\(\int _{ b }^{ \infty }{ \frac { 1 }{ { a }^{ 2 }+{ x }^{ 2 } } dx } ={ \left[ \frac { 1 }{ a } { tan }^{ -1 }\frac { x }{ a } \right] }_{ b }^{ \infty }=\frac { 1 }{ a } { tan }^{ -1 }\infty -\frac { 1 }{ a } { tan }^{ -1 }\frac { b }{ a } =\frac { 1 }{ a } \left[ \frac { \pi }{ 2 } -{ tan }^{ -1 }\frac { b }{ a } \right] \)

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
\(=\frac { \pi }{ 2 } -0-1=\frac { \pi }{ 2 } -1\)

Let I = \(\int ^{a}_{0} \frac{f(x)}{f(x)+f(a-x)}\)    ------- (1)
Applying the formula \(\int ^{a}_{0}\) f(x) dx = \(\int ^{a}_{0}\)f(a-x)dx  in equation (1), we get
 I =  \(\int ^{a}_{0} \frac{f{(a-x)}}{f{(a-x)}+f(a-(a-x))}\) dx
= \(\int ^{a}_{0} \frac{f{(a-x)}}{f{(x)}+f(a-x)}\) dx----- (2)
Adding equations (1) and (2), we get
2I =  \(\int ^{a}_{0} \frac{f{(x)}}{f{(x)}+f(a-x)}\) dx + \(\int ^{a}_{0} \frac{f{(a-x)}}{f{(x)}+f(a-x)}\) dx
= \(\int ^{a}_{0} \frac{f(x) +f{(a-x)}}{f{(x)}+f(a-x)}\) dx
= \(\int ^{a}_{0} dx = a\)
Hence, we get I = \(\frac{a}{2}\)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \frac { sec\ x\ tan\ x }{ 1+{ sec }^{ 2 }x } dx } \)
Put sec x = u. Then, sec x tan x dx = du.
When x = 0, u = sec0 = 1. When x = \(\frac{\pi}{3}, u=sec\frac{\pi}{2}=2\)
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { du }{ 1+{ u }^{ 2 } } ={ [{ tan }^{ -1 }u] }_{ 1 }^{ 2 }={ tan }^{ -1 }1={ tan }^{ -1 } } (2)-\frac { \pi }{ 4 } \)

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

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

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

Given that I =\(\int ^\frac {\pi}{2}_{0} \)( sin²x + cos⁴x)dx =\(\int ^\frac {\pi}{2}_{0} \) sin²x dx+\(\int ^\frac {\pi}{2}_{0} \)cos⁴x dx\(\frac {1}{2} \times \frac {\pi}{2} + \frac {3}{4} \times \frac {1}{2} \times \frac {\pi}{2} = \frac {7\pi}{16} \)

Let f(x) = e^-|-x| = e^-|x| = f(x)
So f (x) is an even function.
Hence, \(\int ^{log 2}_{-log 2} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-x}\) dx
= 2(-e^-x)\(^{log2}_{0}\) = 2 (-e^-log2 + e⁰) = 2 \((-e ^{log \frac{1}{2}} + 1)\)
= 2\((-\frac {1}{2}+1)=1\).

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 \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)

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 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 \(A=4\int _{ 0 }^{ a }{ ydx=4\int _{ 0 }^{ a }{ \frac { b }{ a } } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } \)
\(=\frac { 4b }{ a } { \left[ \frac { x\sqrt { { a }^{ 2 }-{ y }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { x }{ a } \right) \right] }_{ 0 }^{ a }=\frac { 4b }{ a } \times \frac { \pi { a }^{ 2 } }{ 4 } =\pi ab\)

Let I = \(\int ^{\pi}_{-\pi} \frac{cos ^2 x}{1+ a^x} dx\)-- (1) 
Using \(\int ^{b}_{a}\) f(x) dx =\(\int ^{b}_{a}\) f(a+b -x)dx  we get,
I = \(\int ^{\pi}_{-\pi} \frac{cos ^2 (\pi -\pi - x)}{1+ a^{\pi -\pi - x}} dx\)
= \(\int ^{\pi}_{-\pi} \frac{cos ^2 (-x)}{1+ a^{-x}} dx\)
= \(\int ^{\pi}_{-\pi} a^x (\frac{cos ^2 x }{a^x+ 1} )dx\) --- (2)
Adding (1) and (2) we get
2I =  \(\int ^{\pi}_{-\pi} \frac{cos ^2 x }{a^x+ 1}(a^x+ 1) dx\) =  \(\int ^{\pi}_{-\pi} cos ^2 x dx\)
= 2 \(x =\int ^{\pi}_{-\pi} cos ^2 x dx\)  (since cos² x is an even function)
Hence, I =  \(\int ^{\pi}_{0} (\frac{1 + cos2x }{2} )dx\) 
=  \(\frac {1}{2} [ x + \frac {sin 2x}{2}]^{x}_{0}\) 
= \(\frac {1}{2} [\pi]\)
= \(\frac {\pi}{2}\)

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.