Trig integrals: rewrite with an identity first

You can already integrate $\sin(ax + b)$, $\cos(ax + b)$ and $\sec^2(ax + b)$. Those are the standard integrals.

But what about these?

• $\displaystyle\int \sin^2 x\,dx$
• $\displaystyle\int \cos^2 2x\,dx$
• $\displaystyle\int \sin 3x\cos 3x\,dx$

None of them is on the standard list. And you can't use the power rule: $\sin^2 x$ is not "$x$ to a power".

The trick: use an identity to turn the square or product into something on the standard list.

Start from $\cos 2x = 1 - 2\sin^2 x$.

1. Add $2\sin^2 x$ to both sides and subtract $\cos 2x$: $\;2\sin^2 x = 1 - \cos 2x$.
2. Halve: $\;\sin^2 x = \tfrac12(1 - \cos 2x)$.

The same steps on $\cos 2x = 2\cos^2 x - 1$ give $\cos^2 x = \tfrac12(1 + \cos 2x)$.

For the product, halve $\sin 2x = 2\sin x\cos x$.

Watch the signs: $\sin^2$ has a minus, $\cos^2$ has a plus.

1. Rewrite the square or product using the identity.
2. Integrate term by term with the standard results. Remember to divide by the number in front of $x$ inside the bracket: $\int\cos 2x\,dx = \tfrac12\sin 2x$.
3. Add $+\,C$, or put in the limits if there are any.
4. Check by differentiating, if you have time.

The step people forget: the identities for $\sin^2$ and $\cos^2$ have a constant, $\tfrac12$. It integrates to $\tfrac12 x$. Don't drop it.