Question

Evaluate: $\int \frac{{\sin }^{2}x{\cos }^{2}x}{{(\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x{)}^2}dx$

Answer 1

We have,

$\begin{array}{c}\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{{\left({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x\right)}^2}\\ \int \frac{{\sin }^{2}x.{\cos }^{2}xdx}{{\left\{\left({\sin }^{2}x\left({\sin }^{3}x+{\cos }^{3}x\right)+{\cos }^{2}x\left({\sin }^{3}x+{\cos }^{3}x\right)\right)\right\}}^2}\\ \int \frac{{\sin }^{2}x.{\cos }^{2}xdx}{{\left\{\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{3}x+{\cos }^{3}x\right)\right\}}^2}\\ \int \frac{{\sin }^{2}x.{\cos }^{2}xdx}{{\left({\sin }^{3}x+{\cos }^{3}x\right)}^2}\\ Divideby{\cos }^{3}xinnorminatoranddenominatorweget\\ \int \frac{{\sec }^{2}x.{\tan }^{2}xdx}{{\left({\tan }^{3}x+1\right)}^2}\\ Let1+{\tan }^{3}x=t\\ 3{\tan }^{2}x.{\sec }^{2}xdx=dt\\ =\frac{1}{3}\int \frac{dt}{t^2}=\frac{-1}{3}\frac{1}{t}+C\\ =-\frac{1}{3\left(1+{\tan }^{3}x\right)}+C\end{array}$

Hence, this is the answer.