Question

Integral $x\sin xse{c}^{3}xdx$ is equal to

Answer 1

Step 1: Find ‘u’ and ‘v’

Given, $x\sin xse{c}^{3}x$

Now,

We know that , $\int u.vdx=u\int vdx-\int \left\{\frac{du}{dx}\int vdx\right\}dx$

Now ,

$\begin{array}{rcl}\int x\sin xse{c}^{3}xdx& =& \int x.\sin x.\left(\frac{1}{{\cos }^{3}x}\right)dx\\ & =& \int x.\tan x.se{c}^{2}xdx\\ & =& x\int secx\left(secx\tan x\right)dx-\int 1·\left[\int \sec x\left(\sec x\tan x\right)dx\right]dx[u=x,v=\tan x.{\sec }^{2}x]\end{array}$

Step 2: Find the value of $\int secx\left(secx\tan x\right)dx$

Put $secx=t$

$$\begin{gathered} \Rightarrow secx\tan xdx=dt \\ \therefore \int secx\left(secx\tan x\right)dx=\int tdt \\ =\frac{t^2}{2}=\frac{se{c}^{2}x}{2} \\ \therefore \int x\sin xse{c}^{3}xdx=x\left(\frac{se{c}^{2}x}{2}\right)-\int \left(\frac{se{c}^{2}x}{2}\right)dx \\ =x\left(\frac{se{c}^{2}x}{2}\right)-\left(\frac{\tan x}{2}\right)+C\left[\because \int \left(se{c}^{2}x\right)=\tan x+C\right] \end{gathered}$$

Hence , integral of $x\sin xse{c}^{3}xdx$ is $x\left(\frac{se{c}^{2}x}{2}\right)-\left(\frac{\tan x}{2}\right)+C$.