Q.Evaluate: $\int \tan^2 x\,\sec^4 x\,dx$

Concept: U Substitution — rewrite the integral in terms of $u = \tan x$ so that $\sec^2 x\,dx = du$.

Step 1: Factor $\sec^4 x = \sec^2 x \cdot \sec^2 x$. Use the identity $\sec^2 x = 1 + \tan^2 x$ to express everything in $\tan x$:

Step 2: Let $u = \tan x$, so $du = \sec^2 x\,dx$. The integral becomes:

Step 3: Integrate term by term: …

When you see powers of $\tan x$ and $\sec x$ multiplied together, the natural instinct is to look for a substitution. The key is that the derivative of $\tan x$ is $\sec^2 x$, and the derivative of $\sec x$ is $\sec x \tan x$. So if you have an extra $\sec^2 x$ factor, $u = \tan x$ works beautifully; if you have an extra $\sec x \tan x$ factor, $u = \sec x$ is the way.

Here we have $\tan^2 x \sec^4 x$. Notice $\sec^4 x = \sec^2 x \cdot \sec^2 x$. One of those $\sec^2 x$ factors is the derivative of $\tan x$ — that’s our cue.

1. Rewrite the integrand to expose the derivative. Write $\sec^4 x = \sec^2 x \cdot \sec^2 x$. Then the integral becomes

The $\sec^2 x$ at the end is $d(\tan x)/dx$, so we set $u = \tan x$, $du = \sec^2 x\, dx$.

2. Express the remaining $\sec^2 x$ in terms of $u$. Using the identity $\sec^2 x = 1 + \tan^2 x = 1 + u^2$, we have

3. Simplify and integrate. Multiply out: $u^2 (1 + u^2) = u^2 + u^4$. So

4. Substitute back. Since $u = \tan x$, $\int \tan^2 x \sec^4 x \, dx = \frac{1}{3}\tan^3 x + \frac{1}{5}\tan^5 x + C.$ …

Method: Integrating powers of $\tan x$ and $\sec x$

Use this for $\int \tan^m x\,\sec^n x\,dx$ when $n$ (the power of $\sec$) is even. You reserve one $\sec^2 x$ for the differential and convert the rest to $\tan x$.

Steps

Step 1: Split off a $\sec^2 x$.

Write $\sec^n x=\sec^{n-2}x\cdot\sec^2 x$. The $\sec^2 x$ will become $du$.

Step 2: Convert the remaining even secant power using $\sec^2 x=1+\tan^2 x$.

This expresses everything except the reserved $\sec^2 x$ in terms of $\tan x$.

Step 3: Substitute $u=\tan x$, $du=\sec^2 x\,dx$. …

Common Mistakes

Mistake 1: Not reserving a $\sec^2 x$ for $du$.

Why it's wrong: without setting aside one $\sec^2 x$, the substitution $u=\tan x$ has no matching differential. Correct approach: write $\sec^4 x=\sec^2 x\cdot\sec^2 x$ and convert one factor to $1+\tan^2 x$.

Mistake 2: Using the wrong Pythagorean identity.

Why it's wrong: $\sec^2 x=1+\tan^2 x$; writing $\sec^2 x=1-\tan^2 x$ or mixing in $\sin/\cos$ derails the polynomial. Correct approach: memorise $1+\tan^2 x=\sec^2 x$. …