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=\frac{1-\cos2x}2`

`\cos^2x=\frac{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
`(\mathrm d)/(\mathrm dx)\tan x=\sec^2x`.