Definite Integral, Trig Substitution

Question

Evaluate the integral

$$\int _0^a\:\frac{dx}{\left(a^2+x^2\right)^{\frac{3}{2}}}. \quad a>0$$

So this is what I did:

$$x=a\tan\theta, \ dx = a\sec^2\theta d\theta,\\ \int _0^a\frac{a\sec^2\theta }{\left(a^2+a\tan\theta \right)^{\frac{3}{2}}}\,d\theta.$$

I'm not sure if I did the above step correctly because I looked that the next step is supposed to be:

$$\int _0^{\frac{\pi }{4}}\frac{\sec^2\theta }{\left(a^2\left(1+\tan^2\theta \right)\right)^{\frac{3}{2}}}\,d\theta.$$

I am a little confused as to how the bounds changed and how $a\tan\theta$ became $1+\tan^2\theta$.

I think its probably because I made a mistake before but I am not sure what it is.

Any help?

Answer 1

By assuming $a>0$ and letting $x=az$ $$ \int_{0}^{a}\frac{dx}{(a^2+x^2)^{3/2}} = \frac{1}{a^2}\underbrace{\int_{0}^{1}\frac{dz}{(1+z^2)^{3/2}}}_{\text{just a constant}}\stackrel{z\to\tan\theta}{=}\frac{1}{a^2}\int_{0}^{\pi/4}\cos(\theta)\,d\theta=\color{red}{\frac{1}{a^2\sqrt{2}}}. $$

Answer 2

I think your only mistake was forgetting to square the $a\tan \theta$ when you plugged it back in.