Problem : Find $\int \sqrt{\cot x} +\sqrt{ \tan x}\,dx$
My Working :
Let $I_1 = \int \sqrt{\cot x}\,dx$ and $I_2 = \int \sqrt{\tan x}\,dx$
By using integration by parts:
Therefore , $I_1 = \sqrt{\cot x}.\int1 dx - \int\{(d\sqrt{\cot x}\int 1.dx\}$
$\Rightarrow I_1= x\sqrt{\cot x} + 2x\sqrt{\cot x}-2\int \sqrt{\cot x}\,dx$
$\Rightarrow I_1= x\sqrt{\cot x} + 2x\sqrt{\cot x}-2 I_1$
$\Rightarrow 3I_1= x\sqrt{\cot 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...
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 $-\cos x+\sin x=u$