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# antiderivative of sin^2 x (aka (sin x)^2))?

I'm blanking for some reason...

Update:

I know how to find the derivative, just need help with the integral/anti differentiation.

### 1 Answer

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- 1 month agoFavorite Answer
sin(x)^2 * dx

Use the half-angle formula

sin(x/2) = sqrt((1 - cos(x)) / 2)

cos(x/2) = sqrt((1 + cos(x)) / 2)

sin(x)^2 * dx =>

(1 - cos(x)^2) * dx =>

(1 - (1/2) * (1 + cos(2x))) * dx =>

(1/2) * (2 - 1 - cos(2x)) * dx =>

(1/2) * (1 - cos(2x)) * dx

Integrate

(1/2) * (x - (1/2) * sin(2x)) + C =>

(1/4) * (2x - sin(2x)) + C

Derive to make sure we're right

(1/4) * (2 - 2cos(2x)) =>

(1/2) * (1 - cos(2x)) =>

(1/2) * (1 - (cos(x)^2 - sin(x)^2)) =>

(1/2) * (1 - cos(x)^2 + sin(x)^2) =>

(1/2) * (sin(x)^2 + sin(x)^2) =>

(1/2) * 2 * sin(x)^2 =>

sin(x)^2

(1/4) * (2x - sin(2x)) + C

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