Q.Integrate the following function: $\tan^3 2x \sec 2x$

Concept: Sine Double Angle Integration — rewrite $\tan^3 2x \sec 2x$ in terms of $\sin 2x$ and $\cos 2x$, then use substitution.

Step 1: Write the integrand in sine and cosine:

Step 2: Let $u = \cos 2x$, so $du = -2 \sin 2x \, dx$. Then $\sin^3 2x \, dx = \sin^2 2x \cdot \sin 2x \, dx = (1 - \cos^2 2x) \sin 2x \, dx = (1 - u^2) \cdot \left(-\frac{du}{2}\right)$.

Step 3: The integral becomes:

Step 4: Integrate: …

When you see a product of powers of $\tan$ and $\sec$, your first instinct should be to look for a derivative relationship. The derivative of $\sec x$ is $\sec x \tan x$, and the derivative of $\tan x$ is $\sec^2 x$. Here, the presence of $\sec 2x$ multiplied by $\tan^3 2x$ suggests that if we isolate one factor of $\tan 2x \sec 2x$, the rest can be expressed in terms of $\sec 2x$ alone.

Let’s walk through it.

1. Rewrite the integrand to expose the derivative of $\sec 2x$. Notice that $\frac{d}{dx} (\sec 2x) = 2 \sec 2x \tan 2x$. So the factor $\tan 2x \sec 2x$ is almost a derivative — we just need to account for the chain rule factor of 2. Write:

2. Use the Pythagorean identity for $\tan^2$. Recall that $\tan^2 \theta = \sec^2 \theta - 1$. Here $\theta = 2x$, so:

Substituting gives:

3. Substitute $u = \sec 2x$. Then $du = 2 \sec 2x \tan 2x \, dx$, so $\sec 2x \tan 2x \, dx = \frac{du}{2}$. The integrand becomes:

4. Integrate with respect to $u$.

Simplify:

5. Substitute back $u = \sec 2x$. …

Method: Odd power of $\tan$ with a $\sec$ factor — substitute $u=\sec$

For integrands of the form $\tan^{m}(\cdot)\sec(\cdot)$ with $m$ odd, save one $\sec\tan$ pair for the differential, convert the remaining even power of $\tan$ to $\sec$, and substitute $u=\sec(\cdot)$.

Steps

Step 1: Split off a $\sec\tan$ factor.

The $\sec\theta\tan\theta$ is what pairs with $dx$.

Step 2: Convert the leftover $\tan^2$ to $\sec^2-1$.

so the integrand is $(\sec^2\theta-1)\sec\theta\tan\theta$.

Step 3: Substitute $u=\sec\theta$. …

Common Mistakes

Mistake 1: Saving the wrong factor for the differential.

Why it's wrong: with an odd power of $\tan$ and a $\sec$ present, you must reserve $\sec\theta\tan\theta$ (the derivative of $\sec\theta$), not $\sec^2\theta$. Reserving $\sec^2$ here fails because only one $\sec$ is available. Correct approach: write $\tan^3 2x\sec 2x=(\sec^2 2x-1)\sec 2x\tan 2x$ and set $u=\sec 2x$. …