need help with reduction formula

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July 24th 2009, 02:31 PM
justin016

need help with reduction formula

http://calc101.com/Images/deriving_reduction_1_gr_1.gif

http://calc101.com/Images/deriving_reduction_1_gr_3.gif

http://calc101.com/Images/deriving_reduction_1_gr_5.gif
http://calc101.com/Images/deriving_reduction_1_gr_7.gif
http://calc101.com/Images/deriving_reduction_1_gr_9.gif
http://calc101.com/Images/deriving_r...on_1_gr_11.gif

http://calc101.com/Images/deriving_r...on_1_gr_13.gif

for the last part, if I tranpose the last integral to the left side, the constant, should it still be (n-1)? why does it changes to n instead?
July 24th 2009, 10:01 PM
AlephZero

Quote:

Originally Posted by justin016

for the last part, if I tranpose the last integral to the left side, the constant, should it still be (n-1)? why does it changes to n instead?

Moving the rightmost integral in your penultimate line to the left-hand side, you have

$\int \sin^nx\,dx+(n-1)\int\sin^nx\,dx=-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}x\,dx$

$\implies(1+n-1)\int\sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)\int\sin^{n-2}x\,dx$

$\implies n\int\sin^nx\,dx = -\sin^{n-1}x\cos x + (n-1)\int\sin^{n-2}x\,dx$

$\implies \int\sin^nx\,dx = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}\int\sin^{n-2}x\,dx.$

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