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Integral Calculus

Blazing goIITian

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19 Mar 2010 20:55:40 IST

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int(0 to pi/6) of (1-cos3t) sin3t dt =?

None

int(0 to pi/6) of (1-cos3t) sin3t dt =?

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abhishek kalantri

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Joined: 24 Feb 2010

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19 Mar 2010 21:19:15 IST

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it comes out to be 1/6.

ThIrStY of IIT

Blazing goIITian

Joined: 6 Mar 2010

Posts: 380

19 Mar 2010 21:21:04 IST

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I want procedure friend

abhishek kalantri

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19 Mar 2010 21:31:24 IST

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I=int(o to pi/6) (1-cos3t)sin3t= int (0 to pi/6) (1-cos3(pi/6+0-t))(sin3(pi/6+0-t)

adding both,

2I=int(0 to pi/6)(sin3t+cos3t-2sin3tcos3t)

2I= sin3t -cos3t +cos6t (limits 0 to pi/6)

3 3 6

puttind limits we get 2I=1/3

hence I=1/6

» ÆoniuM XEonAX «

Scorching goIITian

Joined: 19 Mar 2010

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19 Mar 2010 21:32:29 IST

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procedure is here

http://www.wolframalpha.com/input/?i=Integrate%5B%5B%5B1-cos%5B3t%5D%5D%2Asin%5B3t%5D%5D%5D

VASU

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30 Mar 2010 22:05:56 IST

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take u=1-cos3t and you get du=3*sin3tand limits from 0 to 1/2 and you get answer 1/24

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