Find the integral of 1/(Sin⁵x+Cos^​5x)

Dear student, 

 $\int \frac{1}{{\sin }^5 x+{\cos }^5 x} dx$ cannot be found upto class 12 level.
We can find       I = $\int \frac{1}{\sin 5x+\cos 5x} dx$
         
Let  $5x=t \Rightarrow 5 dx=dt$ 
      $$\begin{gathered} I=\frac{1}{5}\int \frac{{\displaystyle dt}}{{\displaystyle \sin t+\cos t}} \\ \Rightarrow I=\frac{1}{5}\left[\frac{1}{\sqrt{2}}\int \frac{{\displaystyle 1}}{{\displaystyle \frac{1}{\sqrt{2}}\sin t+\frac{1}{\sqrt{2}}\cos t}} dt\right] \\ =\frac{1}{5}\left[\frac{1}{\sqrt{2}}\int \frac{{\displaystyle 1}}{{\displaystyle \sin \frac{\pi }{4}\sin t+\cos \frac{\pi }{4}\cos t}} dt\right] \\ =\frac{1}{5\sqrt{2}}\left[\int \frac{{\displaystyle 1}}{{\displaystyle \cos \left(t-\frac{\pi }{4}\right)}} dt\right]=\frac{1}{5\sqrt{2}}\left[\int sec \left(t-\frac{\pi }{4}\right) dt\right] \\ =\frac{1}{5\sqrt{2}}\left[\log \left|\tan \left(\frac{\pi }{4}+\frac{t-\frac{\pi }{4}}{2}\right)\right|\right]+C=\frac{1}{5\sqrt{2}}\left[\log \left|\tan \left(\frac{\pi +4t}{8}\right)\right|\right]+C \\ \frac{1}{5\sqrt{2}}\left[\log \left|\tan \left(\frac{\pi +20x}{8}\right)\right|\right]+C \left[\mathrm{since}, t=5x\right] \end{gathered}$$