Question

The integral $\int \frac{{\sin }^{2}x{\cos }^{2}x}{{({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x)}^2}dx$ is equal to:

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

Answer 1

The correct option is C $\frac{-1}{3(1+{\tan }^{3}x)}+C$

$Wehave,I=\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{{\left({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x\right)}^2}=\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{{\left\{\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{3}x+{\cos }^{3}x\right)\right\}}^2}=\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{{\left({\sin }^{3}x+{\cos }^{3}x\right)}^2}$

$Divideby{\cos }^{3}xinnominatoranddenominatorweget=\int \frac{{\sec }^{2}x.{\tan }^{2}x}{{\left({\tan }^{3}x+1\right)}^2}dxLet1+{\tan }^{3}x=t3{\tan }^{2}x{\sec }^{2}xdx=dt$

Therefore,

$=\frac{1}{3}\int \frac{dt}{t^2}=-\frac{1}{3}\frac{1}{t}+C=-\frac{1}{3\left(1+{\tan }^{3}x\right)}+C$

$Whichistherequiredanswer.$