# Trigonometric Substitution

• Mar 8th 2010, 02:17 PM
mj.alawami
Trigonometric Substitution
Question :

$\displaystyle \int x\sqrt{16+x^2}.dx$

Attempt :
$\displaystyle \int 4tan\theta .4sec\theta . 4 sec^2\theta$

How to continue?

• Mar 8th 2010, 02:32 PM
Miss
Quote:

Originally Posted by mj.alawami
Question :

$\displaystyle \int x\sqrt{16+x^2}$

Attempt :
$\displaystyle \int 4tan\theta .4sec\theta . 4 sec^2\theta$

How to continue?

1- You should have "dx" in the integral.

2- $\displaystyle tan(\theta)sec^3(\theta)=\frac{sin(\theta)}{cos^4( \theta)}$. Use $\displaystyle t=cos(\theta)$.

3- No need for the Trigonometric Substitution. Just use $\displaystyle u=16+x^2$.
• Mar 8th 2010, 02:44 PM
Quote:

Originally Posted by mj.alawami
Question :

$\displaystyle \int x\sqrt{16+x^2}$

Attempt :
$\displaystyle \int 4tan\theta .4sec\theta . 4 sec^2\theta$

How to continue?

Hi mj.alawami,

here's another way,

$\displaystyle 16\int{tan\theta\ sec^3\theta}d\theta=\int{u}dv$

Integrating by parts

$\displaystyle u=sec^2\theta\ \Rightarrow\ \frac{du}{d\theta}=\frac{du}{dsec\theta}\frac{dsec \theta}{d\theta}=2sec\theta\ sec\theta\ tan\theta=2sec^2\theta\ tan\theta$

$\displaystyle dv=tan\theta\ sec\theta\ d\theta$

$\displaystyle v=sec\theta$

$\displaystyle uv-\int{v}du=sec^3\theta-2\int{sec^3\theta\ tan\theta} d\theta$

$\displaystyle I=sec^3\theta-2I$

$\displaystyle 3I=sec^3\theta$

$\displaystyle I=\frac{1}{3}sec^3\theta$