# Integration trig help

• April 18th 2013, 02:03 PM
Benja303
Integration trig help
How do i integrate SinxCosX by parts?

I am stuck
• April 18th 2013, 02:22 PM
Plato
Re: Integration trig help
Quote:

Originally Posted by Benja303
How do i integrate SinxCosX by parts?

Forget parts. Not needed.

What is the derivative of $\tfrac{1}{2}\sin^2(x)~?$
• April 18th 2013, 02:27 PM
Benja303
Re: Integration trig help
-1/2 cos^2(x)?

BTW how do i know when to use parts or not?
• April 18th 2013, 02:34 PM
Plato
Re: Integration trig help
Quote:

Originally Posted by Benja303
-1/2 cos^2(x)?
BTW how do i know when to use parts or not?

That is also a possible answer. It is not the derivative as I had in mind.
• April 18th 2013, 03:29 PM
Benja303
Re: Integration trig help
? what other way can u do it?
• April 18th 2013, 03:31 PM
Prove It
Re: Integration trig help
Quote:

Originally Posted by Benja303
How do i integrate SinxCosX by parts?

I am stuck

Plato is right that using Integration by Parts here is overkill, but if you MUST use it...

\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 \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 :)
• April 18th 2013, 04:09 PM
Benja303
Re: Integration trig help
I was not taught these double angle things. Can u give me some insight?