Question

Evaluate:-
$\int \frac{{\tan }^{9}x}{\sin x\cos x}dx=$

• A. $\frac{1}{10}{\tan }^{9}x+c$
• B. $\frac{1}{5}{\tan }^{9}x+c$
• C. $\frac{1}{9}{\tan }^{9}x+c$
• D. none of these

Answer 1

Answer: (C)

$\frac{1}{9}{\tan }^{9}x+c$

We have to find

$\to \int \frac{{\tan }^{9}x}{\sin x\cos x}dx$

$\to \int \frac{{\tan }^{8}x\times sinx}{\sin x\times \cos x\times \cos x}dx$

$\to \int \frac{{\tan }^{8}x}{{\cos }^{2}x}dx$

$\to \int {\tan }^{8}x.{\sec }^{2}xdx$

Put $u=\tan x$

$du={\sec }^{2}xdx$

$\to \int u^{8}du$

$\to \frac{1}{9}{u}^9+c$

$\to \frac{1}{9}(\tan x{)}^9+c$