Question

Solve: \(\displaystyle\int\left[\sqrt{\cot x}+\sqrt{\tan x}\right]dx\).

Hint:

Combine both terms over sqrt(sin x cos x), substitute t = sin x - cos x, and integrate to an arcsine.

Answer 1

Step 1: Combine the two terms over a common form:
\[ \sqrt{\cot x}+\sqrt{\tan x}=\dfrac{\cos x}{\sqrt{\sin x\cos x}}+\dfrac{\sin x}{\sqrt{\sin x\cos x}}=\dfrac{\sin x+\cos x}{\sqrt{\sin x\cos x}} \]

Step 2: Rewrite \(\sin x\cos x\) using the double angle identity:
\(\sin x\cos x=\dfrac{\sin2x}2\), so the integral becomes \(\displaystyle\int\dfrac{\sqrt2(\sin x+\cos x)}{\sqrt{\sin2x}}dx\).

Step 3: Substitute \(t=\sin x-\cos x\), so \(dt=(\cos x+\sin x)\,dx\):
Also \(t^2=(\sin x-\cos x)^2=1-2\sin x\cos x=1-\sin2x\), so \(\sin2x=1-t^2\).
The integral becomes \(\displaystyle\int\dfrac{\sqrt2\,dt}{\sqrt{1-t^2}}\).

Step 4: Integrate using the standard arcsine form:
\[ \sqrt2\int\dfrac{dt}{\sqrt{1-t^2}}=\sqrt2\sin^{-1}(t)+C \]

Final Answer:
\[ \boxed{\sqrt2\,\sin^{-1}(\sin x-\cos x)+C} \]