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Integral Calculus

Blazing goIITian

Joined:	18 Feb 2007
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8 Feb 2008 17:56:28 IST

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integration-NCERT problem-1

None

from NCERT part 2 integration miscelaaneous exercise chapter 7 q26)

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^[pi/4] (sinxcosx) / (sin⁴x+cos⁴x)

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Hari Shankar

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Joined: 28 Feb 2007

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8 Feb 2008 18:22:59 IST

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Pointers: sin⁴x+cos⁴x = (sin²x+cos²x)² - 2sin²xcos²x = 1 - 2sin²xcos²x = 1 -sin²2x/2. = (1+cos²2x)/2. Now let cos2x = t

Priyesh

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8 Feb 2008 18:31:50 IST

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thanks

Nadeem

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8 Feb 2008 18:46:38 IST

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Divide by cos⁴x and put tanx =t . wouldn't this be easier?

Priyesh

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8 Feb 2008 21:40:22 IST

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ya this method would be easier

thanks nadeem

Anjali Singh

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25 Jun 2009 12:59:30 IST

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