Question

How do you find the integral of `sin^2(x)`?

Answer 1

Use trigonometry to rewrite.

Explanation:

When studying trigonometry, some encounter formulas called "Power Reduction Formulas". These are not used for anything -- until one studies integration. Many (most?) students who encounter these formulas forget them before they need them.

`cos(A+B) = cosAcosB-sinAsinB`, so

`cos(2A) = cos^2A - sin^2A` and since `cos^2A = 1-sin^2A`,

`cos(2A) - 1-2sin^2A`.

If we solve for `sin(A)`, we have the half angle formula, but is we stop at solving for `sin^2(A)`, the we have the power reduction formula for `sinx`

`sin^2x = 1/2(1-cos(2x))`

`int sin^2x dx = 1/2 int (1-cos(2x)) dx`

`= 1/2[x-1/2sin(2x)] +C`.

Rewrite the answer to taste.

I like `1/2(x-sinxcosx) +C.`

Additional note

Using the cosine double angle formula in the form

`cos(2A) = 2cos^2A - 1`, we get the power reduction for `cos^2x`

`cos^2x = 1/2(1+cos(2x))`