Question

If $\int \frac{{\cos }^{4}xdx}{{\sin }^{3}x{\left({\sin }^{5}x+{\cos }^{5}x\right)}^{\frac{3}{5}}}=-\frac{1}{K}{\left(1+{\cot }^{P}x\right)}^{\frac{K}{P}}+C$, then the value of $K+P$ is equal to (where $C$ is the constant of integration)

Answer 1

Answer: 7

Let, $I=\int \frac{{\cos }^{4}xdx}{{\sin }^{6}x{\left(1+{\cot }^{5}x\right)}^{\frac{3}{5}}}=\int \frac{{\sec }^{2}xdx}{{\tan }^{6}x{\left(1+{\left(\frac{1}{\tan x}\right)}^5\right)}^{\frac{3}{5}}}$
Substituting, $1+\frac{1}{(\tan x{)}^5}=t$ and $-\frac{5}{\left({\tan }^{6}x\right)}\left({\sec }^{2}x\right)dx=dt$ $I=-\frac{1}{5}\int \frac{dt}{t^{3/5}}=\left(\frac{-1}{5}\right)\frac{t^{\frac{2}{5}}}{\left(\frac{2}{5}\right)}+C$
$=-\frac{1}{2}\int {\left(1+{\cot }^{5}x\right)}^{\frac{2}{5}}+C$
Hence, $K=2,P=5$