<s>[INST] Why is $\int$ $1\over(a^2+b^2)^{3/2}$ $da$ $ = $ $ a\over b^2\sqrt {(a^2+b^2)}$ $$\int\frac{da}{(a^{2}+b^{2})^{3/2}}  =\frac{a}{b^{2}\sqrt{(a^{2}+b^{2})}}.$$
Found this in the solution to a problem in my physics textbook, and left clueless.
 [/INST] Using the trig substitution $a = b\tan \theta$, we get $a^2 + b^2 = b^2\sec^2\theta$ (since $\tan^2\theta + 1 = \sec^2\theta$), so $(a^2 + b^2)^{3/2} = b^3\sec^3\theta$. Since $da = b\sec^2\theta\, d\theta$, we obtain
$$\int \frac{1}{(a^2 + b^2)^{3/2}}\, da = \int \frac{1}{b^3\sec^3\theta} b\sec^2\theta\, d\theta = \frac{1}{b^2}\int \frac{d\theta}{\sec \theta} = \frac{1}{b^2}\int \cos \theta\, d\theta = \frac{1}{b^2}\sin\theta + C = \frac{a}{b^2\sqrt{a^2 + b^2}} + C.$$
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