How to Solve a Tricky Integration Using Prosthaphaeresis Formulas?

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The discussion centers on solving the integral \(\int_0^h \sin[\beta(h-z)] \cos(\beta z \cos \theta) \; dz\) using prosthaphaeresis formulas. The initial approach involved substitution \(U=\beta(h-z)\), which proved ineffective. The recommended solution utilizes the prosthaphaeresis identity \(\sin\theta\cos\phi=\frac{\sin(\theta+\phi)+\sin(\theta -\phi)}{2}\) to simplify the integral effectively. This method provides a clear pathway to resolving the integration challenge presented by the user, Alan.

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I need help how to solve this integration:

[tex]\int_0^h sin[\beta(h-z)] \; cos(\beta\;z \; cos \theta)\; dz[/tex]

I tried substitution of [itex]\;U=\beta(h-z)\;[/itex] and going nowhere. [itex]\; cos(A+B)\;[/itex] type of method won't work either because you can't get A and B out of this. Please help.

Thanks

Alan
 
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yungman said:
I need help how to solve this integration:

[tex]\int_0^h sin[\beta(h-z)] \; cos(\beta\;z \; cos \theta)\; dz[/tex]

I tried substitution of [itex]\;U=\beta(h-z)\;[/itex] and going nowhere. [itex]\; cos(A+B)\;[/itex] type of method won't work either because you can't get A and B out of this. Please help.

Thanks

Alan

Use one of the prosthaphaeresis formulas:

[tex]\sin\theta\cos\phi=\frac{\sin(\theta+\phi)+\sin(\theta -\phi)}{2}[/tex]
 
LCKurtz said:
Use one of the prosthaphaeresis formulas:

[tex]\sin\theta\cos\phi=\frac{\sin(\theta+\phi)+\sin(\theta -\phi)}{2}[/tex]

Sorry to acknowledge so late, took me a while to work it out.

Thanks
 

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