Question

$\int x\sin x{\sec }^{3}xdx=$

• A. $\frac{1}{2}[{\sec }^{2}x-\tan x]+c$
• B. $\frac{1}{2}[x{\sec }^{2}x-\tan x]+c$
• C. $\frac{1}{2}[x{\sec }^{2}x+\tan x]+c$
• D. $\frac{1}{2}[{\sec }^{2}x+\tan x]+c$

Answer 1

The correct option is A $\frac{1}{2}[{\sec }^{2}x-\tan x]+c$

Consider the following integral .

$I=\int xsinxse{c}^{3}xdx=\underset{}{\overset{}{\int }}x\tan x{\sec }^{2}x$

let $t=\tan x$ and differentiate both side w.r.t x, we get.

$\frac{dt}{{\sec }^{2}x}=dx$

$=\underset{}{\overset{}{\int }}\frac{t{\tan }^{-1}t{\sec }^{2}x}{{\sec }^{2}x}dt$

$=\underset{}{\overset{}{\int }}t{\tan }^{-1}tdt$

$={\tan }^{-1}t\underset{}{\overset{}{\int }}tdt-\frac{1}{2}\underset{}{\overset{}{\int }}\frac{1}{1+t^2}{t}^{2}dt$

$={\tan }^{-1}t\underset{}{\overset{}{\int }}tdt-\frac{1}{2}\left(\underset{}{\overset{}{\int }}\frac{t^2+1-1}{1+t^2}dt\right)$

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

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

Hence, this is the correct answer.

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