\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\ tanxdx } \)..(i)
Apply the property  \(\int _{ 0 }^{ a }{ f(x) } =\int _{ 0 }^{ a }{ f(a-x)dx, } \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\quad tan\left( \frac { \pi }{ 2 } -x \right) dx } \)  ...(ii)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\quad cotxdx } \)
by adding eqn. (i) and (ii)
\(2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (log\quad tanx+log\quad cotx)xdx } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log(tanx\times cotx } )dx\)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log1dx } =0\)
\(\Rightarrow 2I=0\)
\(\Rightarrow I=0\)

f(x)={\(-(x-5),\quad if\quad x<5\\ +(x-5),\quad if\quad x>5\)
\(I=-\int _{ 2 }^{ 5 }{ (x-5)dx } +\int _{ 5 }^{ 8 }{ (x-5) } dx\)
\(\Rightarrow I=-\left| \frac { { x }^{ 2 } }{ 2 } \right| ^{ 5 }_{ 2 }+5\left| x \right| ^{ 8 }_{ 5 }-\left| 5x \right| ^{ 8 }_{ 5 }\)
\(\Rightarrow I=-\frac { 1 }{ 2 } \left[ 25-4 \right] +5(5-2)+\frac { 1 }{ 2 } (64-25)-5(8-5)\)
\(\Rightarrow I=-\frac { 1 }{ 2 } (21)+5(3)+\frac { 1 }{ 2 } (39)-5(3)\)
\(\Rightarrow I=-\frac { 21 }{ 2 } +15+\frac { 39 }{ 2 } -15\)
\(=\frac { -21+39 }{ 2 } =\frac { 18 }{ 2 } =9\)
\(\Rightarrow I=9\)

\(\left[ \frac { 3x^{ 3 } }{ 3 } \right] ^{ 1 }_{ 0 }=8\)
\(\Rightarrow \left[ x^{ 3 } \right] ^{ a }_{ 0 }=8\)
\(\Rightarrow { a }^{ 3 }=8\)
\(\Rightarrow a=2\)

\(\\ \int _{ 0 }^{ x }{ t } sintdt=f(x),\)
\(\therefore \ f'(x)=xsinx\)

Put,  \(1+{ x }^{ 2 }=t\)
\(\Rightarrow 2xdx=dt\)
\(\Rightarrow xdx=\frac { dt }{ 2 } \)
\(\therefore \int _{ 2 }^{ 4 }{ \frac { x }{ { x }^{ 2 }+1 } } dx=\int _{ 2 }^{ 4 }{ \frac { dt }{ 2t } =\frac { 1 }{ 2 } \left[ logt \right] ^{ 4 }_{ 2 } } \)
\(\frac { 1 }{ 2 } \left[ log(1+{ x }^{ 2 }) \right] ^{ 4 }_{ 2 }=\frac { 1 }{ 2 } log\frac { 17 }{ 5 } \)

\(\int { \frac { dx }{ 1+{ e }^{ x } } } \)
Multiply by e^-x to numerator and denominator
\(\int { \frac { { e }^{ -x } }{ { e }^{ -x }({ e }^{ x }+1) } } dx\quad =\int { \frac { { e }^{ -x } }{ 1+{ e }^{ -x } } } dx\)
Put  \(1+{ e }^{ -x }=t\)
\(\Rightarrow { e }^{ -x }dx=-dt\)
\(\Rightarrow { e }^{ -x }=\frac { dt }{ dx } \)
\(\Rightarrow { e }^{ -x }dx=-dt\)
\(\Rightarrow -\int { \frac { dt }{ t } = } -log\left| t \right| \)
\(\Rightarrow -log(1+{ e }^{ -x })=-log\left( 1+\frac { 1 }{ { e }^{ x } } \right) \)
\(=-log\left( 1+\frac { 1 }{ e^{ x } } \right) +C\)

\(\int { \frac { dx }{ 1+sinx } } \times \frac { (1-sinx }{ (1-sinx) } \)
\(=\int { \frac { (1-sinx) }{ 1-sin^{ 2 }x } } dx\)
\(=\int { \frac { dx }{ { cos }^{ 2 }x } -\int { \frac { sinx }{ { cos }^{ 2 }x } dx } } \)
\(=\int { { sec }^{ 2 }xdx } -\int { tanx\quad secx\quad dx } \)
\(=tanx-secx+C\)

\(\int \frac{\sin ^{2} x-\cos ^{2} x}{\sin x \cos x} d x=\int(\tan x-\cot x) d x=\log |\sec x|-\log |\sin x|+C\)

\(={sin\ 4x\over8}+{sin\ 2x\over4}+c\)

Put \(1+{ e }^{ x }=L\)
\(\Rightarrow { e }^{ x }dx=dL\)
\(I=\int { \frac { { e }^{ x } }{ 1+{ e }^{ x } } } dx=\int { \frac { dL }{ L } } \) 
\(=\log { \left| 1+{ e }^{ x } \right| } +c\) 

= x-2 tan^-1x + c

\(\int \frac{d x}{x \cos ^{2}(1+\log x)} =\int \sec ^{2} t d t \)
= tan t+C
= tan |1 + log x| + c

\(\int \frac{1+\tan x}{1-\tan x} d x =\int \frac{\cos x+\sin x}{\cos x-\sin x} d x \)
\(=-\int \frac{1}{t} d t \)
\(=-\log |t|+C \)
\(=-\log |\cos x-\sin x|+C\)

\(\int \frac{x^{2}}{1+x^{3}} d x =\frac{1}{3} \int \frac{1}{t} d t=\frac{1}{3} \log |t|+C \)
\(=\frac{1}{3} \log \left|1+x^{3}\right|+C\)

Note that
\(\int \sqrt{x^{2}+2 x+5} d x=\int \sqrt{(x+1)^{2}+4} d x\)
Put x + 1 = y, so that dx = dy. Then
\(\int \sqrt{x^{2}+2 x+5} d x =\int \sqrt{y^{2}+2^{2}} d y \)
\(=\frac{1}{2} y \sqrt{y^{2}+4}+\frac{4}{2} \log \left|y+\sqrt{y^{2}+4}\right|+C \quad[\text { using } 7.6 .2 \text { (ii) }] \)
\(=\frac{1}{2}(x+1) \sqrt{x^{2}+2 x+5}+2 \log \left|x+1+\sqrt{x^{2}+2 x+5}\right|+C \)

\(\text {(i) We have } \int { (\sin { x } } +\cos { x } )dx\)
\(=\int { \sin { xdx+ } } \int { \cos { x } } dx\)
\(=-\cos { x } +\sin { x } +c\)
\(\text {(ii) We have } \int { co\sec { x } ( } co\sec { x } +\cot { x } )dx\)
\(=\int { { co\sec { x } }^{ 2 } } x\quad dx+\int { co\sec { x } } \cot { x\quad dx } \)
\( =-\cot { x } -co\sec { x } +c\)
\(\text {(iii) We have} \int { \frac { 1-\sin { x } }{ \cos ^{ 2 }{ x } } } dx=\frac { 1 }{ \cos ^{ 2 }{ x } } dx-\int { \frac { \sin { x } }{ \cos ^{ 2 }{ x } } } dx\)
\(=\int { { sec }^{ 2 } } xdx- \int { \tan { x } \sec { x } } dx\)
\(=\tan { x } -\sec { x } +c\)

Here \(f\left( x \right) =\sin ^{ 5 }{ x } \) is an odd function of x.
ஃ \(\int _{ -\pi /2 }^{ -\pi /2 }{ \sin ^{ 5 }{ xdx } } =0\).

Put \({ x }^{ 2 }\) = t so that 2xdx = dt i.e. xdx = \(\frac {dt }{ 2 } \)
When x = 1, t = 1. when x = 2, t = 4.
\(I=5\int _{ 1 }^{ 4 }{ \sqrt { 5-t } } \frac { dt }{ 2 } =\frac { 5 }{ 2 } \int _{ 1 }^{ 4 }{ { (5-t) }^{ 1/2 }dt } \)
\(=\frac { 5 }{ 2 } { \left[ \frac { { (5-t) }^{ 1/2 } }{ \frac { 3 }{ 2 } -1 } \right] }_{ 1 }^{ 4 }\)
\(=-\frac { 5 }{ 3 } \left[ { 1-4 }^{ 3/2 } \right] =-\frac { 5 }{ 3 } [1-8]=\frac { 35 }{ 3 } \)

\(={1\over3}(x^3+1)log(1+x)-{x^3\over9}+{x^2\over6}-{x\over3}+c\)

\({1\over2\sqrt2}tan^{-1}({x^2-4\over2\sqrt2 x})+c\)

\(I=\int _{ 0 }^{ \pi }{ \frac { x }{ { a }^{ 2 }\cos ^{ 2 }{ x } +{ b }^{ 2 }\sin ^{ 2 }{ x } } } dx\) ..(i)
Apply property
\(\Rightarrow I=\int _{ 0 }^{ \pi }{ \frac { \left( \pi -x \right) }{ { a }^{ 2 }\cos ^{ 2 }{ x } +{ b }^{ 2 }\sin ^{ 2 }{ x } } } dx\)
By adding eqn. (i) and (ii),
\(2I=\pi \int _{ 0 }^{ \pi }{ \frac { 1 }{ { a }^{ 2 }\cos ^{ 2 }{ x } +{ b }^{ 2 }\sin ^{ 2 }{ x } } } dx\)
\(\Rightarrow I=\pi \int _{ 0 }^{ \pi /2 }{ \frac { \sec ^{ 2 }{ x } }{ { a }^{ 2 }+{ b }^{ 2 }\tan ^{ 2 }{ x } } } dx\)
Dividing Nr. & Dr. by cos²x
Put \(b\tan { x } =t\Rightarrow \sec ^{ 2 }{ x } dx=\frac { 1 }{ b } dt\)
Where x = 0, t = 0;
and \(x=\frac { \pi }{ 2 } ,\)
\(t=\infty ;\)
\(\Rightarrow I=\frac { \pi }{ b } \int _{ 0 }^{ \infty }{ \frac { dt }{ { a }^{ 2 }+{ t }^{ 2 } } } \)
\(\Rightarrow =\frac { \pi }{ b } \left[ \frac { 1 }{ a } \tan ^{ -1 }{ \frac { t }{ a } } \right] _{ 0 }^{ \infty }\)
\(\Rightarrow I=\frac { \pi }{ ab } \left( \frac { \pi }{ 2 } -0 \right) \)
\(=\frac { { \pi }^{ 2 } }{ 2ab } \)

\(I=\int { \frac { 1 }{ { cos }^{ 4 }x+{ sin }^{ 4 }x } } dx\)
Dividing numerator and denominator by cos⁴x
\(=\int { \frac { sec^{ 4 }x }{ 1+tan^{ 4 }x } } dx\)
\(=\int { \frac { (1+tan^{ 2 }x)sec^{ 2 }x }{ 1+tan^{ 4 }x } } dx\)
Putting, tanx=t
\(\Rightarrow sec^{ 2 }xdx=dt\)
\(=\int { \frac { ({ t }^{ 2 }+1)dt }{ { t }^{ 4 }+1 } } \)
\(=\int { \frac { 1+\frac { 1 }{ { t }^{ 2 } } }{ { t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } } } dt\)     [dividing by t²]
\(=\int { \frac { dz }{ { z }^{ 2 }+(\sqrt { 2 } )^{ 2 } } } \) ,  where \(t-\frac { 1 }{ t } =z\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { z }{ \sqrt { z } } \right) +C\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { { t }^{ 2 }-1 }{ \sqrt { 2t } } \right) +C\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { { tan }^{ 2 }x-1 }{ \sqrt { 2 } tanx } \right) +C\)