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Evaluate the indefinite integral.

by CellCoree
Tags: evaluate, indefinite, integral
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Sep8-04, 10:35 PM
P: 42
[tex] \int sin(7x)sin(13x)dx [/tex]

i just need to know how to start this, i cant use u-du...
i cant borrow anything....
so how would i start this?
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Sep8-04, 10:55 PM
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Is the integrand the same as
[tex]\frac {\cos {6x} - \cos {20x}}{2}[/tex]
Sep8-04, 11:01 PM
P: 157
One way is to look at the back of your physics or calculus book for a table of indefinite integrals. You should be able to solve it by the equations in the table.

Sep8-04, 11:03 PM
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PF Gold
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P: 16,091
Evaluate the indefinite integral.

As tide is suggesting, one way is to use your trig identities. (Sigh, nobody learns their trig identities these days) The product of two sines, two cosines, or a sine and a cosine can always be converted into a sum of two sines, or a sum of two cosines. The identities can probably be found in your calc or precalc book.

Another approach, if you've learned the necessities, is to replace the sine function with its definition in terms of complex exponentials.

Another way is to use the angle sum identities. Your problem is that 7x and 13x have different coefficients, so break 13x into 7x + 6x and use the sum of angles identity for sine. Rinse, and repeat until you can do all of the resulting integrals. One thing you should always try and find, when dealing with complicated integrals, is an operation that changes a piece of the integral from something you can't handle into something you can.
Sep9-04, 01:14 AM
HW Helper
P: 10,757
Quote Quote by Tide
Is the integrand the same as
[tex]\frac {\cos {6x} - \cos {20x}}{2}[/tex]

You know, [tex]cos(\alpha +\beta )= cos(\alpha )cos(\beta ) - sin(\alpha )sin(\beta) \mbox { and } cos(\alpha -\beta )= cos(\alpha )cos(\beta ) + sin(\alpha )sin(\beta)\mbox. \\ [/tex]
[tex]\alpha = 13x \mbox{ and } \beta = 7x [/tex]....


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