Q.Integrate the following function: $\frac{\sec^2 x}{\sqrt{\tan^2 x+4}}$

Concept: U Substitution — the numerator $\sec^2 x$ is the derivative of $\tan x$, so we set $u = \tan x$.

Step 1: Let $u = \tan x$. Then $du = \sec^2 x \, dx$.

Step 2: The integral becomes

Step 3: This is a standard form:

with $a = 2$. …

Why this approach works

When you see $\sec^2 x$ multiplied by something involving $\tan x$, your first instinct should be: substitute $u = \tan x$. Why? Because the derivative of $\tan x$ is $\sec^2 x$, which sits right there in the numerator. That means $du = \sec^2 x \, dx$, and the whole integral collapses into something much simpler.

The denominator $\sqrt{\tan^2 x + 4}$ becomes $\sqrt{u^2 + 4}$ after substitution. That’s a classic form: $\int \frac{du}{\sqrt{u^2 + a^2}}$, which integrates to $\sinh^{-1}(u/a) + C$ (or equivalently $\log|u + \sqrt{u^2 + a^2}| + C$). So the problem is really just a disguised standard integral.

Step-by-step solution

1. Set up the substitution

Let $u = \tan x$. Then differentiate:

The integral becomes:

2. Recognise the standard form

The denominator is $\sqrt{u^2 + 2^2}$. This matches the formula:

Here $a = 2$, so:

3. Substitute back

Replace $u$ with $\tan x$: …

Method: Substitute $u=\tan x$ to reach a standard $\sqrt{u^2+a^2}$ form

A $\sec^2 x\,dx$ next to a function of $\tan x$ under a radical is a direct substitution: $u=\tan x$ turns it into a standard square-root integral.

Steps

Step 1: Substitute $u=\tan x$.

The $\sec^2 x\,dx$ in the numerator becomes $du$ exactly.

Step 2: Rewrite the integral.

Step 3: Apply the standard result and restore. …

Common Mistakes

Mistake 1: Overlooking that $\sec^2 x$ is the derivative of $\tan x$.

Why it's wrong: without pairing $\sec^2 x\,dx$ with $du$, the integrand looks unrelated to a standard form. Correct approach: substitute $u=\tan x$ so $du=\sec^2 x\,dx$ and the integral becomes $\int\frac{du}{\sqrt{u^2+4}}$.

Mistake 2: Using the arcsine form because of the square root. …