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I'm trying to do the following:

$$\int \cos(wt)w\sin(wt+\phi) \,\mathrm{d}t$$

I can't figure what to do here. I can't do integration by parts because of the $\phi$. Any ideas?

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3  
begin by using the addition identity on the sine function. Then this integral will yield to the usual techniques. –  ncmathsadist Feb 6 '12 at 1:45
4  
Note that $\cos(a) \sin(b) = (\sin(a+b)-\sin(a-b))/2$. –  Robert Israel Feb 6 '12 at 1:51
    
got it, that works –  mugetsu Feb 6 '12 at 1:54
1  
You could express it all with complex exponential functions. The integrand is $\frac{w}{4i}(e^{iwt}+e^{-iwt})(e^{\phi}e^{iwt}-e^{-\phi}e^{-iwt})$. At the end this might leave you with the issue of how to translate back to real functions, but that is not insurmountable. –  alex.jordan Feb 6 '12 at 1:57

1 Answer 1

FWIW

$$\eqalign{ & \int {w\cos wt\sin \left( {wt + \phi } \right)dt} = \cr & \int {w\cos wt\left( {\sin wt\cos \phi + \cos wt\sin \phi } \right)dt} = \cr & \cos \phi \int {w\cos wt\sin wtdt} + \sin \phi \int {w{{\cos }^2}wtdt} = \cr & \frac{{\cos \phi }}{2}\int {w\sin 2wtdt} + \frac{{\sin \phi }}{2}\int {w\left( {1 + \cos 2wt} \right)dt} = \cr & w\left\{ {\frac{{\sin \phi }}{2}\left( {\frac{{2t + \sin 2wt}}{2}} \right) - \frac{{\cos \phi }}{2}\left( {\frac{{\sin 2wt}}{2}} \right)} \right\} + C \cr} $$

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