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

Answer 1

Answer: (D)

$\frac{-1}{3(1+{\tan }^{3}x)}+C$

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

$=\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{\left[\right({\sin }^{2}x+{\cos }^{2}x\left)\right({\sin }^{3}x+{\cos }^{3}x){]}^2}$

$=\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{({\sin }^{3}x+{\cos }^{3}x{)}^2}$

$=\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{{\cos }^{6}x({\tan }^{3}x+1{)}^2}$

$=\int \frac{{\tan }^{2}x{\sec }^{2}x}{({\tan }^{3}x+1{)}^2}$

Put ${\tan }^{3}x+1=t$

$\therefore 3{\tan }^{2}x{\sec }^{2}xdx=dt$

$=\int \frac{dt}{3t^2}$

$=\frac{-1}{3t}+1$

$=\frac{-1}{3({\tan }^{3}x+1)}+c$