Question

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

Answer 1

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

$=\int \frac{{\sin }^{2}x.{\cos }^{2}x}{{\{\left[{\sin }^{2}x[{\sin }^{3}x.{\cos }^{3}x]\right]+{\cos }^{2}x[{\sin }^{3}x+{\cos }^{3}x]\}}^2}dx$

$\int \frac{{\sin }^{2}x.{\cos }^{2}x}{{\left\{({\sin }^{2}x+{\cos }^{2}x)[{\sin }^{3}x+{\cos }^{3}x]\right\}}^2}dx$

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

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

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

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

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

$\therefore \int \frac{1}{3}\frac{dt}{t^2}=\frac{-1}{3}\frac{1}{t}+c$

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