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

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

Show that \[
\tan^4x=\tan^2x \sec^2x- \sec^2x+1
\]

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,

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