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We have to evaluate the following integral:

$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$

I tried this:

I multiplied both the numerator and denominator by $\sec x$
And substituted $\tan x = t$.

But after that I got stuck.

The book where this is taken from gives the following as the answer: $$\ln(1+t)-\frac14\ln(1+t^4)+\frac1{2\sqrt2}\ln\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}-\frac12\tan^{-1}t^2+c$$ where $t=\sqrt{\cot x}$

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    Are you sure it is not a definite integral? – 2017-01-08
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    @Rohan It doesn't have limits, so it must be indefinite – 2017-01-08
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    The indefinite integral is not nice to calculate, the answer if far too large, and involves many elliptic functions. Even Wolfram alpha won't calculate it: http://www.wolframalpha.com/input/?i=int+%28sqrt%28sin%28x%29%29+%2F%28+sqrt%28sin%28x%29%29+%2B+sqrt%28cos%28x%29%29%29%29+dx+ – 2017-01-08
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    It's possible with elementary functions. – 2017-01-08
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    @Rohan the answer given in my book as http://i.imgur.com/qHhVFpe.jpg – 2017-01-08
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    And if it's a definite integral from $0$ to $\frac{\pi}{2}$, then it is a very popular question and already has an answer [here](http://math.stackexchange.com/questions/167409/evaluating-displaystyle-int-0-frac-pi2-frac-sqrt-sin-x-sqrt-s) and [here](http://math.stackexchange.com/questions/530779/integral-of-int-0-pi-2-frac-sqrt-sin-x-sqrt-sin-x-sqrt-cos-x) – 2017-01-08
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    @SakethMalyala how ? – 2017-01-08
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    See also [Evaluation of $ \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$](http://math.stackexchange.com/q/875014). Found [using Approach0](https://approach0.xyz/search/?q=%24%5Cint%20%5Cfrac%7B%5Csqrt%7B%5Csin%20x%7D%7D%7B%5Csqrt%7B%5Csin%20x%7D%20%2B%20%5Csqrt%7B%5Ccos%20x%7D%7D%20dx%24&p=1). – 2017-01-08

3 Answers 3

5

$\displaystyle \mathcal{I} = \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx = \int \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$

substitute $\tan x= t^2$ and $\displaystyle dx = \frac{1}{1+t^4}dt$

$\displaystyle \mathcal{I}= \int\frac{t}{(1+t)(1+t^4)}dt = \frac{1}{2}\int\frac{\bigg((1+t^4)+(1-t^4)\bigg)t}{(1+t)(1+t^4)}dt$

$\displaystyle = \frac{1}{2}\int\frac{t}{1+t}dt+\frac{1}{2}\int\frac{(t-t^2)(1+t^2)}{1+t^4}dt$

$\displaystyle = \frac{1}{2}\int \frac{(1+t)-1}{1+t}dt+\frac{1}{2}\int \frac{t+t^3-(t^2-1)-t^4-1}{1+t^4}dt$

$\displaystyle =-\frac{t}{2}+\frac{1}{2}\ln|t+1|+\frac{1}{4}\int\frac{2t}{1+t^4}+\frac{1}{2}\int\frac{t^3}{1+t^4}dt-\frac{1}{2}\int \frac{t^2-1}{1+t^4}dt-\frac{1}{2}t+\mathcal{C}$

all integrals are easy except $\displaystyle \mathcal{J} = \int\frac{t^2-1}{1+t^4}dt = \int\frac{1-t^{-2}}{\left(t+t^{-1}\right)^2-2}dt = \int\frac{(t-t^{-1})'}{(t-t^{-1})^2-2}dt$

4

We rationalise the denominator to get $$I =\int \frac {\sin x-\sqrt {\cos x\sin x}}{\sin x-\cos x} dx $$ Writing everything in terms of $\cot x $, we get $$I =\int \mathrm{csc}^2 x\left(\frac {\sqrt {\cot x}-1}{\cot^3 x-\cot^2 x+\cot x-1}\right) dx $$ Now substituting $u=\cot x $ and further $v=\sqrt {u} $ gives us $$I = -\int\frac {2v }{v^5+ v^4 + v+1} dv $$ Performing a partial fraction decomposition we have $$I = \frac {2}{4-4\sqrt {2}}\int \frac {v +\sqrt {2}-1}{v^2 +\sqrt {2}v +1} dv +\frac {2}{4+4\sqrt {2}}\int \frac {v-\sqrt {2}-1}{v^2-\sqrt {2 }v+1} dv+\int \frac {1}{1+v} dv =I_1 +I_2 +I_3$$ Hope you can take it from here.

3

HINT multiply nominator and denominator by $\frac{1}{\sqrt{sin(x)}}$, then $t = \sqrt{cot(x)} $ after all you'll have $\displaystyle\frac{2t}{(t^4 + 1)(t+1)}$

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    After that does we have to substitue t = sin k – 2017-01-08
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    @koolman I have no this substitution in my hint. – 2017-01-08
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    I mean what we have to after$\displaystyle\frac{2t}{(t^4 + 1)(t+1)}$ – 2017-01-08
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    @koolman we could represent this as sum of polynomials with denominators $t^4 + 1$ and $t + 1$ – 2017-01-08