- #1

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## Homework Statement

How do you integrate a product of two trigonometric functions of x, when the argument is different, ie:

what is the integral of

sin x cos 3x

I ought to know this, but don't seem to be able to do it.

Thanks in adv!

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- Thread starter electric.avenue
- Start date

- #1

- 8

- 0

How do you integrate a product of two trigonometric functions of x, when the argument is different, ie:

what is the integral of

sin x cos 3x

I ought to know this, but don't seem to be able to do it.

Thanks in adv!

- #2

AlephZero

Science Advisor

Homework Helper

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- 297

sin(a+b) = sin a cos b + cos a sin b

sin(a-b) = sin a cos b - cos a sin b

so

sin(a+b) + sin(a-b) = 2 sin a cos b

- #3

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choose one part as v

choose the other as du

{-integral sigh

{vdu=v*u -{dv*u

- #4

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

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To see what I mean with an "integrand loop", I'll take an example:

We are to compute:

[tex]I=\int_{0}^{1}e^{x}\cos(x)dx[/tex]

we use partial integration to write this as:

[tex]I=\int_{0}^{1}e^{x}\cos(x)dx=e^{x}\cos(x)\mid^{x=1}_{x=0}+\int_{0}^{1}e^{x}\sin(x)dx=(e^{x}\cos(x)+e^{x}\sin(x))\mid_{x=0}^{x=1}-\int_{0}^{1}e^{x}\cos(x)dx=(e^{x}\cos(x)+e^{x}\sin(x))\mid_{x=0}^{x=1}-I[/tex]

Therefore, we have gained the equation for I:

[tex]I=(e^{x}\cos(x)+e^{x}\sin(x))\mid_{x=0}^{x=1}-I[/tex]

which is easily solved for I.

- #5

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Thanks for all the help and advice, everyone. I got an A in maths, btw.

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