Explained: "Find the antiderivative of (sin⁡x)2"

Adela Brown

Answered question

2021-12-16

Find the antiderivative of

{(\sin x)}^{2}

eskalopit

Beginner2021-12-17Added 31 answers

\int \sin^{2} x d x

has no clear solutions.
Instead of integrating

\sin^{2} x

, you can integrate

\frac{1}{2} (1 - \cos 2 x)

\int \frac{1}{2} (1 - \cos 2 x) d x

= \frac{1}{2} \int 1 - \cos 2 x d x

Now, use the sum rule to split this into:

\frac{1}{2} \int 1 d x - \frac{1}{2} \int \cos 2 x d x

\frac{1}{2} x - \frac{1}{2} \int \cos 2 x d x

\int \cos 2 x d x = \frac{1}{2} \sin 2 x

So, we have

\frac{1}{2} x - \frac{1}{2} \int \cos 2 x d x

\frac{1}{2} x - \frac{1}{2} (\frac{1}{2} \sin 2 x) + C = \frac{1}{2} x - \frac{1}{4} \sin 2 x + C

Wendy Boykin

Beginner2021-12-18Added 35 answers

\int {(\sin (x))}^{2} d x

\int \sin^{2} (x) d x

Rewrite using trig identities

\int \frac{1 - \cos (2 x)}{2} d x

Take the constant out

= \frac{1}{2} \times \int 1 - \cos (2 x) d x

Apply the sum rule

= \frac{1}{2} (\int 1 d x - \int \cos (2 x) d x)

As we know that

\int 1 d x = x

\int \cos (2 x) d x = \frac{1}{2} \sin (2 x)

, we have

= \frac{1}{2} (x - \frac{1}{2} \sin (2 x))

= \frac{1}{2} (x - \frac{1}{2} \sin (2 x)) + C

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