If f ( x ) = ∫ cosec 5 x d x , then f π 4 =

Given:

$f\left(x\right)=\int {cosec}^{5}x dx$

        $=\int \underset{\text{I}}{\left({cosec}^{3}x\right)}\underset{\text{II}}{\left({cosec}^{2}x\right)}dx$

$={cosec}^{3}x\left(-\cot x\right)-\int \left\{\left(-3{cosec}^{3}x\cot x\right)\left(-\cot x\right)\right\}dx$

$=-{cosec}^{3}x\cot x-\int 3{cosec}^{3}x{\cot }^{2}xdx$

$=-{cosec}^{3}x\cot x-\int 3{cosec}^{3}x\left({cosec}^{2}x-1\right)dx$

$=-{cosec}^{3}x\cot x-3\int {cosec}^{5}xdx+3\int {cosec}^{3}xdx$

$=-{cosec}^{3}x\cot x-3f\left(x\right)+3\int {cosec}^{3}xdx$

$\Rightarrow 4f\left(x\right)=-{cosec}^{3}x\cot x+3\int {cosec}^{3}xdx$

$\Rightarrow 4f\left(x\right)=-{cosec}^{3}x\cot x+3I ...\left(i\right)$

Now,

$I=\int {cosec}^{3}xdx$

  $=\int \left(cosecx\right)\left({cosec}^{2}x\right)dx$

$=\left(cosecx\right)\int \left({cosec}^{2}x\right)dx-\int \left\{\left(-cosecx \cot x\right)\left(-\cot x\right)\right\}dx$

$=-cosecx \cot x-\int \left(cosecx {\cot }^{2}x\right)dx$

$=-cosecx \cot x-\int cosecx\left({cosec}^{2}x-1\right)dx$

$=-cosecx \cot x-\int {cosec}^{3}xdx+\int \mathrm{cose}\text{c}xdx$

$\Rightarrow I=-cosecx \cot x-I+\int \mathrm{cose}\text{c}x dx$

$\Rightarrow 2I=\left(-cosecx \cot x\right)+\log \left|cosecx-\cot x\right|+C$

$\Rightarrow I=\frac{1}{2}\left\{\left(-cosecx \cot x\right)+\log \left|cosecx-\cot x\right|+C\right\}$

Putting in $\left(i\right)$, we get

$\Rightarrow 4f\left(x\right)=-{cosec}^{3}x \cot x+\frac{3}{2} \left\{\left(-cosecx \cot x\right)+\log \left|cosecx-\cot x\right|\right\}+c$

Putting $x=\frac{\mathrm{\pi }}{4}$ in above equation, we get

$\Rightarrow 4f\left(\frac{\mathrm{\pi }}{4}\right)=-{\left\{cosec\left(\frac{\mathrm{\pi }}{4}\right)\right\}}^3\cot \left(\frac{\mathrm{\pi }}{4}\right)+\frac{3}{2}\left[\left\{-cosec\left(\frac{\mathrm{\pi }}{4}\right)\cot \left(\frac{\mathrm{\pi }}{4}\right)\right\}+\log \left|cosec\left(\frac{\mathrm{\pi }}{4}\right)-\cot \left(\frac{\mathrm{\pi }}{4}\right)\right|\right]+c\text{'}$

 $\Rightarrow 4f\left(\frac{\mathrm{\pi }}{4}\right)=-2\sqrt{2}+\frac{3}{2}\left\{-\sqrt{2}+\log \left|\sqrt{2}-1\right|\right\}+c\text{'}$

$\Rightarrow 4f\left(\frac{\mathrm{\pi }}{4}\right)=-\frac{7\sqrt{2}}{2}+\frac{3}{2}\left(\log \left|\sqrt{2}-1\right|\right)+c\text{'}$

$\Rightarrow 4f\left(\frac{\mathrm{\pi }}{4}\right)=-\frac{7\sqrt{2}}{2}+\frac{3}{2}\left(\log \left|\frac{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)}\right|\right)+c\text{'}$

$\Rightarrow 4f\left(\frac{\mathrm{\pi }}{4}\right)=-\frac{7\sqrt{2}}{2}+\frac{3}{2}\left\{\log \left(\frac{1}{\sqrt{2}+1}\right)\right\}+c\text{'}$

$\Rightarrow 4f\left(\frac{\mathrm{\pi }}{4}\right)=-\frac{7\sqrt{2}}{2}-\frac{3}{2}\left\{\log \left(\sqrt{2}+1\right)\right\}+c\text{'}$

$\Rightarrow f\left(\frac{\mathrm{\pi }}{4}\right)=-\frac{1}{8}\left[7\sqrt{2}+3\left\{\log \left(\sqrt{2}+1\right)\right\}\right]+c$.