the right circular conical frustum is obtained by revolving the line y = x between x = a and x = b around the x - axis

\(\therefore\) Height of the frustum h = b - a
\(\therefore\)Volume \(=\pi \int _{ a }^{ b }{ { x }^{ 2 }dx } =\pi { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ a }^{ b }\)
\(=\frac { \pi }{ 3 } [{ b }^{ 3 }-{ a }^{ 3 }]\)
\(=\frac { \pi }{ 3 } (b-a)({ b }^{ 2 }+ab+{ a }^{ 2 })\)
Now, substitute h = b - a, r = a and R = b we get Volume of the conical frustum
\(\frac { \pi }{ 3 } [h({ R }^{ 2 }+rR+{ r }^{ 2 })]\)
Given h = 2 m, r = 1 m, R = 2 m we get
Required volume \(=\frac { \pi }{ 3 } [2(4+2+1)]\)
\(=\frac { \pi }{ 3 } (14)\)
\(=\frac { 14\pi }{ 3 } \)

Equation of the given circle is x² + y² = 16 ...(1)
and the parabola is y² = 6x. ...(2)
Substituting (2) in (1) we get,
x² + 6x - 16 = 0 \(\Rightarrow\) (x + 8) (x - 2) = 0
\(\Rightarrow\) (x-4) (x-2) = 0 \(\Rightarrow\) x = 2, 4
\(\Rightarrow\) x = -8, 2

\(\therefore\) Required area = 2
\(\\ \\ \\ \\ \\ \\ \\ \int _{ 0 }^{ 2 }{ Area\ below\ the\ parabola } +\int _{ 2 }^{ 4 }{ Area\ below\ the\ circle } \)
\(=2\left[ \int _{ 0 }^{ 2 }{ \sqrt { 6x } } dx+\int _{ 2 }^{ 4 }{ \sqrt { 16-{ x }^{ 2 } } dx } \right] \)
\(=2\left[ { \left( \frac { \sqrt { 6 } .{ x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right) }_{ 0 }^{ 2 }+{ \left( \frac { x }{ 2 } \sqrt { 16-{ x }^{ 2 } } +\frac { 16 }{ 2 } { sin }^{ -1 }\left( \frac { x }{ 4 } \right) \right) }_{ 0 }^{ 4 } \right] \)
\(=2\left[ \frac { 2 }{ 3 } \sqrt { 6. } 2\sqrt { 2 } +8{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right] \)
\(=2\left[ \frac { 4\sqrt { 12 } }{ 3 } +8\left( \frac { \pi }{ 2 } \right) -\sqrt { 12 } -8\left( \frac { \pi }{ 6 } \right) \right] \)
\(=2\left[ \frac { 4\sqrt { 12 } }{ 3 } +4\pi -2\sqrt { 3 } -\frac { 4\pi }{ 3 } \right] \)
\(=2\left[ \frac { 8\sqrt { 3 } -6\sqrt { 3 } }{ 3 } +\frac { 12\pi -4\pi }{ 3 } \right] \)
\(=\left[ \frac { 2\sqrt { 3 } }{ 3 } +\frac { 8\pi }{ 3 } \right] =2\times \frac { 2 }{ 3 } \left[ \sqrt { 3 } +4\pi \right] \)
\(=\frac { 4 }{ 3 } \left[ 4\pi +\sqrt { 3 } \right] \)sq.units

Equation of the given curves are y = sin x  ..(1)
Y = cos x ...(2)
from (1) and (2), sin x = cos x
y = sin x

x	0	\(\pi\)/2
y	0	1

y = cos x

x	0	\(\pi\)/2
y	1	0

\(\Rightarrow x=\frac { \pi }{ 4 } \)

\(\therefore \) Required area = \(2\int _{ \frac { \pi }{ 4 } }^{ \frac { 3\pi }{ 4 } }{ (sinx-cosx)dx } \)
[\(\because\) the area is symmetrical about X - axis]
\(=2{ \left[ -cosx-sinx \right] }_{ \frac { \pi }{ 4 } }^{ \frac { 3\pi }{ 4 } }\)
\(=-2\left[ \left( cos\frac { 3\pi }{ 4 } +sin\frac { 3\pi }{ 4 } \right) -\left( cos\frac { \pi }{ 4 } +sin\frac { \pi }{ 4 } \right) \right] \)
= -2\(\left[ \left( -\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \right) -\left( \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \right) \right] \)
[cos 135^o = cos(180^o- 45) = -cos 45^o sin135^o = sin(180^o- 45) = -sin 45^o]
\(=-2\left[ \frac { -2 }{ \sqrt { 2 } } \right] =\frac { 4 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 4\sqrt { 2 } }{ 2 } =2\sqrt { 2 } \) 

The ellipse is symmetric about both major and minor axes. It is sketched. So, viewing in the positive direction of y -axis, the required area A is four times the area of the region bounded by the portion of the ellipse in the first quadrant 
\(\left( y=\frac { b }{ a } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } ,0

x -axis, x = 0 and x = a .
Hence, by taking vertical strips, we get \(A=4\int _{ 0 }^{ a }{ ydx=4\int _{ 0 }^{ a }{ \frac { b }{ a } } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } \)
\(=\frac { 4b }{ a } { \left[ \frac { x\sqrt { { a }^{ 2 }-{ y }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { x }{ a } \right) \right] }_{ 0 }^{ a }=\frac { 4b }{ a } \times \frac { \pi { a }^{ 2 } }{ 4 } =\pi ab\)

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

x	0	\(\infty\)
t	0	\(\infty\)

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

I =\(\int ^{1}_{0} (tan ^{-1} x + tan ^{-1}(1-x))\) dx
=\(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-x)) dx\)
= \(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-(1-x)) dx\),  Since\(\int ^{a}_{0} f (x) dx = \int ^{a}_{0} f(a-x) dx \)
=\(\int ^{1}_{0} tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1} x dx\)
= 2\(\int ^{1}_{0} tan^{-1} x dx\)
= \([2 \int udv]^1_0\) ,  Where u = tan^-1 x and dν = dx
= 2\([uν - \int udv]^1_0,\) applying integration by parts
= 2\((x tan^{-1} x - \int x \frac{dx}{1+x^2})^{1}_{0}\)
= 2\((x tan^{-1} x - \frac {1}{2} log (1+x^2))^{1}_{0}\)
= \(\frac{\pi}{2} \)- log 2

Let us put I = \(\int ^\frac{\pi}{4}_{0}\) log(1 + tan x) dx
Applying the property \(\int ^{a}_{0}\) f(x) dx =\(\int ^{a}_{0}\)f(a-x) dx in equation (1), we get
I = \(\int ^\frac{\pi}{4}_{0}\) log \([1+tan (\frac{\pi}{4}-x)]\)dx = \(\int ^\frac{\pi}{4}_{0}\) log\([1 + \frac{tan \frac {\pi}{4}- tan x}{1 + tan {\frac {\pi}{4} tan x}}]\) dx
=\(\int ^\frac{\pi}{4}_{0}\) log \([1 + \frac {1 - tan x}{1 + tan x}]\)dx = \(\int ^\frac{\pi}{4}_{0}\) log\([\frac {1+tan x +1 - tan x}{1 + tan x}]\) dx
=\(\int ^\frac{\pi}{4}_{0}\) log \([\frac{2}{1+tanx}]\) dx = \(\int ^\frac{\pi}{4}_{0}\) [log 2 - log (1+tan x)] dx
= log 2  \(\int ^\frac{\pi}{4}_{0}\) dx - \(\int ^\frac{\pi}{4}_{0}\)log (1+tan x)] dx
= \(\frac{\pi}{4}\)log 2 - I
So, we get 2I = \(\frac{\pi}{4}\)log 2.
Hence, we get I = \(\frac{\pi}{8}\)log 2.