If $f\left(x\right)=\int \cos e{c}^{5}xdx$ then $f\left(\frac{{\displaystyle \pi }}{{\displaystyle 4}}\right)=$

$f\left(x\right)=\int \cos e{c}^{5}xdx$

$I_n=\int \cos e{c}^{n}dx=\frac{-\cos e{c}^{n-2}x\cot x}{n-1}+\frac{n-2}{n-1}{I}_{n-2}$

$n=5\Rightarrow \int \cos e{c}^{5}dx=\frac{-\cos e{c}^{3}x cotx}{4}+\frac{3}{4}{I}_3$

$=\frac{-\cos e{c}^{3}x\cot x}{4}+\frac{3}{4}\left[\frac{-\cos ecx.\cot x}{2}+\frac{1}{2}{I}_1\right]$

$=\frac{-\cos e{c}^{3}x\cot x}{4}-\frac{3}{8}\cos ecx.\cot x+\frac{3}{8}\left[\log |\cos ecx-\cot x|\right]+c$

$f\left(\frac{{\displaystyle \pi }}{{\displaystyle 4}}\right)=-\frac{{\left(\sqrt{2}\right)}^3\left(1\right)}{4}-\frac{3}{8}\left(\sqrt{2}\right)\left(1\right)+\frac{3}{8}\left[\log |\sqrt{2}-1|\right]+c$

$=\frac{-1}{8}[4\sqrt{2}+3\sqrt{2}-3\log (\sqrt{2}-1)]+c$

$$\begin{gathered} =\frac{-1}{8}[7\sqrt{2}-3\log (\sqrt{2}-1)]+c \\ =\frac{-1}{8}[7\sqrt{2}+3\log (\sqrt{2}+1)]+c \end{gathered}$$