Q.Integrate the following function: $\tan^2 (2x - 3)$

The key idea is to rewrite $\tan^2$ using the identity $\tan^2 \theta = \sec^2 \theta - 1$, then integrate term by term.

Step 1: Let $u = 2x - 3$, so $du = 2\,dx$, hence $dx = \frac{du}{2}$.

Step 2:

Step 3: Integrate:

Step 4: Substitute back $u = 2x - 3$: …

The key insight here is that $\tan^2$ is not directly integrable in elementary form — but its close relative $\sec^2$ is. The identity $\tan^2\theta = \sec^2\theta - 1$ is the bridge. Once you see that, the problem becomes a sum of two simple integrals: one of $\sec^2$ (which gives $\tan$) and one of a constant.

Let’s walk through it.

1. Set up the substitution. The argument $2x-3$ suggests a linear substitution. Let $u = 2x - 3$. Then $du = 2\,dx$, so $dx = \frac{du}{2}$. The integral becomes:

2. Apply the Pythagorean identity. Recall the fundamental identity:

This is derived from $\sin^2 u + \cos^2 u = 1$ divided by $\cos^2 u$. It’s the single most useful trick for integrating squares of tangent.

Substituting:

3. Integrate each term.

   • The integral of $\sec^2 u$ is $\tan u$ (this is a standard derivative result in reverse).
   • The integral of $1$ with respect to $u$ is $u$.

   So:

4. Back-substitute. Replace $u$ with $2x - 3$:

Simplify the second term:

So the expression becomes:

$\frac12 \tan(2x - 3) - x + \frac32 + C.$ …

Method: Use $\tan^2\theta=\sec^2\theta-1$, then integrate a linear argument

Use this for $\tan^2$ of a linear expression, e.g. $\tan^2(2x-3)$: there is no direct formula for $\tan^2$, so convert with a Pythagorean identity first.

Steps

Step 1: Replace $\tan^2$ using the identity.

Step 2: Integrate each piece with the linear-argument rule. …

Common Mistakes

Mistake 1: Trying to integrate $\tan^2(2x-3)$ directly.

Why it's wrong: there is no direct antiderivative for $\tan^2$; you must convert first. Correct approach: use $\tan^2\theta=\sec^2\theta-1$.

Mistake 2: Forgetting the $\dfrac12$ factor for the linear argument. …