Question

How do i find the antiderivative of sin^2(x, cos^2(x and tan^2(x? i know how to antidifferentiate sinx, cosx, and tanx, but these just stump me.. thank you for the help!?

Answer 1

For $\sin^2x$ and $\cos^2x$, it's helpful to remember the half-angle identities:

$\sin^2x=\dfrac{1-\cos2x}2$
$\cos^2x=\dfrac{1+\cos2x}2$

So

$\displaystyle\int\sin^2x\,\mathrm dx=\frac12\int(1-\cos2x)\,\mathrm dx=\frac x2-\frac{\sin2x}4=C$
$\displaystyle\int\cos^2x\,\mathrm dx=\frac12\int(1+\cos2x)\,\mathrm dx=\frac x2+\frac{\sin2x}4=C$

For $\tan^2x$, the Pythagorean identity suffices:

$\sin^2x+\cos^2x=1\implies \tan^2x+1=\sec^2x$

which means

$\displaystyle\int\tan^2x\,\mathrm dx=\int(\sec^2x-1)\,\mathrm dx=\tan x-x+C$

since $\dfrac{\mathrm d}{\mathrm dx}\tan x=\sec^2x$.