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How do you integrate even powers of sine and cosine? For example, how is $\int\sin^2(x)\ dx$ solved? What about $\int\cos^2(x)\ dx$ or $\int\sin^2(x)cos^2(x)\ dx$ or $\int\sin^2(x)cos^4(x)\ dx$?

I know that when there is an odd function you can use the identity property - but here they are even. A similar question would be solving tan/sec integrals when they are odd.

I want to show my steps, but I am not even sure where to begin. The only thing I can say is that I saw mention of something called the power reduction formula when reading through other questions - but I never learned it.

Burt
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    $\cos (2 x) = 2 \cos^2 x -1$. Solve for $\cos^2 x$ and integrate easily. Or $\cos(2x) = 1 - \sin^2 x$, solve for $\sin^2 x$ and integrate easily. Surely these are in your calculus textbook. – GEdgar Dec 27 '19 at 02:10

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These are easily integrable using the identities

$$\sin^2(x) = 1-\cos^2(x)$$ $$\cos^2(x) = 1-\sin^2(x)$$ $$\cos^2(x) = \dfrac{\cos(2x)+1}{2}$$ $$\sin^2(x) = \dfrac{1-\cos(2x)}{2}$$

David P
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As you read, you'll want to use "power reduction," which comes from the cosine double angle formula: $$\int \sin^2(x) \, dx = \int \frac{1 - \cos(2x)}{2} \, dx. $$ After we use this identity, we are left with two more familiar integrals: $$\int \frac{1 - \cos(2x)}{2} \, dx = \int \frac{1}{2} \, dx - \int \frac{\cos(2x)}{2} \, dx = \frac12 x - \frac{\sin(2x)}{4}. $$

The integral $\int \cos^2(x) dx$ words out similarly.

For $\int \sin^2(x) \cos^2(x) \, dx$, try replacing both $\sin^2(x)$ and $\cos^2(x)$ with the above double angle identities, which should leave an integral containing only powers of $\cos(2x)$.

Sam Freedman
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Use the identities $\cos 2x = 2\cos ^2 x-1 $ and $\cos 2x=1-2\sin^2 x$ solve for $\sin^2,\cos^2$

IrbidMath
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