Question

Integrate $\frac{{\sin }^{3}x+{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}$

Answer 1

We have,

$I=\int \frac{{\sin }^{3}x+{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}dx$

$I=\int \frac{{\sin }^{3}x}{{\sin }^{2}x{\cos }^{2}x}dx+\int \frac{{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}dx$

$I=\int \frac{\sin x}{{\cos }^{2}x}dx+\int \frac{\cos x}{{\sin }^{2}x}dx$

$I=I_1+I_2$ ……. (1)

Let $I_1=\int \frac{\sin x}{{\cos }^{2}x}dx$

Put $t=\cos x$

$\frac{dt}{dx}-\sin x$

$-dt=\sin xdx$

Therefore,

$I_1=-\int \frac{1}{t^2}dt$

$I_1=\frac{1}{t}+C$

On putting the value of $t$, we get

$I_1=\frac{1}{\cos x}+C$

Now,

$I_2=\int \frac{\cos x}{{\sin }^{2}x}dx$

Let $u=\sin x$

$du=\cos xdx$

Therefore,

$I_2=\int \frac{1}{u^2}du$

$I_2=-\frac{1}{u}+C$

Put the value of $u$, we get

$I_2=-\frac{1}{\sin x}+C$

From equation (1),

$I=\frac{1}{\cos x}-\frac{1}{\sin x}+C$

$I=\sec x-cosecx+C$

Hence, this is the answer.