Question

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

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

Answer 1

Answer: (B)

$I=\int \frac{{\sin }^{2}x\cdot {\cos }^{2}xdx}{{\left\{\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{3}x+{\cos }^{3}x\right)\right\}}^2}$
Dividing the numerator and denominator by ${\cos }^{6}x$
$\Rightarrow I=\int \frac{{\tan }^{2}x{\sec }^{2}xdx}{{\left(1+{\tan }^{3}x\right)}^2}$
Let, ${\tan }^{3}x=z$
$\Rightarrow 3{\tan }^{2}x\cdot {\sec }^{2}xdx=dz$
$I=\frac{1}{3}\int \frac{dz}{z^2}=\frac{-1}{3z}+C$
$=\frac{-1}{3\left(1+{\tan }^{3}x\right)}+C$