Question

The value of $\int \frac{{\sin }^{2}x{\cos }^{2}x}{({\sin }^{3}x+{\cos }^{3}x{)}^2}dx$ is?

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

Answer 1

Answer: (B)

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

Let $I=\int \frac{{\sin }^{2}x{\cos }^{2}x}{({\sin }^{3}x+{\cos }^{3}x{)}^2}dx$
$=\int \frac{{\tan }^{2}x{\sec }^{2}x}{(1+{\tan }^{3}x{)}^2}dx$
[dividing numerator and denominator by ${\cos }^{6}x$]
$=\int \frac{{\tan }^{2}x{\sec }^{2}x}{(1+{\tan }^{3}x{)}^2}dx$

Put $\tan x=u⟹{\sec }^{2}xdx=du$

$⟹I=\int \frac{u^2}{(1+u^3{)}^2}du$

Put $u^3=v⟹3u^{2}du=dv$

$⟹I=\int \frac{dv}{3(1+v{)}^2}dv$

$=\frac{1}{3}\left(\frac{(1+v{)}^{-1}}{-1}\right)+C$

$=-\frac{1}{3(1+v)}+C$

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