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?

  • 1
    $\begingroup$ $\int_0^a\cdots\,dx$ means that the value of $x$ goes from $0$ to $a$. If $x=a\tan\theta$ this means that the value of $\theta$ goes from $0$ to $\pi/4$. So you get $\int_0^{\pi/4}\cdots\,d\theta$. $\endgroup$
    – David
    Feb 13, 2018 at 5:11

2 Answers 2


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


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

  • 1
    $\begingroup$ Once you square it, the terms inside the parentheses in the denominator become $a^2+a^2\tan^2\theta=a^2(1+\tan^2\theta)$, as shown by the next step. $\endgroup$ Feb 13, 2018 at 5:13
  • $\begingroup$ If this answer was helpful, feel free to accept it! $\endgroup$ Feb 13, 2018 at 5:20
  • $\begingroup$ Just did. I appreciate the help! $\endgroup$
    – sktsasus
    Feb 13, 2018 at 5:23
  • $\begingroup$ Many thanks. Happy to help! $\endgroup$ Feb 13, 2018 at 5:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.