Question

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

Answer 1

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

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

$\Rightarrow \int \frac{\sin xdx}{{\cos }^{2}x(1-{\cos }^{2}x)}$

$\cos x=t$

$dt=-\sin xdx$

$\Rightarrow \int \frac{-dt}{t^2(1-t^2)}$

$\Rightarrow \int -\frac{dt}{t^2(1-t)(1+t)}$

$\Rightarrow \frac{-1}{t^2(1-t)(1+t)}=\frac{A}{t^2}+\frac{B}{1-t}+\frac{C}{1+t^2}$

$-1=A(1-t^2)+B(1+t)t^2+C(1-t)t^2$

Comparing coefficients

$-1=A+0+0\Rightarrow A=-1$

$0=-A+B+C$

$0=B-C\Rightarrow B=C$

$0=-(-1)+B+B$

$B=-1/2=C$

$\Rightarrow \int -\frac{1}{x^2}dt-\int \frac{dt}{2(1-t)}-\int \frac{dt}{2(1+t)}$

$\Rightarrow t^{-1}+\frac{ln(1-t)}{2}-\frac{ln(1+t)}{2}+c$.