Question

Integrate : $\int \frac{{\sin }^{3}x-{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}dx$

Answer 1

$\int \frac{{\sin }^{3}x-{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}dx$
by simplifying,

$\frac{{\sin }^{3}x}{{\sin }^{2}x{\cos }^{2}x}-\frac{{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}$

$=\frac{\sin x}{\cos x}.\frac{1}{\cos x}-\frac{\cos x}{\sin x}.\frac{1}{\sin x}=\tan x\cdot \sec x-\cot x\cdot cosecx$

$⟹\frac{{\sin }^{3}x-{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}=\tan x\cdot \sec x-\cot x\cdot cosecx....\left(1\right)$

integrating equation (1),

$\int \sec x\cdot \tan xdx-\int cosecx\cdot \cot xdx$

Therefore,

$\int \frac{{\sin }^{3}x-{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}=\sec x+cosecx+C$ $\{\because \int secx.tanx=secx$ and,

$\int cosecx.cotx=-cosecx\}$