$$\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}\mathrm dx.$$

What we did:

we tried using $t=\tan (\frac x2)$ and also dividing both numerator and denominator by $\sqrt {\cos x}$, eventually using the second method we got to this: $\displaystyle \int \frac {2t+2}{t^2+2t-1}-\frac {2}{t^2+2t-1} +\frac {\sqrt{2t(1-t^2)}}{t^2+2t-1} $, for which we know how to solve the first and second integral but not the third...


  • $\begingroup$ Are you sure you need to find the anti-derivative, or do you have integration limits? $\endgroup$ – Daniel R Oct 18 '13 at 8:44
  • 3
    $\begingroup$ Are you sure that it is not definite integral ? $\endgroup$ – lab bhattacharjee Oct 18 '13 at 8:47
  • $\begingroup$ OP, this integral can be done in elementary terms, but it's gonna cost you in blood and tears. Are you absolutely certain you want the whole antiderivative, and not a definite integral perhaps? $\endgroup$ – David H Oct 18 '13 at 9:03
  • $\begingroup$ Yeah, We wanted it from 0 to $\frac \pi2$. I didn't know it mattered that much. Thx $\endgroup$ – jreing Oct 18 '13 at 15:08

Alternatively, if we call our integral $I$ and perform a dummy variable substitution, it all falls out rather nicely! $$I = \int_{0}^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x \ \overset{x \mapsto \frac{\pi}{2} - x}= \ \int_{0}^{\pi/2} \dfrac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x $$ Adding the two integrals together gives $$ 2I = \int_{0}^{\pi/2} \dfrac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x = \int_{0}^{\pi/2} \text{d}x = \dfrac{\pi}{2} \implies \ I = \dfrac{\pi}{4}$$

| cite | improve this answer | |
  • 2
    $\begingroup$ Nice answer! +1 before I sleep. Goodnite! $\endgroup$ – Anastasiya-Romanova 秀 Sep 4 '14 at 19:03

You mentioned dividing numerator and denominator by $\sqrt{\cos x}$.

$$\int\frac{\sqrt{\tan x}}{\sqrt{\tan x}+1}dx$$

Did you consider the substitution $x=\tan^{-1}u^2$?



I don't remember offhand how to factor $u^4+1$ into quadratics (probably from using the fourth roots of $-1$), but you should be able to use partial fractions from there.

| cite | improve this answer | |
  • $\begingroup$ We started out working on your suggestion, but really our integral is from (0 to $\frac \pi2$ ) - does that affect your substitution somehow or is it still legal doing so? $\endgroup$ – jreing Oct 19 '13 at 8:32
  • $\begingroup$ @user1685224 My guess is there are no major complications. We may want to keep in mind, though, the domain of the original function: sine and cosine must be positive. What I've technically done is $u=\sqrt{\tan x}$, hence replacing $\sqrt{\tan x}$ with u. It just makes it a lot easier to see what happens as done above. I'm starting to think things would have been easier had I done $u=-\sqrt{\tan x}$. I may try that out on paper, then expand my answer here. $\endgroup$ – Mike Oct 19 '13 at 20:09
  • $\begingroup$ @user1685224 Oops. the other substitution doesn't help. The wrong signs change. You're still left with needing to factor $u^4+1$. $\endgroup$ – Mike Oct 19 '13 at 20:16
  • $\begingroup$ An easier way would be to try to decompose $\tfrac{u^2}{u^4+1}$, dividing top and bottom by $u^2$, and using this trick yields $$\begin{align} \frac{1}{\dfrac1{x^2}+x^2}&=\frac{1}{\left(\dfrac{1}x+x\right)^2-2} \\&=\dfrac{1} {\left (\dfrac{1}{x}+x+\sqrt{2}\right)\left(\dfrac{1}{x}+x-\sqrt{2}\right)}\\&= \dfrac{A}{\dfrac{1}{x}+x+\sqrt{2}}+\dfrac{B}{\dfrac{1}{x}+x-\sqrt{2}}\\&=\ldots \end{align}$$ $\endgroup$ – Workaholic Dec 31 '15 at 19:49
  • $\begingroup$ @Workaholic - For this question Glasser's Master Theorem comes into play as well: en.wikipedia.org/wiki/Glasser%27s_master_theorem $\endgroup$ – user150203 Dec 28 '18 at 9:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.