# Why does this type of substitution work in integration?

How does one integrate $du / (a^2 + u^2)^{3/2}$?

We want to express $(a^2 + u^2)^{3/2}$ as something without square roots. We want to use some form of the Pythagorean trigonometric identity $\sin^2 x + \cos^2 x = 1$. Multiplying each side by $\frac{a^2}{\cos^2 x}$, we get $a^2 \tan^2 x + a^2 = a^2 \sec^2 x$, which is in the desired form. of (sum of two squares) = (something squared).

This suggests that we should use the substitution $u^2 = a^2 \tan^2 x$. Equivalently, we substitute $u = a \tan x$ and $du = a \sec^2 x dx$. Then $$\int \frac{du}{(a^2 + u^2)^{3/2}} = \int \frac{a \sec^2 x \, dx}{(a^2 + a^2 \tan^2 x)^{3/2}}.$$ Applying the trigonometric identity considered above, this becomes $$\int \frac{a \sec^2 x \, dx}{(a^2 \sec^2 x)^{3/2}} = \int \frac{dx}{a^2 \sec x} = \frac{1}{a^2} \int \cos x \, dx,$$ which can be easily integrated as $$=\frac{1}{a^2} \sin x.$$ Since we set $u = a \tan x$, we substitute back $x = \tan^{-1} (\frac ua)$ to get that the answer is $$=\frac{1}{a^2} \sin \tan^{-1} \frac{u}{a}.$$ Since $\sin \tan^{-1} z = \frac{z}{\sqrt{z^2 + 1}}$, this yields the desired result of $$=\frac{u/a}{a^2 \sqrt{(u/a)^2 + 1}} = \frac{u}{a^2 (a^2 + u^2)^{1/2}}.$$

I have a doubt regarding the above method of substitution.I agree that "we set $u = a \tan x$" but why should "we substitute back $x = \tan^{-1} (\frac ua)$"?

I mean $x$ could even be $n\pi+\tan^{-1} (\frac ua)$? This is my doubt.

Another doubt.$\sin \tan^{-1} \frac{u}{a}=\sin (\sin^{-1} \pm (\frac{u/a} {\sqrt{(u/a)^2 + 1}}))$.So how do we decide to take the + sign or the - sign?

Why does these type of substitution work ?

It actually does not matter. If you consider $x=n\pi+\tan^{-1} (\frac ua)$, then the $n\pi$ part will go to the integration constant part and make a new integration constant. That's it. As far as an indefinite integral is concerned, the above substitution is well justified.
• I like your explaination to the question.But I have another doubt regarding the approach.$\sin \tan^{-1} \frac{u}{a}=\sin (\sin^{-1} \pm (\frac{u/a} {\sqrt{(u/a)^2 + 1}}))$.So how do we decide to take the + sign or the - sign? – user251680 Aug 24 '16 at 17:24
• @MathJack Depends on the domain of $u$. It will be different for different $u$. I think the $\pm$ should be added. Or as most solutions assume by convention, neglect the negative solution and take the positive one. – SchrodingersCat Aug 24 '16 at 17:28
• So you mean in case of definite integration suppose $u$ is from -2 to -1 then taking the positive sign will give wrong answer?Right? – user251680 Aug 24 '16 at 17:31