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I'm trying to solve this integral with trigonometric substitution but am having a ton of trouble: $$\int\limits_{0}^{a}{\frac{dx}{(a^2+x^2)^{\frac{3}{2}}}}$$

I tried $x=a\tan{\theta}$ and thus $dx=a\sec^2{\theta}$ but I cannot get anywhere.

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    $\begingroup$ Hint: Try using substitution $x=a\sinh{t}.$ $\endgroup$ Feb 6 '16 at 22:06
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Assume, all the variables are real and positive so $a>0$:

$$\int_{0}^{a}\frac{1}{\left(a^2+x^2\right)^{\frac{3}{2}}}\space\text{d}x=$$


Substitute $u=\arctan\left(\frac{x}{a}\right)$ and $\text{d}x=a\sec^2(u)\space\text{d}u$;

This gives a new lower bound $u=\arctan\left(\frac{0}{a}\right)=0$ and

upper bound $u=\arctan\left(\frac{a}{a}\right)=\arctan\left(1\right)=\frac{\pi}{4}$:


$$a\int_{0}^{\frac{\pi}{4}}\frac{\cos(u)}{a^3}\space\text{d}u=$$ $$\frac{1}{a^2}\int_{0}^{\frac{\pi}{4}}\cos(u)\space\text{d}u=$$ $$\frac{1}{a^2}\left[\sin(u)\right]_{0}^{\frac{\pi}{4}}=$$ $$\frac{1}{a^2}\left(\sin\left(\frac{\pi}{4}\right)-\sin\left(0\right)\right)=$$ $$\frac{1}{a^2}\left(\frac{1}{\sqrt{2}}-0\right)=$$ $$\frac{\frac{1}{\sqrt{2}}}{a^2}=\frac{1}{a^2\sqrt{2}}$$

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Following your approach, let $x = a\tan \theta$. Then $dx = a \sec^{2} \theta$ and using $\frac{1}{\sec \theta} = \cos \theta$ and $\tan^{2} \theta + 1 = \sec^{2} \theta$, we have:

$I = \int_{0}^{a} \frac{1}{\left(a^{2} + x^{2} \right)^{\frac{3}{2}}} dx = \int_{0}^{\frac{\pi}{4}} \frac{1}{a^{3}\sec^{3} \theta} \cdot a\sec^{2} d \theta = \frac{1}{a^{2}} \int_{0}^{\frac{\pi}{4}} \cos \theta d\theta = \frac{\sqrt{2}}{2a^{2}}$

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