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Trigonometric Integrals

shamieh

Active member
Sep 13, 2013
539
Quick question.

\(\displaystyle \int sin^{4}x dx\)

so I know:

\(\displaystyle \frac{1}{2} \int 1 - 2cos2x + \frac{1}{2}(1 + cos4x)dx\)

So here I first brought out the 1/2 because it's a constant and it's nasty.

so now I have

\(\displaystyle \frac{1}{4} \int 1 - 2cos2x + 1 + cos4x dx\)

so...Just as I brought 1/2 out can I now precede to take the 2 constant out that is in front of 2cos2x? thus turning it into:

\(\displaystyle \frac{1}{2} \int 1 - cos2x + 1 + cos4x dx\) ?

- - - Updated - - -

Ah wait, I think I forgot to square the denominator of (1 - cos2x/2)^2 because it was sin^4
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
A factor brought out in front of the integral must be a factor of the entire integrand. For example:

\(\displaystyle \sin^4(x)=\frac{1}{8}\left(3-4\cos(2x)+\cos(4x) \right)\)

And so we may write:

\(\displaystyle \int \sin^4(x)\,dx=\frac{1}{8}\int 3-4\cos(2x)+\cos(4x)\,dx\)
 

shamieh

Active member
Sep 13, 2013
539
A factor brought out in front of the integral must be a factor of the entire integrand. For example:

\(\displaystyle \sin^4(x)=\frac{1}{8}\left(3-4\cos(2x)+\cos(4x) \right)\)

And so we may write:

\(\displaystyle \int \sin^4(x)\,dx=\frac{1}{8}\int 3-4\cos(2x)+\cos(4x)\,dx\)
OH I see, so you can't bring the 3 out because it's not a factor.
 

shamieh

Active member
Sep 13, 2013
539
\(\displaystyle \frac{1}{4} \int 1 - 2cos2x + \frac{1}{2}(1 + cos4x) \ dx\)

So now I have

\(\displaystyle \frac{1}{4} \int \frac{3}{2} - 2cos2x + \frac{1}{2} + \frac{1}{2}cos4x) \ dx\)

which is \(\displaystyle \frac{1}{4} \int \frac{3}{2} - 2cos2x + \frac{1}{2}cos4x \ dx\)


Figured out what to do. Thanks for the help. Integrated by parts the 2cos2x and the 1/2cos4x which were extremely similar so basically did it in my head to get the 1/8.
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You really don't need integration by parts to integrate something of the form:

\(\displaystyle \int \cos(ax)\,dx\) where $a$ is a non-zero real constant. Observing that:

\(\displaystyle \frac{d}{dx}\left(\frac{1}{a}\sin(ax) \right)=\cos(ax)\)

We may then simply write:

\(\displaystyle \int \cos(ax)\,dx=\frac{1}{a}\sin(ax)+C\)