Question

Solve $\int \frac{1}{\sin x{\cos }^{3}x}dx$

Answer 1

Consider the given integral.

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

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

$I=\int \frac{\sin xdx}{(1-{\cos }^{2}x){\cos }^{3}x}$

Let $t=\cos x$

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

$-dt=\sin xdx$

Therefore,

$I=-\int \frac{dt}{(1-t^2)t^3}$

$I=\int \frac{dt}{(t^2-1)t^3}$

$I=\int \frac{dt}{(t+1)(t-1)t^3}$

$I=\int \frac{dt}{2(t+1)}+\int \frac{dt}{2(t-1)}-\int \frac{dt}{t}-\int \frac{dt}{t^3}$

$I=\frac{1}{2}\ln (t+1)+\frac{1}{2}\ln (t-1)-\ln \left(t\right)+\frac{1}{2t^2}+C$

On putting the value of $t$, we get

$I=\frac{1}{2}\ln (\cos x+1)+\frac{1}{2}\ln (\cos x-1)-\ln \left(\cos x\right)+\frac{1}{2{\left(\cos x\right)}^2}+C$

Hence, this is the answer.