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Problem : Find $\int \sqrt{\cot x} +\sqrt{ \tan x}dx$

My Working :

Let $I_1 = \sqrt{\cot x}dx$ and $I_2 = \sqrt{\tan x}dx$

By using integration by parts:

Therefore , $I_1 = \sqrt{\cot x}.\int1 dx - \int\{(d\sqrt{\cot x}\int 1.dx\}$

$\Rightarrow I_1= \sqrt{\cot x}x + 2x\sqrt{\cot x}-2\int \sqrt{\cot x}dx$

$\Rightarrow I_1= \sqrt{\cot x}x + 2x\sqrt{\cot x}-2 I_1$

$\Rightarrow 3I_1= \sqrt{\cot x}x + 2x\sqrt{\cot x} = 3x\sqrt{\cot x} $

$\Rightarrow I_1 = x\sqrt{\cot x}$

Similarly we can find $I_2 = \int\sqrt{\tan x}dx$

$I_2 = x\sqrt{\tan x}$

Please suggest whether this is wrong or correct...

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marked as duplicate by Mark Bennet, Shuhao Cao, Micah, Amzoti, Start wearing purple Jun 2 '13 at 17:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I don't think so, what is $T_1$ and how did you get it? What is that last result $3T_1=x\sqrt{\cot\,x}$? If you're saying $I_1=\int\sqrt{\cot\,x}\,dx=x\,\sqrt{\cot\,x}$, I don't think that's okay, I'm doing it and it seems to get quite long and can't be much simplified. Sorry I can't comment, hope this gets the question clearer. – user80668 Jun 2 '13 at 14:28
What is the derivative of $\sqrt{\cot x}?$ – lab bhattacharjee Jun 2 '13 at 14:57
up vote 2 down vote accepted

HINT: $$\sqrt{\cot x}+\sqrt{\tan x}=\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}$$

Now as $\int (\sin x+\cos x)dx=-\cos x+\sin x$

and $(-\cos x+\sin x)^2=1-2\sin x\cos x\implies \sin x\cos x=\frac{1-(-\cos x+\sin x)^2}2$

So, $$\sqrt{\cot x}+\sqrt{\tan x}=\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}=\sqrt2\frac{\sin x+\cos x}{\sqrt{1-(-\cos x+\sin x)^2}}$$

Put $-\cos x+\sin x=u$

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