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Integral of sin^6(x)dx

by shft600
Tags: integral, sin6xdx
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shft600
#1
Apr12-10, 06:40 PM
P: 24
Evaluate the following integral:
I=int{sin6(x)dx

This form of question has shown up on both midterms, and is on my final assignment of the semester and most likely the final exam as well, and I still haven't figured out how to solve them. Could somebody point me in the right direction? Should I be integrating by parts? U-substitution? I just don't know how to deal with the exponenet. I tried expanding it into the integral of (sin2x)3, but ended up with a page and of a half of work, and the wrong answer. I know the answer is 5x/16-15/64(sin2x)+3/64(sin4x)-1/192(sin6x), but as I said have no idea how to get there...
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Mark44
#2
Apr12-10, 06:51 PM
Mentor
P: 21,311
Use the identity
[tex] sin^2(x) = \frac{1 - cos(2x)}{2}[/tex]
The right side above is easy to integrate.
[tex] sin^4(x) = \left(\frac{1 - cos(2x)}{2}\right)^2 = (1/4)(1 - 2cos(2x) + cos^2(2x)[/tex]

The first two terms on the far right side are easy to integrate, but the last one requires a little more work, this time using the identity cos2(2x) = (1/2)(1 + cos(4x)).

Do you see how this idea could be extended?
shft600
#3
Apr12-10, 08:19 PM
P: 24
I can get up to the point that
=(1/8) [ x - (3/2)sin(2x) + (3/2)( x + (1/4)sin(4x)) - int(cos3(2x)dx ]
but here I get stuck again on the new integral again,

I tried
int{ cos3(2x)
int{ cos2(2x) cos(2x) dx
int{ (1-sin2(2x)) (cos(2x)) dx, u=sin(2x), du=2cos(2x)
then,
1/2 int{ (1-u2)(du)
1/2 [ int{du - int{u2du ]
1/2 (u - u3/3)
1/2 (sin(2x) - sin3(2x)/3)
sin(2x)/2 - sin3(2x)/6

did I do something wrong in this part?

shft600
#4
Apr12-10, 09:40 PM
P: 24
Integral of sin^6(x)dx

anybody? I haven't made any progress on this last integral yet... I found something online that claimed a rule that cos(3x)=4cos3(x)-3sin(x), so I manipulated it into cos3(2x)=(1/4)(cos(6x)+3sin(2x)), and then by integrating the right side of that I got (1/24)(sin(6x)-3/8(cos(2x))), which is almost the right answer EXCEPT for the negative should be a positive, and cosine a sine. The rule was just stated on yahoo answers though, so I can't tell if I did something wrong or if the other guy mistyped his rule, can anybody verify that?

int{ cos3(2x) = int{ cos6x/4 + int{3sin(2x)/4
= 1/4 int{ cos6x dx + 3/4 int{ sin(3x)dx
= 1/24(sin6x) - 3/8(cos2x)

where I need
= 1/24(sin6x) + 3/8(sin2x)
Mark44
#5
Apr12-10, 11:03 PM
Mentor
P: 21,311
That formula you found online is incorrect: it should be cos(3x) = 4cos3(x) - 3cos(x).

You can get it by working with cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x), and using the double angle formulas and the identity sin2(x) + cos2(x) = 1.
Antiphon
#6
Apr12-10, 11:29 PM
P: 1,781
You need a CRC handbook.

The foundation of good engineering career is to understand as much technology as you can and to have an excellent reference library for the parts you haven't comitted to memory. Which is most everything. Don't skimp on building your reference library.
shft600
#7
Apr13-10, 12:02 AM
P: 24
excellent, I actually just found that mistake as I refreshed this page

but now I've taken a look at the next question and am even more confused...

I = int{ sin7x cos-3/5x dx

would this be an improper integral type question or can I use the same ideas on the last question?
shft600
#8
Apr13-10, 12:15 AM
P: 24
never mind, that ended up being much easier than I expected


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