Question

The value of $\int \frac{{\cos }^{4}x}{{\sin }^{3}x{\left({\sin }^{5}x+{\cos }^{5}x\right)}^{\frac{3}{5}}}dx$ is

• A. $\frac{1}{3}{\left(2+{\tan }^{5}x\right)}^{\frac{2}{5}}+c$
• B. $\frac{1}{3}{\left(1+{\tan }^{5}x\right)}^{\frac{3}{5}}+c$
• C. $\frac{-1}{2}{\left(1+{\cot }^{5}x\right)}^{\frac{2}{5}}+c$
• D. $\frac{1}{2}{\left(1+{\cot }^{4}x\right)}^{\frac{3}{5}}+c$

Answer 1

Answer: (C)

$=\int \frac{{\cos }^{4}xdx}{{\sin }^{3}x{\left({\sin }^{5}x+{\cos }^{5}x\right)}^{3/5}}=\int \frac{{\cos }^{4}xdx}{{\sin }^{6}x{\left(1+{\cot }^{5}x\right)}^{3/5}}=\int \frac{{\sec }^{2}xdx}{{\tan }^{6}x{\left(1+\frac{1}{{\tan }^{5}x}\right)}^{3/5}}$
$\text{ Put }1+\frac{1}{{\tan }^{5}x}=t$
$\frac{-5}{{\tan }^{6}x}\cdot {\sec }^{2}xdx=dt$
$I=\int \frac{dt}{-5t^{3/5}}=\frac{-1}{5}\int t^{\frac{-3}{5}}dt=\frac{-1}{5}\cdot \frac{t^{\frac{-3}{5}+1}}{\frac{-3}{5}+1}+c$
$=\frac{-1}{5}\cdot \frac{5}{2}\cdot t^{\frac{2}{5}}+c=\frac{-1}{2}\cdot {\left(1+\frac{1}{{\tan }^{5}x}\right)}^{\frac{2}{5}}+c$