# Difficult integral with trigonometric substitution: $\int\frac{\sqrt{1-x}}{\sqrt x}dx$

This is the integral I am talking about

$$\int\frac{\sqrt{1-x}}{\sqrt x}dx$$

As you can tell I tried substitution $u = \sqrt x$, and from there I went to $u = \sin{\theta}$ like here.

$$\newcommand{\dd}{\; \mathrm{d}} \int \frac{\sqrt{1-x}}{\sqrt x} \dd x= \begin{vmatrix} u=\sqrt x \\ \dd u = \frac1{2\sqrt x}\end{vmatrix} = 2\int \sqrt{1-u^2} \dd u = \begin{vmatrix} u=\sin\theta \\ \theta = \arcsin u \\ \dd u = \cos\theta \dd\theta \end{vmatrix} = 2\int \sqrt{1-\sin^2\theta} \cos\theta \dd\theta = 2\int \cos^2\theta \dd\theta= 2\int \frac{1+\cos2\theta}2 \dd\theta = \int \dd\theta + \int\cos2\theta \dd\theta = \theta + \frac12 \sin2\theta + C = \arcsin u + \frac12 \sin(2\arcsin u) + C = \arcsin \sqrt x + \frac12 \sin(2\arcsin \sqrt x) + C$$

Anyways, you can see I entered this answer and was wrong and I have no idea why. I even tried to go back and substitute in cosine instead of sine since in this case they were equivalent and I still got the answer wrong, any help is appreciated thank you!

## 1 Answer

Consdering, just as you apparently did, $$I=\int \frac{\sqrt{1-x}}{\sqrt{x}}\,dx$$ First $$x=u^2\implies dx=2u\,du\implies I=2\int \sqrt{1-u^2}\,du$$ Second $$u=\sin(t)\implies du=\cos(t)\,dt \implies I=2\int \cos^2(t) dt=\int (1+\cos(2t))\,dt$$ $$I=t+\frac{1}{2} \sin (2 t)+C$$ Now, back to $x$ $$t=\sin^{-1}(u)=\sin^{-1}(\sqrt x) \implies I=\sin ^{-1}(\sqrt x)+\frac 12\sin ^{-1}\left(2\sin ^{-1}(\sqrt x)\right)+C$$ which could simplify to $$I=\sqrt{(1-x) x}+\sin ^{-1}\left(\sqrt{x}\right)+C$$

• Hey thank you so much for the reply and the help! (This answer was correct so thank you for that as well!) but do you mind explaining what identities you used when substituting back in x and simplifying? – Aleksandr Diamond Feb 12 '17 at 5:10
• @AleksandrDiamond It may help to write the integral as $t+\sin t\cos t+C$ before back-substitution. – Mike Feb 12 '17 at 9:29