Thread: Trig Substitution

Evaluate: int. 1/[x sqrt.(9+x^2)] dx
I am having trouble integrating it toward the end. Thanks for the help.

Hello, nystudent2729!

$\displaystyle \int \frac{dx}{x\sqrt{9+x^2}}$

Let: $\displaystyle x \,=\,3\tan\theta \quad\Rightarrow\quad dx \,=\,3\sec^2\!\theta\,d\theta$


Substitute: .$\displaystyle \int\frac{3\sec^2\!\theta\,d\theta}{3\tan\theta\cd ot3\sec\theta} \;=\;\frac{1}{3}\int\frac{\sec\theta}{\tan\theta}\ ,d\theta \;=\;\frac{1}{3}\int\frac{\frac{1}{\cos\theta}}{\f rac{\sin\theta}{\cos\theta}}\,d\theta $

. . . . . . . $\displaystyle = \;\frac{1}{3}\int\frac{d\theta}{\sin\theta} \;=\;\frac{1}{3}\int \csc\theta\,d\theta \qquad\hdots\;\text{Got it?}$