Question

Write a value of $\int \frac{\sin x}{{\cos }^{3}x}dx$.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\sin x}{{\cos }^{3}x}dx \\ \mathrm{Let}\cos x=t \\ \Rightarrow -\sin xdx=dt \\ \Rightarrow \sin xdx=-dt \\ \therefore I=-\int \frac{dt}{t^3} \\ =-\int t^{-3}dt \\ =-\left[\frac{t^{-3+1}}{-3+1}\right]+C \\ =\frac{1}{2t^2}+C \\ =\frac{1}{2{\cos }^{2}x}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }t\mathit{=}\cos \mathit{}x\right) \\ =\frac{1}{2}{\sec }^{2}x+C\mathit{}\mathit{} \end{gathered}$$