Question

Evaluate: $\int \frac{1}{\sin x{\cos }^{2}x}dx$

Answer 1

$\int \frac{1}{\sin x{\cos }^{2}x}dx$

$=\int \frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x{\cos }^{2}x}dx$ [using ${\sin }^{2}x+{\cos }^{2}x=1$]

$=\int \frac{\sin x}{{\cos }^{2}x}dx+\int \frac{1}{\sin x}dx$

for the first integral, put $\cos x=t$

$\Rightarrow -\sin xdx=dt$

$\Rightarrow dx=\frac{-1}{\sin x}dt$

$=\int \frac{-dt}{t^2}+\int cosecxdx$

$=\frac{-(t{)}^{-2+1}}{-2+1}+(-ln|cosecx+\cot x|)$

$=\frac{1}{t}-(ln|cosecx+cotx|)$

$=\frac{1}{\cos x}-ln|cosecx+cotx|$

$=\sec x-ln|cosecx+\cot x|$