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    hi guys,
    Im stuck on this would appreciate any help.
    How do the highlighted steps follow on?Name:  Untitled.jpg
Views: 289
Size:  68.6 KB
    THANKS
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    The idea is to get all of the integrals on the RHS in terms of sine. When you derive the reduction formula, you will always find that one of the integrals on the RHS is of the same form as the original integral; you can just move that back to the LHS. The integral that you are left with on the RHS will have a lower exponent than the original. Once you have reduced enough, you can then integrate.
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    just out of interest what is that other picture just to the right. it seems colourfull
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    (Original post by 041087)
    hi guys,
    Im stuck on this would appreciate any help.
    How do the highlighted steps follow on?Name:  Untitled.jpg
Views: 289
Size:  68.6 KB
    THANKS
    The two red ones? All they've done is multiplied out the brackets and then separated it into two integrals in the one step.
 
 
 
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