Q.Integrate the following function: $\int \sec x (\sec x + \tan x) dx$

The key idea is to rewrite the integrand in terms of $\sec^2 x$ and $\sec x \tan x$, which are standard derivatives.

First, expand the product:

Now integrate term by term:

$\int \sec^2 x \, dx = \tan x + C, \quad \int \sec x \tan x \, dx = \sec x + C.$ …

Concept and Intuition

When you see an integral like $\int \sec x (\sec x + \tan x) \, dx$, your first instinct might be to multiply it out and then stare at the result. That’s exactly what we’ll do, but with a purpose.

The expression $\sec x (\sec x + \tan x)$ expands to $\sec^2 x + \sec x \tan x$. Now, here’s the beautiful part: both of these terms have well-known antiderivatives. The derivative of $\tan x$ is $\sec^2 x$, and the derivative of $\sec x$ is $\sec x \tan x$. So integrating each term separately gives us back the original functions, plus the constant of integration.

This is a classic case where the integrand is already set up as a sum of derivatives. No substitution, no trick — just recognition.

Step-by-Step Solution

1. Expand the integrand Multiply out the expression:

So the integral becomes:

2. Split the integral The sum rule for integrals lets us break this into two separate integrals:

3. Integrate each term
   • The antiderivative of $\sec^2 x$ is $\tan x$, because $\frac{d}{dx}(\tan x) = \sec^2 x$. …

Method: Expand a trig product into recognisable derivatives

Use this when a trig product has no direct formula but expands into terms you already know as derivatives — here $\sec x(\sec x+\tan x)$.

Steps

Step 1: Multiply out the product.

Step 2: Recognise each piece as a standard derivative.

Recall $\dfrac{d}{dx}(\tan x)=\sec^2 x$ and $\dfrac{d}{dx}(\sec x)=\sec x\tan x$, so each term integrates back to a known function. …

Common Mistakes

Mistake 1: Not expanding the product first.

Why it's wrong: $\sec x(\sec x+\tan x)$ has no direct antiderivative until you expand it to $\sec^2 x+\sec x\tan x$. Correct approach: multiply out, then integrate each standard form.

Mistake 2: Confusing the antiderivatives of $\sec^2 x$ and $\sec x\tan x$. …