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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?

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    $\begingroup$ $\int_0^a\cdots\,dx$ 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

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

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

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    $\begingroup$ 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

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