Question

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

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

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

$\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{({\sin }^{5}x+co{s}^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x{)}^2}$

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

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

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

Divide by ${\cos }^{3}x$ in numerator and denominator we get

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

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

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

$=\frac{1}{3}\int \frac{dt}{t^2}$

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

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

Where c is the constant of integration.