Q.Integrate the following function: $\frac{\sqrt{\tan x}}{\sin x \cos x}$

The key idea is to rewrite the integrand in terms of $\tan x$ and then use U Substitution.

First, simplify the expression:

Since $\sec^2 x = 1 + \tan^2 x$, but here we notice a cleaner path: $\frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$, but substitution is simpler.

Rewrite directly: …

When you see a messy mix of $\tan x$, $\sin x$, and $\cos x$, your first instinct should be to simplify the algebra. The denominator $\sin x \cos x$ is a classic hint — it’s closely related to the derivative of $\tan x$.

Recall that $\frac{d}{dx}(\tan x) = \sec^2 x = \frac{1}{\cos^2 x}$. That doesn’t look like $\sin x \cos x$ at first, but we can connect them. Also, $\sqrt{\tan x}$ suggests that if we set $u = \tan x$, then $\sqrt{u}$ appears, and the derivative $du = \sec^2 x \, dx$ might clean up the denominator.

Let’s rewrite the integrand to make this substitution obvious.

1. Rewrite the integrand in terms of $\tan x$ and $\sec x$. We have $\frac{\sqrt{\tan x}}{\sin x \cos x}$. Multiply numerator and denominator by $\frac{1}{\cos^2 x}$ (a common trick):

Why? Because $\frac{\sin x}{\cos x} = \tan x$, and $\frac{1}{\cos^2 x} = \sec^2 x$. So the denominator becomes $\tan x$, and the numerator picks up a $\sec^2 x$.

2. Simplify the fraction. Now we have $\frac{\sqrt{\tan x} \cdot \sec^2 x}{\tan x} = \frac{\sec^2 x}{\sqrt{\tan x}}$, since $\frac{\sqrt{\tan x}}{\tan x} = \frac{1}{\sqrt{\tan x}}$. So the integral becomes:

3. Spot the substitution. The derivative of $\tan x$ is $\sec^2 x$. That’s exactly the numerator! Let $u = \tan x$, then $du = \sec^2 x \, dx$. The integral transforms beautifully: …

Method: Rewrite a sin/cos integrand to expose $\tan x$, then substitute

This method handles trigonometric integrals where a $\sqrt{\tan x}$ (or a power of $\tan x$) sits over a product like $\sin x\cos x$ — the goal is to reshape the integrand until the derivative of $\tan x$ appears.

Steps

Step 1: Force $\sec^2 x$ to surface by dividing top and bottom by $\cos^2 x$.

A denominator of $\sin x\cos x$ is the telltale sign. Multiplying numerator and denominator by $\dfrac{1}{\cos^2 x}$ turns $\dfrac{1}{\sin x\cos x}$ into $\dfrac{\sec^2 x}{\tan x}$, because $\dfrac{\sin x}{\cos x}=\tan x$ and $\dfrac{1}{\cos^2 x}=\sec^2 x$. This is legal since you multiply by $1$.

Step 2: Simplify the powers of $\tan x$. …

Common Mistakes

Mistake 1: Trying to integrate $\dfrac{\sqrt{\tan x}}{\sin x\cos x}$ directly without reshaping it.

Why it's wrong: as written there is no visible $u$ whose derivative is present, so students stall or guess. Correct approach: multiply top and bottom by $\dfrac{1}{\cos^2 x}$ to convert the integrand into $\dfrac{\sec^2 x}{\sqrt{\tan x}}$, which makes $u=\tan x$ obvious.

Mistake 2: Forgetting that $\displaystyle\int u^{-1/2}\,du=2u^{1/2}$, not $\tfrac12 u^{1/2}$. …