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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Feb 2008 13:31:47 IST
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how to find the ingral of sin^nx and cos ^nx where n=1,2,3,............ also  [  sinx cos^3x]dx
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28 Feb 2008 13:51:32 IST
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general solution for any cosnx:
integrating by parts as cos x and cosn-1x...
In=sin x cos n-1x -  (n-1)cosn-2x. sin x* sin x dx
= sin x cosn-1x+(n-1)cosn-2x(1-cos2x)dx
In=sin x cosn-1x+(n-1) In-2-(n-1)In
or
nIn-(n-1)In-2= sin x cosn-1x +c
now put value of n and step down to 5,3 and so on to get answer
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28 Feb 2008 13:53:00 IST
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the figure that cant be seen is integral sign!
  cos3x sin x dx
=( cot3x)cosec2x dx
cot x=t
-cosec2x dx=dt
so -t3/2dt
=- 2/5 t5/2
or
-2/5(cot x)5/2+c
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28 Feb 2008 14:06:35 IST
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Yup Akhil is correct
Similary you can find that Integral of sin n x
If I n = sin n x
then we can similarly find that n In - (n-1) I n-2 + cos x sin n-1 x = 0
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