9
$\begingroup$

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 $t=\sqrt{\cot x}$

$\endgroup$
8
  • $\begingroup$ Are you sure it is not a definite integral? $\endgroup$
    – user371838
    Jan 8 '17 at 6:00
  • $\begingroup$ @Rohan It doesn't have limits, so it must be indefinite $\endgroup$ Jan 8 '17 at 6:01
  • $\begingroup$ 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: wolframalpha.com/input/… $\endgroup$ Jan 8 '17 at 6:02
  • $\begingroup$ It's possible with elementary functions. $\endgroup$ Jan 8 '17 at 6:04
  • $\begingroup$ @Rohan the answer given in my book as i.imgur.com/qHhVFpe.jpg $\endgroup$
    – Koolman
    Jan 8 '17 at 6:04
8
$\begingroup$

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

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

$\endgroup$
1
  • 1
    $\begingroup$ if we substitute $\tan x= t^2$ then $\displaystyle dx = \frac{2t}{1+t^4}dt$ $\endgroup$
    – Jane
    Jun 9 '20 at 13:46
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ After that does we have to substitue t = sin k $\endgroup$
    – Koolman
    Jan 8 '17 at 6:10
  • $\begingroup$ @koolman I have no this substitution in my hint. $\endgroup$
    – openspace
    Jan 8 '17 at 6:12
  • $\begingroup$ I mean what we have to after$\displaystyle\frac{2t}{(t^4 + 1)(t+1)}$ $\endgroup$
    – Koolman
    Jan 8 '17 at 6:14
  • $\begingroup$ @koolman we could represent this as sum of polynomials with denominators $t^4 + 1$ and $t + 1$ $\endgroup$
    – openspace
    Jan 8 '17 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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