Question

The value of $\int x\sin x{\sec }^{3}xdx$ is

• 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

Answer: (B)

$\frac{1}{2}[x{\sec }^{2}x-\tan x]+c$

$\int x\sin x{\sec }^{3}xdx=\int x\sin x\frac{1}{{\cos }^{3}x}dx$
$=\int x\tan x\cdot {\sec }^{2}xdx$
Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
and $x={\tan }^{-1}t$
Then, it reduces to
$\int {\tan }^{-1}t\cdot tdt=\frac{t^2}{2}{\tan }^{-1}t-\int \frac{t^2}{2(1+t^2)}dt$
$=\frac{x{\tan }^{2}x}{2}-\frac{1}{2}t+\frac{1}{2}{\tan }^{-1}t+c$
$=\frac{x({\sec }^{2}x-1)}{2}-\frac{1}{2}\tan x+\frac{1}{2}x+c$
$=\frac{1}{2}[x{\sec }^{2}x-\tan x]+c$