Question

$\int \frac{{\sin }^{2}x}{{\cos }^{4}x}dx=$

(a) $\frac{1}{3}{\tan }^{2}x+C$

(b) $\frac{1}{2}{\tan }^{2}x+C$

(c) $\frac{1}{3}{\tan }^{3}x+C$

(d) none of these

Answer 1

(c) $\frac{1}{3}{\tan }^{3}x+C$

$$\begin{gathered} \mathrm{Let}I=\int \frac{{\sin }^{2}xdx}{{\cos }^{4}x} \\ =\int \frac{{\sin }^{2}x}{{\cos }^{2}x}\times \frac{1}{{\cos }^{2}x}dx \\ =\int {\tan }^{2}x·{\sec }^{2}xdx \\ \mathrm{Let}\tan x=t \\ \Rightarrow {\sec }^{2}xdx=dt \\ \therefore I=\int t^2·dt \\ =\frac{t^3}{3}+C \\ =\frac{{\tan }^{3}x}{3}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }\mathit{}t=\tan \mathit{}x\right) \end{gathered}$$