Question

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

Answer 1

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