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Mathematics 21 Online
OpenStudy (anonymous):

Find the integral of cos^2 x dx from 0 to 2pi.

OpenStudy (anonymous):

hint: cos^2(x) = (1/2) (1 - cos(2x))

OpenStudy (anonymous):

And do I take the integral of that? How?

OpenStudy (anonymous):

opps *plus*

OpenStudy (anonymous):

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

OpenStudy (anonymous):

But how do you integrate that?

OpenStudy (anonymous):

well, we know:$$\int\frac12(1+\cos2x)\ \mathrm{d}x=\frac12\int(1+\cos2x)\ \mathrm{d}x=\frac12\left(\int1\ \mathrm{d}x+\int\cos 2x\ \mathrm{d}x\right)$$

OpenStudy (anonymous):

I trust you can simplify both of these simpler integrals! for the latter, think back to u-substitution

OpenStudy (anonymous):

Thanks. I got it.

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