Q.Integrate the following function: $\tan^4 x$

The key idea is to rewrite $\tan^4 x$ using the identity $\tan^2 x = \sec^2 x - 1$, then integrate term by term.

Step 1: Write $\tan^4 x = \tan^2 x \cdot \tan^2 x = (\sec^2 x - 1) \tan^2 x$.

Step 2: Expand: $(\sec^2 x - 1) \tan^2 x = \sec^2 x \tan^2 x - \tan^2 x$. …

Why this approach works

When you see a high power of $\tan x$, the natural reflex is to look for a way to reduce it. The identity $\tan^2 x = \sec^2 x - 1$ is your best friend here, because $\sec^2 x$ is the derivative of $\tan x$. That means if you can peel off one factor of $\tan^2 x$ and rewrite it as $\sec^2 x - 1$, the remaining expression becomes something you can integrate by substituting $u = \tan x$.

The trick is to do this repeatedly until you're left with only powers of $\tan x$ and a single $\sec^2 x$ factor to absorb into the differential.

Step-by-step solution

1. Rewrite the integrand using the identity. Start with $\tan^4 x = \tan^2 x \cdot \tan^2 x$. Replace one $\tan^2 x$ with $\sec^2 x - 1$:

2. Handle the $\tan^2 x$ term similarly. The second term, $-\tan^2 x$, can be rewritten again using the same identity:

So the whole integrand becomes:

3. Set up the substitution. Let $u = \tan x$. Then $du = \sec^2 x \, dx$. This is perfect because we have a $\sec^2 x$ factor sitting next to $\tan^2 x$ in the first term. The integral splits into three parts:

4. Integrate each term.
   • For the first term: $\int \tan^2 x \sec^2 x \, dx = \int u^2 \, du = \frac{1}{3} u^3 = \frac{1}{3} \tan^3 x$. …

Method: Reduce a power of $\tan$ with $\tan^2=\sec^2-1$ repeatedly

Even powers of $\tan x$ have no direct antiderivative; peel off a $\tan^2 x$ and replace it with $\sec^2 x-1$ each pass, generating a $\sec^2 x$ (easy) and a lower power of $\tan$ (recurse).

Steps

Step 1: Factor out one $\tan^2 x$.

Step 2: Integrate the $\tan^2 x\sec^2 x$ term by substitution.

Let $u=\tan x$, $du=\sec^2 x\,dx$, so $\displaystyle\int\tan^2 x\sec^2 x\,dx=\int u^2\,du=\tfrac13\tan^3 x$. …

Common Mistakes

Mistake 1: Writing $\displaystyle\int\tan^4 x\,dx=\frac{\tan^5 x}{5}$.

Why it's wrong: the power rule needs a $\sec^2 x$ differential factor that is not present. Correct approach: reduce with $\tan^2 x=\sec^2 x-1$, giving $\tan^2 x\sec^2 x-\tan^2 x$.

Mistake 2: Forgetting the leftover $\int\tan^2 x\,dx$ still needs reducing. …