Question

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

• A. $\frac{1}{(1+{\cot }^{2}x)}+c$
• B. $-\frac{1}{3(1+{\tan }^{3}x)}+c$
• C. $\frac{{\sin }^{3}x}{1+{\cos }^{3}x}+c$
• D. $-\frac{{\cos }^{3}x}{3(1+{\sin }^{3}x)}+c$

Answer 1

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

$I=\int \frac{{\sin }^{2}x{\cos }^{2}x}{({\sin }^{3}x+{\cos }^{3}x{)}^2}dx$
Dividing both denominator and numerator by ${\cos }^{6}x$, we get
$=\int \frac{{\tan }^{2}x{\sec }^{2}x}{{(1+{\tan }^{3}x)}^2}dx$
Substitute
$1+{\tan }^{3}x=t\Rightarrow 3{\tan }^{2}x{\sec }^{2}xdx=dt$
The integral becomes
$\frac{1}{3}\int \frac{dt}{t^2}=-\frac{1}{3}\times \frac{1}{t}+C=-\frac{1}{3(1+{\tan }^{3}x)}+C$