Integrating Trigonometric Functions Videos

Hint 1: If you let \(u=t\), what would \(dv\) have to be? Can you integrate it?

\(dv=\csc^2(t) dt\) so \(v=-cot(t)\)

\(uv-\int v du = -t \cot(t) - \int cot(t)\)
If you know how to integrate tan(t), you should be able to figure out how to integrate cot(t).

let \(u=x\) and \( dv=\sin x \cos x dx\)
Then to find v, you need to integrate \(\int \sin x \cos x dx\), which you should now know how to do!

Either \( \dfrac{1}{2} x \sin^2(x) - \int \dfrac{1}{2} \sin^2(x) dx\)
OR you might have gotten \( -\dfrac{1}{2} x \cos^2(x) + \int \dfrac{1}{2} \cos^2(x) dx\)
OR, if you are very good with your trig identities, you might have gotten \( -\dfrac{1}{4} x \cos(2x) + \int \dfrac{1}{2} \cos(2x) dx\)
All of which will give you a correct answer, and after you plug in the endpoints will all give you the same number.