Question

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

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

Answer 1

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

$\begin{array}{c}\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{{\left({\sin }^{2}x\left({\sin }^{3}x+{\cos }^{3}x\right)+{\cos }^{2}x\left({\sin }^3+cs{o}^{3}x\right)\right)}^2}\\ \Rightarrow \int \frac{{\sin }^2{\cos }^{2}x}{\left({\sin }^2+{\cos }^2\right){\left({\sin }^3+{\cos }^{3}x\right)}^2}\\ \Rightarrow \int \frac{{\sin }^{2}x{\cos }^{2}xdx}{{\left({\sin }^{3}x+{\cos }^{3}x\right)}^2}dividedby{\cos }^{3}x\\ \Rightarrow \int \frac{{\sec }^{2}x{\tan }^{2}x}{{\left({\tan }^{3}x+1\right)}^2}dx\left[\begin{array}{c}1+{\tan }^{3}x=t\\ 3{\tan }^{3}x{\sec }^{2}xdx=dt\end{array}\right]\\ \Rightarrow \frac{1}{3}\int \frac{dt}{t^2}=-\frac{1}{3}\frac{1}{t}+c\\ \Rightarrow -\frac{1}{3\left(1+{\tan }^{3}x\right)}+c\end{array}$