How do i integrate SinxCosX by parts?

I am stuck

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- Apr 18th 2013, 02:03 PMBenja303Integration trig help
How do i integrate SinxCosX by parts?

I am stuck - Apr 18th 2013, 02:22 PMPlatoRe: Integration trig help
- Apr 18th 2013, 02:27 PMBenja303Re: Integration trig help
-1/2 cos^2(x)?

BTW how do i know when to use parts or not? - Apr 18th 2013, 02:34 PMPlatoRe: Integration trig help
- Apr 18th 2013, 03:29 PMBenja303Re: Integration trig help
? what other way can u do it?

- Apr 18th 2013, 03:31 PMProve ItRe: Integration trig help
Plato is right that using Integration by Parts here is overkill, but if you MUST use it...

$\displaystyle \displaystyle \begin{align*} I &= \int{\sin{(x)}\cos{(x)}\,dx} \\ I &= \sin^2{(x)} - \int{\cos{(x)}\sin{(x)}\,dx} \\ I &= \sin^2{(x)} - I \\ 2I &= \sin^2{(x)} \\ I &= \frac{1}{2}\sin^2{(x)} + C \end{align*}$

I personally prefer to avoid integration by parts where possible. This is easier integrated if you change the integrand using a double angle formula.

$\displaystyle \displaystyle \begin{align*} \int{\sin{(x)}\cos{(x)}\,dx} &= \int{\frac{1}{2}\sin{(2x)}\,dx} \\ &= -\frac{1}{4}\cos{(2x)} + C \\ &= -\frac{1}{4} \left[ 1 - 2\sin^2{(x)} \right] + C \\ &= \frac{1}{2}\sin^2{(x)} - \frac{1}{4} + C \end{align*}$

As you can see, these answers differ only by a constant, and since the constants are arbitrary anyway, they can be considered to be equivalent :) - Apr 18th 2013, 04:09 PMBenja303Re: Integration trig help
I was not taught these double angle things. Can u give me some insight?