How could we integrate $\tan^4x$?

How could we integrate $\tan^4x$?

Question

Given that $\dfrac{d}{dx}(\tan x)=\sec^2 x,$ differentiate $\tan^3x$ with respect to $x$.

Show that $\tan^4x=\tan^2x \sec^2x- \sec^2x+1$ and hence evaluate $\int_0^{\pi/4}\tan^4x \, dx.$

Solution

Using the chain rule, we have that $\frac{d}{dx}\tan^3 x = 3 \tan^2 x \times \frac{d}{dx}\tan x= 3\tan^2 x \sec^2x.$

We can write

$$\begin{align*} \tan^4x&=\tan^2x\tan^2x\\ &=\tan^2x(\sec^2x-1)\\ &=\tan^2x\sec^2x-\tan^2 x\\ &=\tan^2x\sec^2x-(\sec^2x-1)\\ &=\tan^2x\sec^2x-\sec^2x+1. \end{align*}$$

…and hence evaluate $\int_0^{\pi/4}\tan^4x \, dx.$

Using the previous parts,

$$\begin{align*} \int_0^{\pi/4} \tan^4x \, dx&= \int_0^{\pi/4}\tan^2x\sec^2x-\sec^2x+1 \, dx \\ &=\left[\frac{1}{3}\tan^3x-\tan x+x\right]^{\pi/4}_0 \\ &=\frac{1}{3}-1+\frac{\pi}{4} \\ &=\frac{\pi}{4}-\frac{2}{3}. \end{align*}$$