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

$\bf{My\; Try::}$ Given $\displaystyle \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx = \int \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$

Now Let $\displaystyle \tan x = t^2\;,$ Then $\sec^2 xdx = 2tdt$ or $\displaystyle dx = \frac{2t}{1+t^4}dt$

So Integral is $\displaystyle \int\frac{t}{1+t}\cdot \frac{2t}{1+t^4}dt = 2\int\frac{t^2}{(1+t)\cdot (1+t^4)}dt$

Now How can I solve after that

Help me

Thanks

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  • $\begingroup$ Split the last integral into partial fractions $\to \int {1\over 1 + t} - {(1+t)(t-1)^2\over 1 + t^4} dt$ $\endgroup$
    – Winther
    Jul 22, 2014 at 18:12
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    $\begingroup$ Hint: $x^4+1=(x^2+x\sqrt2+1)(x^2-x\sqrt2+1)$. $\endgroup$
    – Lucian
    Jul 22, 2014 at 18:41
  • $\begingroup$ Use Lucien hint and completing the square in each term and then use partial fraction $\endgroup$
    – Mohammad W. Alomari
    Jul 22, 2014 at 21:25
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    $\begingroup$ I'm voting to reopen because the proposed duplicate concerns a definite integral, which allows slick solutions (by abusing symmetry). This question mainly focus on how to solve the indefinite integral. $\endgroup$
    – YuiTo Cheng
    Feb 28, 2020 at 11:45
  • $\begingroup$ This question isn't a duplicate of this. $\endgroup$
    – Zacky
    Apr 5, 2020 at 18:05

1 Answer 1

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By collecting all the suggestions, you should be able to prove that: $$\frac{2t}{(t+1)(t^4+1)}=\frac{t}{1+t^4}+\frac{t^3}{1+t^4}+\frac{1-t^2}{1+t^4}-\frac{1}{1+t},$$ and since $(1+t^4)=(1+\sqrt{2}t+t^2)(1-\sqrt{2}t+t^2)$, it follows that: $$\int\frac{2t\,dt}{(t+1)(t^4+1)}=\frac{1}{2}\arctan t^2+\frac{1}{4}\log(1+t^4)+\frac{1}{2\sqrt{2}}\log\frac{1+\sqrt{2}t+t^2}{1-\sqrt{2}t+t^2}-\log(1+t).$$ $\\$

Addendum

Suggestion for evaluating $\displaystyle{\int \frac{1 - t^2}{t^4 + 1}\,\mathrm{d}t}$: Write $$ \begin{aligned} \int- \frac{t^2 - 1}{t^4 + 1}\,\mathrm{d}t &= -\int \frac{1 - \dfrac{1}{t^2}}{t^2 + \dfrac{1}{t^2}}\,\mathrm{d}t \\ &=-\int \frac{1 - \dfrac{1}{t^2}}{\left(t + \dfrac{1}{t}\right)^2 - 2}\,\mathrm{d}t \end{aligned} $$

Now, set $\displaystyle{u = t + \frac{1}{t}}$ and $\mathrm{d}u = \left(1 - \dfrac{1}{t^2}\right)\,\mathrm{d}t$: $$ \begin{aligned} -\int \frac{\mathrm{d}u}{u^2 - 2} &= \dfrac{1}{2\sqrt{2}}\ln\left|\frac{u + \sqrt{2}}{u - \sqrt{2}} \right| + C \\ &=\frac{1}{2\sqrt{2}}\ln\left|\frac{t^2 + t\sqrt{2} +1}{t^2 - t\sqrt{2}+1} \right| + C \end{aligned} $$ Note that $t^2 + t\sqrt{2}+1 = \left(t + \dfrac{1}{\sqrt{2}}\right)^{\!2} + \dfrac{1}{2} >0$ and $t^2 - t\sqrt{2}+1 = \left(t - \dfrac{1}{\sqrt{2}}\right)^{\!2} + \dfrac{1}{2} >0$, then we can get rid of the absolute value bars: $$\int \frac{1-t^2}{t^4 + 1}\,\mathrm{d}t = \frac{1}{2\sqrt{2}}\ln\!\left(\dfrac{t^2 + t\sqrt{2}+1}{t^2 - t\sqrt{2}+1}\right) + C $$

This is a very famous technique.

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