Question

Find the value of \(\displaystyle\int\left(\sqrt{\cot x}+\sqrt{\tan x}\right)dx\).

Hint:

Combine both terms over √(sinx cosx), substitute t = sinx − cosx so dt matches the numerator.

Answer 1

Step 1: Combining under a common expression:
\(\sqrt{\cot x}+\sqrt{\tan x}=\sqrt{\dfrac{\cos x}{\sin x}}+\sqrt{\dfrac{\sin x}{\cos x}}=\dfrac{\cos x+\sin x}{\sqrt{\sin x\cos x}}\) (combining over a common denominator \(\sqrt{\sin x\cos x}\)).

Step 2: Choosing the substitution:
Let \(t=\sin x-\cos x\), so \(dt=(\cos x+\sin x)\,dx\) — this matches the numerator exactly.

Step 3: Rewriting the denominator in terms of t:
\(t^2=(\sin x-\cos x)^2=1-2\sin x\cos x\Rightarrow \sin x\cos x=\dfrac{1-t^2}{2}\).

Step 4: Substituting into the integral:
\(\displaystyle\int\dfrac{dt}{\sqrt{(1-t^2)/2}}=\sqrt2\int\dfrac{dt}{\sqrt{1-t^2}}=\sqrt2\sin^{-1}t+C\).

Step 5: Back-substituting:
Since \(t=\sin x-\cos x\).

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