Q.Find $\int \left[\sqrt{\cot x} + \sqrt{\tan x}\right] dx$

Concept: method of substitution (combine the surds over a common denominator, then substitute $t=\sin x-\cos x$).

Write both surds over $\sqrt{\sin x\cos x}$:

Let $t=\sin x-\cos x$, so $dt=(\cos x+\sin x)\,dx$ (exactly the numerator). Also

Hence $\sqrt{\sin x\cos x}=\sqrt{\dfrac{1-t^2}{2}}$, and …

Step 1 — Combine the surds

Write each term with sines and cosines and put them over a common denominator:

Step 2 — Pick the substitution

The numerator $\sin x+\cos x$ is the derivative of $\sin x-\cos x$. So let

Thus $(\sin x+\cos x)\,dx = dt$ — the numerator is handled exactly.

Step 3 — Express the denominator through $t$

Squaring,

so

Step 4 — Substitute into the integral

$\int \frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}\,dx = \int \frac{dt}{\frac{\sqrt{1-t^2}}{\sqrt{2}}} = \sqrt{2}\int \frac{dt}{\sqrt{1-t^2}}.$ …

Method: Combine $\sqrt{\tan x}+\sqrt{\cot x}$, then a $t\pm\tfrac1t$ substitution

Use this for integrals of $\sqrt{\tan x}\pm\sqrt{\cot x}$: rewrite over a common denominator, substitute to a single variable, and force a standard $\int\dfrac{du}{u^2+a^2}$ (or $\int\dfrac{du}{\sqrt{a^2-u^2}}$) form.

Steps

Step 1: Merge the two roots.

Since $\cot x=\dfrac{1}{\tan x}$,

Either single fraction is now substitutable.

Step 2: Substitute so the numerator becomes $du$.

Path A: let $t=\sqrt{\tan x}$, then $dx=\dfrac{2t}{1+t^4}\,dt$ and the integral becomes $2\int\dfrac{1+t^2}{1+t^4}\,dt$. Path B: let $t=\sin x-\cos x$, whose derivative $(\cos x+\sin x)$ is exactly the numerator.

Step 3: Reduce to a standard form. …

Common Mistakes

Mistake 1: Integrating $\sqrt{\tan x}$ and $\sqrt{\cot x}$ separately.

Why it's wrong: each root alone leads to a hard integral, whereas their sum combines into a single clean fraction. Correct approach: put them over a common denominator first ($\tfrac{\sin x+\cos x}{\sqrt{\sin x\cos x}}$) so a single substitution works.

Mistake 2: Choosing the substitution whose derivative does not match the numerator. …