Kindly considwer the following

 

$I={\displaystyle \underset{0}{\overset{\pi /4}{\int }}\frac{xdx}{si{n}^4\left(2x\right)+co{s}^4\left(2x\right)}}$

           Let 2x = t then   $dx=\frac{1}{2}dt$

$I={\displaystyle \underset{}{\overset{}{\int }}\frac{\frac{t}{2}\cdot \frac{1}{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{tdt}{si{n}^{4}t+co{s}^{4}t}dt}$            

$I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\left(\frac{\pi }{2}-t\right)dt}{si{n}^{4}t+co{s}^{4}t}dt}$

$2I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\pi }{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$2I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\pi }{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$2I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{si{n}^{4}tdt}{ta{n}^{4}t+1}}$            

Let tan t = y then

$2I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\left(1+y^2\right)dy}{1+y^4}}$             

$=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}-2+2}dy}$

$=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\left(1+\frac{1}{y^2}\right)dy}{2+{\left(y-\frac{1}{y}\right)}^2}}$             

Let $y-\frac{1}{y}=u$  

$2I=\frac{\pi }{8}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{du}{2+u^2}}$

$=\frac{\pi }{8\sqrt[]{2}}{\left[ta{n}^{-1}\frac{4}{\sqrt[]{2}}\right]}_{-\infty }^{\infty }$                  

$I=\frac{{\pi }^2}{16\sqrt[]{2}}$

$k=\frac{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]-\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}$            

k = 3

$\underset{x\to 0}{lim}{\left[si{n}^2\left(\frac{\pi }{2-3x}\right)\right]}^{se{c}^2\left(\frac{\pi }{2-5x}\right)}$

$e^{\underset{x\to 0}{lim}\left[si{n}^2\left(\frac{\pi }{2-3x}\right)-1\right]se{c}^2\left(\frac{\pi }{2-5x}\right)}$

$=e^{\underset{x\to 0}{lim}\frac{-si{n}^2\left(\frac{-3\pi x}{2\left(2-3x\right)}\right)}{si{n}^2\left(\frac{-5\pi x}{2\left(2-5x\right)}\right)}}=e^{-\frac{9}{25}}$

I = ∫ (e²? + 2e? - e? - 1)e^ (e? +e? ) dx
I = ∫ (e²? + e? - 1)e^ (e? +e? ) dx + ∫ (e? - e? )e^ (e? +e? ) dx
I = ∫ (e? + 1 - e? )e^ (e? +e? ) dx + e^ (e? +e? )
(e? - e? + 1)dx = du
I = e^ (e? +e? ) + e^ (e? +e? ) = e^ (e? +e? ) (e? + 1) then g (x) = e? + 1
g (0) = 2