Question

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

Hint:

Write both terms over a common denominator using sinx and cosx, then substitute \(t=\sin x-\cos x\).

Answer 1

Step 1: Understanding the Concept:
Write \(\sqrt{\tan x}\) and \(\sqrt{\cot x}\) in terms of sine and cosine so the two terms combine into one fraction.
This makes it possible to use a single substitution for the whole integral.

Step 2: Combine the terms:
\[ \sqrt{\tan x}+\sqrt{\cot x}=\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}=\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}} \]
Since \(\sin x\cos x=\dfrac{\sin2x}{2}\), rewrite the denominator.
\[ \sqrt{\tan x}+\sqrt{\cot x}=\frac{\sqrt2(\sin x+\cos x)}{\sqrt{\sin2x}} \]

Step 3: Choose the substitution:
Let \(t=\sin x-\cos x\), so \(dt=(\cos x+\sin x)dx\).
Also \(t^2=1-\sin2x\), so \(\sin2x=1-t^2\).
The integral becomes:
\[ \int\sqrt2\,\frac{dt}{\sqrt{1-t^2}} \]

Step 4: Integrate:
Use the standard result \(\displaystyle\int\dfrac{dt}{\sqrt{1-t^2}}=\sin^{-1}t+C\).
\[ \int\left[\sqrt{\tan x}+\sqrt{\cot x}\right]dx=\sqrt2\sin^{-1}t+C \]

Final Answer:
Substitute back \(t=\sin x-\cos x\). \[ \boxed{\int\left[\sqrt{\tan x}+\sqrt{\cot x}\right]dx=\sqrt2\sin^{-1}(\sin x-\cos x)+C} \]