# Integration by trigonometric substitution

• Sep 5th 2010, 09:59 PM
acevipa
Integration by trigonometric substitution
Could someone check whether I've got the correct answer for the following integrals?

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

I used the trigonometric substitution $\displaystyle x=4\tan\theta$ and got this as my answer

$\displaystyle dx=4\sec^2\theta \ d\theta$

$\displaystyle \Rightarrow\displaystyle\int{\dfrac{dx}{x^2\sqrt{x ^2+16}}=\int{\frac{4\sec^2\theta}{16\tan^2\theta\s qrt{16\tan^2\theta+16}}} \ d\theta$

$\displaystyle =\displaystyle\int{\frac{4\sec^2\theta}{16\tan^2\t heta\cdot 4\sec\theta}} \ d\theta$

$\displaystyle =\displaystyle\int{\frac{\sec\theta}{16\tan^2\thet a}} \ d\theta$

$\displaystyle =\displaystyle\frac{1}{16}\int{\frac{\cos\theta}{\ sin^2\theta}} \ d\theta$

Let $\displaystyle u=\sin\theta\Rightarrow du=\cos\theta \ d\theta$

$\displaystyle =\displaystyle\frac{1}{16}\int{\frac{1}{u^2}} \ du$

$\displaystyle =\displaystyle -\frac{1}{16u}+C$

$\displaystyle -\dfrac{1}{16\sin\theta} + C$

$\displaystyle \tan\theta=\dfrac{x}{4}\Rightarrow \sin\theta=\dfrac{x}{\sqrt{x^2+16}}$

$\displaystyle =-\dfrac{\sqrt{x^2+16}}{16x}+C$
• Sep 5th 2010, 10:07 PM
sa-ri-ga-ma