Let \(f(x)=\log _e\left(\frac{2-x}{2+x}\right)\)
\(\therefore f(-x)=\log _e\left(\frac{2+x}{2-x}\right)=-\log _e\left(\frac{2-x}{2+x}\right)=-f(x)\)
So, f(x) is an odd function.
\(\therefore  \int_{-1}^1 \log _e\left(\frac{2-x}{2+x}\right) d x=0\)

\( \text { Let } I=\int \frac{\sin 2 x}{\sqrt{9-\cos ^4 x}} d x \)
\( =\int \frac{\sin 2 x}{\sqrt{(3)^2-\left(\cos ^2 x\right)^2}} d x\)
\(\text { Put } \cos ^2 x=t \Rightarrow-2 \cos x \sin x d x=d t \)
\(\Rightarrow \sin 2 x d x=-d t \)
\(\therefore \quad I=-\int \frac{d t}{\sqrt{3^2-t^2}}=-\sin ^{-1}\left(\frac{t}{3}\right)+C\)
\( I=-\sin ^{-1}\left(\frac{\cos ^2 x}{3}\right)+C\)

Let \(I=\int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x\) 
\(\Rightarrow I=\mid \int_{0}^{\pi / 2} \frac{\tan ^{7}(\pi / 2-x)}{\cot ^{7}(\pi / 2-x)+\tan ^{7}(\pi / 2-x)} d x\) 
\(\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) 
\(\Rightarrow I=\int_{0}^{\pi / 2} \frac{\cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x\)  ...(ii)
\(\left[\because \tan \left(\frac{\pi}{2}-x\right)=\cot x \text { and } \cot \left(\frac{\pi}{2}-x\right)=\tan x\right]\) 
On adding Eqs. (i) and (ii), we get
\(2 I=\int_{0}^{\pi / 2} \frac{\tan x+\cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x\) 
\(=\int_{0}^{\pi / 2} 1 d x=[x]_{0}^{\pi / 2}=\frac{\pi}{2}-0=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}\) 

Use the formula,\(1+\sin 2 x=(\cos x+\sin x)^{2}\) 
x + C

Let \(I=\int(a x+b)^{3} d x \)
Put \(a x+b=t \Rightarrow a d x=d t \Rightarrow d x=\frac{1}{a} d t \)
\(=\frac{(a x+b)^{4}}{4 \cdot a}+C=\frac{(a x+b)^{4}}{4 a}+C \quad[\because t=a x+b]\)

\(\\ \int _{ 0 }^{ 1 }{ \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } } dx\)
Let,  \(\\ { e }^{ x }=y\)
\(\Rightarrow { e }^{ x }dx=dy\)
when, \(x=0; y=1\) and  \(x=1; y=e\)
\(\therefore I=\int _{ 1 }^{ e }{ \frac { dy }{ 1+{ y }^{ 2 } } ={ tan }^{ -1 } } y\)
\(\\ I=\left[ tan^{ -1 }y \right] ^{ e }_{ 1 }=\left[ { tan }^{ -1 }{ e }^{ x } \right] ^{ 1 }_{ 0 }\)
\(\\ ={ tan }^{ -1 }{ e }^{ 1 }-{ tan }^{ -1 }{ e }^{ 0 }={ tan }^{ -1 }e-\frac { \pi }{ 4 } \)

\(I=\int _{ 0 }^{ 1 }{ { e }^{ x^{ 2 } } } xdx\)
Let,  \(\Rightarrow xdx=\frac { 1 }{ 2 } dt\)
Put,  \({ x }^{ 2 }=t\)
\(\Rightarrow xdx=\frac { 1 }{ 2 } dt\)
Also when \(x=0 \Rightarrow t=0\) and when \(x=1 \Rightarrow t=1\)
\(\therefore \ I=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ e^{ t } } dt\)
\(\Rightarrow I=\frac { 1 }{ 2 } \left[ e^{ t } \right] ^{ 1 }_{ 0 }=\frac { 1 }{ 2 } (e-1)\)

\(I=\int _{ 0 }^{ 1 }{ \frac { tan^{ -1 }x }{ 1+x^{ 2 } } } dx\)
\(t={ tan }^{ -1 }x\)
\(\Rightarrow dt=\frac { 1 }{ 1+{ x }^{ 2 } } dx\)
Put, \(x=0 \Rightarrow t=0\)
\(x=1 \Rightarrow t=\frac { \pi }{ 4 } \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tdt } =\left[ \frac { { t }^{ 2 } }{ 2 } \right] ^{ \frac { \pi }{ 4 } }_{ 0 }\)
\(=\frac { 1 }{ 2 } \frac { { \pi }^{ 2 } }{ 16 } =\frac { { \pi }^{ 2 } }{ 32 } \)

\(\int { { e }^{ x }secx } dx+\int { { e }^{ x } } log\left| secx+tanx \right| dx\)
\(=\int { { e }^{ x } } secxdx+log\left| secx+tanx \right| { e }^{ x }dx-\int { \left( \frac { d }{ dx } log\left| secx+tanx \right| \right) } \int { { e }^{ x } } dx\)
\(=\int { { e }^{ x }secx } dx+log\left| secx+tanx \right| \times { e }^{ x }-\int { \frac { 1 }{ (secx+tanx) } } \times (secxtanx+{ sec }^{ 2 }x){ e }^{ x }dx\)
\(=\int { { e }^{ x } } secxdx+{ e }^{ x }log\left| secx+tanx \right| -\int { \frac { secx(tanx+secx) }{ (secx+tanx) } } { e }^{ x }dx\)
\(=\int { { e }^{ x } } sexdx+{ e }^{ x }log\left| secx+tanx \right| -\int { { e }^{ x }secxdx } \)
\(={ e }^{ x }log\left| secx+tanx \right| +C\)

\(\int \frac{\sin ^{2} x}{\cos ^{2} x} d x=\int \tan ^{2} x d x=\int\left(\sec ^{2} x-1\right) d x=\tan x-x+C \)

Let  \((x+2)=A \frac{d}{d x}\left(2 x^{2}+6 x+5\right)+B\) 
\(\Rightarrow (x+2)=A(4 x+6)+B \)
\(\Rightarrow (x+2)=4 A x+(6 A+B)\) 
On equating the coefficients of x and constant term
from both sides, we get 1 = 4A and 6A + B = 2
\( \Rightarrow A=\frac{1}{4} \text { and } 6 \cdot \frac{1}{4}+B=2 \Rightarrow A=\frac{1}{4} \text { and } B=2-\frac{3}{2} \)
\(\Rightarrow A=\frac{1}{4} \text { and } B=\frac{1}{2} \) 
\(\therefore\) From Eq. (i), we get \((x+2)=\frac{1}{4}(4 x+6)+\frac{1}{2}\) 
Now, the given integral can be written as
\( I =\int \frac{(x+2)}{2 x^{2}+6 x+5}=\int \frac{\frac{1}{4}(4 x+6)+\frac{1}{2}}{2 x^{2}+6 x+5} d x \)
\(=\frac{1}{4} \int \frac{4 x+6}{2 x^{2}+6 x+5} d x+\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5} \)
\(\Rightarrow I =\frac{1}{4} I_{1}+\frac{1}{2} I_{2} ...(ii) \) 
where, \(I_{1}=\int \frac{4 x+6}{2 x^{2}+6 x+5} d x\) 
and \(I_{2}=\int \frac{d x}{2 x^{2}+6 x+5}\) 
Now, consider \(I_{1}=\int \frac{4 x+6}{2 x^{2}+6 x+5} d x\) 
\(\therefore \ I_{1}=\int \frac{d t}{t}=\log |t|+C_{1}=\log \left|2 x^{2}+6 x+5\right|+C_{1}\) 
and  \( I_{2} =\int \frac{d x}{2 x^{2}+6 x+5}=\frac{1}{2} \int \frac{d x}{x^{2}+3 x+\frac{5}{2}} \)
\( =\frac{1}{2} \int \frac{d x}{\left.x\right|^{2}+2 \cdot \frac{3}{2} \cdot x+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+\frac{5}{2}} \) 
\(\left[\text { adding and subtracting }\left(\frac{3}{2}\right)^{2} \text { from denominator }\right]\) 
\( =\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{1}{2} \cdot \frac{1}{\left(\frac{1}{2}\right)} \tan ^{-1}\left(\frac{x+\frac{3}{2}}{\frac{1}{2}}\right)+C_{2} \)
\(=\frac{1}{2} \cdot 2 \tan ^{-1}(2 x+3)+C_{2}=\tan ^{-1}(2 x+3)+C_{2} \) 
\( =\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{1}{2} \cdot \frac{1}{\left(\frac{1}{2}\right)} \tan ^{-1}\left(\frac{x+\frac{3}{2}}{\frac{1}{2}}\right)+C_{2} \)
\(\left[\because \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right] \)
\( =\frac{1}{2} \cdot 2 \tan ^{-1}(2 x+3)+C_{2}=\tan ^{-1}(2 x+3)+C_{2} \) 
Now, substitute the value of I₁ and I₂ in Eq. (ii). Then,
\(I =\frac{1}{4} \log \left|2 x^{2}+6 x+5\right|+\frac{1}{2} \tan ^{-1}(2 x+3)+\frac{1}{4} C_{1}+\frac{1}{2} C_{2} \)
\(=\frac{1}{4} \log \left|2 x^{2}+6 x+5\right|+\frac{1}{2} \tan ^{-1}(2 x+3)+C \) 
where,\(C=\frac{1}{4} C_{1}+\frac{1}{2} C_{2}\)

\(I=\int _{ 2 }^{ 5 }{ \left[ \left| x-2 \right| +\left| x-3 \right| +\left| x-5 \right| \right] } dx\)
\(=\int _{ 2 }^{ 3 }{ f(x)dx+\int _{ 3 }^{ 5 }{ f(x)dx, } } \)
where \(f(x)=\left| x-2 \right| +\left| x-3 \right| +\left| x-5 \right| \)
 =\(\{\)\(x-2-x+3-x+5=6-x,\quad if\quad 2\le x\le 3\\ x-2+x-3-x+5=x,\quad if\quad 3\le x\ge 5\)
Now, from (i) we have 
\(I=\int _{ 2 }^{ 3 }{ (6-x)dx+ } \int _{ 3 }^{ 5 }{ xdx } \)
\(=\left[ 6x-\frac { { x }^{ 2 } }{ 2 } \right] ^{ 3 }_{ 2 }+\left[ \frac { { x }^{ 2 } }{ 2 } \right] ^{ 5 }_{ 3 }\)
\(=\left( 18-\frac { 9 }{ 2 } \right) -(12-2)+\frac { 1 }{ 2 } (25-9)\)
\(=\frac { 23 }{ 2 } \)

\(I=\int { \tan ^{ 8 }{ x } } (\sec ^{ 2 }{ x } )\)
\(=\int { \tan ^{ 8 }{ x } } (\sec ^{ 2 }{ x } )\sec ^{ 2 }{ x } dx\)
\(=\int { \tan ^{ 8 }{ x } } (1+\tan ^{ 2 }{ x } )\sec ^{ 2 }{ x } dx\)
\(=\int { \tan ^{ 8 }{ x } } \sec ^{ 4 }{ x } dx+ \int { \tan ^{ 10 }{ x } \sec ^{ 2 }{ x } } dx.\)
\(Put\ \tan { x } =t\) so that \(\sec ^{ 2 }{ x } dx=dt\)
\(\therefore I= \int { { t }^{ 8 } } dt+\int { { t }^{ 10 } } dt=\frac { { t }^{ 9 } }{ 9 } +\frac { { t }^{ 11 } }{ 11 } +c\)
\(I= \frac { \tan ^{ 11 }{ x } }{ 11 } +\frac { \tan ^{ 9 }{ x } }{ 9 } +C.\)

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

\({1\over\sqrt2}tan^{-1}({tan\ x - 1 \over \sqrt{2tan\ x}})+{1\over2\sqrt2} \ log|{tan\ x-\sqrt{2tan\ x}+1\over tan\ x + \sqrt{2tan\ x}+1}|+c\)

Let  \(I=\int \frac{(2 x-5) e^{2 x}}{(2 x-3)^3} d x=\int \frac{(2 x-3-2) e^{2 x}}{(2 x-3)^3} d x \)
\(=\int \frac{e^{2 x}}{(2 x-3)^2} d x-2 \int \frac{e^{2 x}}{(2 x-3)^3} d x \)
\(=\int e_{11}^{2 x}(2 x-3)^{-2} d x-2 \int e^{2 x}(2 x-3)^{-3} d x\)
\(=\left[\begin{array}{l} (2 x-3)^{-2} \int e^{2 x} d x \\ -\int\left\{\frac{d}{d x}(2 x-3)^{-2} \int e^{2 x} d x\right\} d x \end{array}\right] -2 \int e^{2 x}(2 x-3)^{-3} d x\)
[using integration by parts] 
\( =(2 x-3)^{-2} \frac{e^{2 x}}{2}-\int-2(2 x-3)^{-3} \times 2 \times \frac{e^{2 x}}{2} d x -2 \int e^{2 x}(2 x-3)^{-3} d x\)
\(=\frac{e^{2 x}(2 x-3)^{-2}}{2}+2 \int e^{2 x}(2 x-3)^{-3} d x-2 \int e^{2 x}(2 x-3)^{-3} d x\)
\(=\frac{e^{2 x}(2 x-3)^{-2}}{2}+C\)

Let \(I=\int \frac{x^2+x+1}{(x+2)\left(x^2+1\right)} d x\)
By partial fraction,
\(\frac{x^2+x+1}{\left(x^2+1\right)(x+2)}=\frac{A}{(x+2)}+\frac{B x+C}{x^2+1} \)
\(\Rightarrow x^2+x+1=A\left(x^2+1\right)+(B x+C)(x+2)\)
Putting x = -2,
\(4-2+1=A(5)+0 \Rightarrow 5 A=3 \Rightarrow A=\frac{3}{5}\)
Putting x = 0
0+0+1=A(0+1)+(0+C)(0+2)
\(\Rightarrow 1=A+2 C \Rightarrow 1=\frac{3}{5}+2 C \Rightarrow 2 C=\frac{2}{5} \Rightarrow C=\frac{1}{5}\)
and putting  x=1 ,
1+1+1=2 A+(B+C)(3)
\(\Rightarrow \quad 3=2 A+3(B+C)\)
\(\Rightarrow \quad 3=2\left(\frac{3}{5}\right)+3\left(B+\frac{1}{5}\right)\)
\(\Rightarrow  3-\frac{6}{5}=3\left(B+\frac{1}{5}\right)\)
\(\Rightarrow \frac{9}{5}=3\left(B+\frac{1}{5}\right) \Rightarrow \frac{3}{5}-\frac{1}{5}=B \Rightarrow B=\frac{2}{5}\)
\(\text { Thus, } \frac{x^2+x+1}{(x+2)\left(x^2+1\right)}=\frac{3}{5(x+2)}+\frac{\left(\frac{2}{5} x+\frac{1}{5}\right)}{\left(x^2+1\right)}\)
Now, \( \int \frac{x^2+x+1}{(x+2)\left(x^2+1\right)} d x\)
\(=\int \frac{3}{5(x+2)} d x+\frac{1}{5} \int \frac{2 x}{x^2+1} d x+\frac{1}{5} \int \frac{d x}{x^2+1}\)
\(=\frac{3}{5} \log |x+2|+\frac{1}{5} \log \left|x^2+1\right|+\frac{1}{5} \tan ^{-1} x+C\)

Let \(I=\int_{0}^{\pi} x \log |\sin x| d x\)   ...(i)  
\(\Rightarrow I=\int_{0}^{\pi}(\pi-x) \log |\sin (\pi-x)| d x\) 
\(=\int_{0}^{\pi}(\pi-x) \log |\sin x| d x\)   ..(ii)
On adding Eqs. (i) and (ii), we get
\(2 I=\pi \int_{0}^{\pi} \log |\sin x| d x\) 
\(\Rightarrow 2 I=2 \pi \int_{0}^{\pi / 2} \log |\sin x| d x\)    ...(iii)
\(\left[\because \int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x, \text { if } f(2 a-x)=f(x)\right]\) 
\(\Rightarrow I=\pi \int_{0}^{\pi / 2} \log |\sin x| d x\)  ...(iv)
\(\Rightarrow I=\pi \int_{0}^{\pi / 2} \log |\sin (\pi / 2-x)| d x\) 
\(\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) 
\(=\pi \int_{0}^{\pi / 2} \log |\cos x| d x\)   ...(v)
On adding Eqs. (iv) and (v), we get
\(2 I=\pi \int_{0}^{\pi / 2}(\log |\sin x|+\log |\cos x|) d x\) 
\(\Rightarrow 2 I I=\pi \int_{0}^{\pi / 2} \log |\sin x \cos x| d x\) 
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2} \log \left|\frac{2 \sin x \cos x}{2}\right| d x\) 
[multiply by 2 from numerator and denominator]
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2}(\log |\sin 2 x|-\log 2) d x\) 
\(2 I=\pi \int_{0}^{\pi / 2} \log |\sin 2 x| d x-\pi \int_{0}^{\pi / 2} \log 2 d x\) 
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2} \log |\sin 2 x| d x-\pi \log 2[x]_{0}^{\pi / 2}\) 
Now, put \(2 x=t \Rightarrow d x=\frac{1}{2} d t\) 
Lower limit When \(x=0, \text { then } t=0\) 
Upper limit When \(x=\frac{\pi}{2}, \text { then } t=\pi\) 
\(\therefore \ 2 I=\frac{\pi}{2} \int_{0}^{\pi} \log |\sin t| d t-\frac{\pi^{2}}{2} \log 2\) 
\(\Rightarrow 2 I=\frac{\pi}{2} \int_{0}^{\pi} \log |\sin x| d x-\frac{\pi^{2}}{2} \log 2\) 
\(\Rightarrow 2 I=I-\frac{\pi^{2}}{2} \log 2\) [from ii]
\(\therefore I=-\frac{\pi^{2}}{2} \log 2=\frac{\pi^{2}}{2} \log \left(\frac{1}{2}\right)\)

\(\text { Let } I=\int \frac{1}{\sin ^{4} x+\sin ^{2} x \cos ^{2} x+\cos ^{4} x} d x\) 
On dividing numerator and denominator by cos 4 x, we get
\( I =\int \frac{\sec ^{4} x}{\tan ^{4} x+\tan ^{2} x+1} d x \)
\(=\int \frac{\left(\sec ^{2} x\right)\left(\sec ^{2} x\right)}{\tan ^{4} x+\tan ^{2} x+1} d x \) 
Now, put \(\tan x=t \Rightarrow \sec ^{2} x d x=d t\) 
\(\left[\because \sec ^{2} x=1+\tan ^{2} x=1+t^{2}\right]\) 
\(\therefore I \neq \int \frac{1+t^{2}}{t^{4}+t^{2}+1} d t \quad[\because t=\tan x]\) 
\(I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)+C\)

\(\int _{ 1 }^{ 4 }{ \left( { x }^{ 2 }-x \right) dx } =\int _{ 1 }^{ 4 }{ { x }^{ 2 }dx } -\int _{ 1 }^{ 4 }{ { x }dx } ={ I }_{ 1 }-{ I }_{ 2 }\)
Let \({ I }_{ 1 }=\int _{ 1 }^{ 4 }{ { x }^{ 2 }dx } \)  [See Imp.keys and formulae]
Here  \(h-\frac { b-a }{ n } =\frac { 4-1 }{ n } =\frac { 3 }{ n } \)
\(=\left( 4-1 \right) \underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ f\left( 1 \right) +f\left( 1+\frac { 3 }{ n } \right) +......+n\ terms \right] \)
\(= 3\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ { 1 }^{ 2 }+{ \left( 1+\frac { 3 }{ n } \right) }^{ 2 }+\left( 1+\frac { 6 }{ n } \right) ^{ 2 }+......+\ n\ terms \right] \)
\(= 3\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ \left( { 1 }^{ 2 }+{ 1 }^{ 2 }+....n\ terms \right) +\left\{ \left( \frac { 3 }{ n } \right) ^{ 2 }+{ \left( \frac { 6 }{ n } \right) }^{ 2 }+.....+\left( n-1 \right) \ terms \right\} +\left\{ 2\times \frac { 3 }{ n } +2\times \frac { 6 }{ n } +.....+\left( n-1 \right) terms \right\} \right] \)
\(=3\underset { n\rightarrow \infty}{ lim } \frac { 1 }{ n } \left[ n+\frac { 9 }{ { n }^{ 2 } } \frac { \left( n-1 \right) n\left( 2n-1 \right) }{ 6 } +\frac { 6 }{ n } \left( \frac { n\left( n+1 \right) }{ 2 } \right) \right] \)
\(=3\left[ 1+2\times \frac { 9 }{ 6 } +3 \right] =3\times 7=21\)
and \({ I }_{ 2 }=\int _{ 1 }^{ 4 }{ xdx } \)
\(=\left( 4-1 \right) \underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ f\left( 1 \right) +f\left( 1+\frac { 3 }{ n } \right) +......+n\ terms \right] \)
\(= 3\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ { 1 }^{ 2 }+{ \left( 1+\frac { 3 }{ n } \right) }^{ 2 }+\left( 1+\frac { 6 }{ n } \right) ^{ 2 }+......+\ n\ terms \right] \)
\(= 3\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ n+\frac { { n }^{ 2 } }{ 2 } +\frac { 3n }{ 2 } \right] \)
\(= 3\underset { n\rightarrow \infty }{ lim } \left[ 1+\frac { 3 }{ 2 } \right] \)
\(= 3\left[ 1+\frac { 3 }{ 2 } \right] =\frac { 15 }{ 2 } \)
\(\therefore \int _{ 1 }^{ 4 }{ \left( { x }^{ 2 }-x \right) } .dx={ I }_{ 1 }-{ I }_{ 2 }=21-\frac { 15 }{ 2 } =\frac { 27 }{ 2 } .\)

\(I=\int { \frac { 1 }{ { sin }^{ 4 }x+{ sin }^{ 2 }x{ cos }^{ 2 }x+{ cos }^{ 4 }x } } dx\)
\(=\int { \frac { ({ tan }^{ 2 }x+1){ sec }^{ 2 }x }{ { tan }^{ 4 }x+{ tan }^{ 2 }x+1 } } dx\)
[ Dividing N & D by cos⁴x]
\(=\int { \frac { { t }^{ 2 }+1 }{ { t }^{ 4 }+{ t }^{ 2 }+1 } } dt\) where tanx=t
\(=\int { \frac { 1+\frac { 1 }{ { t }^{ 2 } } }{ { t }^{ 2 }+\frac { 1 }{ t^{ 2 } } +1 } } dt\quad \left[ dividing\quad N\quad \& \quad D\quad by\quad { t }^{ 2 } \right] \)
Putting  \(t-\frac { 1 }{ t } =z\quad so\quad that\quad \left( 1+\frac { 1 }{ { t }^{ 2 } } \right) dt=dz\)
and  \({ t }^{ 2 }\frac { 1 }{ t^{ 2 } } ={ z }^{ 2 }+2\)
\(\therefore \ I=\int { \frac { dz }{ { z }^{ 2 }+(\sqrt { 3 } )^{ 2 } } } \)
\(=\frac { 1 }{ \sqrt { 3 } } { tan }^{ -1 }\frac { z }{ \sqrt { 3 } } +C\)
\(=\frac { 1 }{ \sqrt { 3 } } tan^{ -1 }\frac { z }{ \sqrt { 3 } } +C\)
\(=\frac { 1 }{ \sqrt { 3 } } { tan }^{ -1 }\left( \frac { { t }^{ 2 }-1 }{ \sqrt { 3t } } \right) +C\)
\(=\frac { 1 }{ \sqrt { 3 } } { tan }^{ -1 }\left( \frac { { tan }^{ 2 }x-1 }{ \sqrt { 3 } tanx } \right) +C\)

\(I=\int { \frac { 1 }{ { cos }^{ 4 }x+{ sin }^{ 4 }x } } dx\)
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 \(t-\frac { 1 }{ t } =z\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { z }{ \sqrt { z } } \right) +C\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { { t }^{ 2 }-1 }{ \sqrt { 2t } } \right) +C\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { { tan }^{ 2 }x-1 }{ \sqrt { 2 } tanx } \right) +C\)