Question

$\int \frac{{\sin }^{6}x}{{\cos }^{8}x}dx=$

(a) tan 7x + C

(b) $\frac{{\tan }^{7}x}{7}+C$

(c) $\frac{\tan 7x}{7}+C$

(d) sec⁷ x + C

Answer 1

(b) $\frac{{\tan }^{7}x}{7}+C$

$$\begin{gathered} \mathrm{Let}I=\int \frac{{\sin }^{6}x}{{\cos }^{8}x}dx \\ =\int \frac{{\sin }^{6}x}{{\cos }^{6}x}\times \frac{1}{{\cos }^{2}x}dx \\ =\int {\tan }^{6}x·{\sec }^{2}xdx \\ \mathrm{Putting}\tan x=t \\ \Rightarrow {\sec }^2\mathrm{x}\mathit{}dx=dt \\ \therefore I=\int t^6·\mathrm{dt} \\ =\frac{t^7}{7}+C \\ =\frac{{\tan }^{7}x}{7}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }\mathit{}t=\tan \mathit{}x\right) \end{gathered}$$