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Q.
The value of $\int \frac{\cos ^4 x}{\sin ^3 x\left(\sin ^5 x+\cos ^5 x\right)^{\frac{3}{5}}} d x$ is
Integrals
Solution:
$=\int \frac{\cos ^4 x d x}{\sin ^3 x\left(\sin ^5 x+\cos ^5 x\right)^{3 / 5}}=\int \frac{\cos ^4 x d x}{\sin ^6 x\left(1+\cot ^5 x\right)^{3 / 5}}=\int \frac{\sec ^2 x d x}{\tan ^6 x\left(1+\frac{1}{\tan ^5 x}\right)^{3 / 5}}$
$\text { Put }1+\frac{1}{\tan ^5 x }= t $
$ \frac{-5}{\tan ^6 x } \cdot \sec ^2 x d x = dt $
$I =\int \frac{ dt }{-5 t ^{3 / 5}}=\frac{-1}{5} \int t ^{\frac{-3}{5}} dt =\frac{-1}{5} \cdot \frac{ t ^{\frac{-3}{5}+1}}{\frac{-3}{5}+1}+ c$
$=\frac{-1}{5} \cdot \frac{5}{2} \cdot t ^{\frac{2}{5}}+ c =\frac{-1}{2} \cdot\left(1+\frac{1}{\tan ^5 x }\right)^{\frac{2}{5}}+ c$