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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: integration-NCERT problem-1
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[Post New]8 Feb 2008 17:56:28 IST
    Subject: integration-NCERT problem-1
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priyesh (1586)

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from NCERT part 2 integration miscelaaneous exercise chapter 7 q26)
 
 
 
[0][pi/4] (sinxcosx) / (sin4x+cos4x)
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[Post New]8 Feb 2008 18:22:59 IST
    Subject: Re:integration-NCERT problem-1
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hsbhatt (3699)

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Pointers: sin4x+cos4x = (sin2x+cos2x)2 - 2sin2xcos2x = 1 - 2sin2xcos2x = 1 -sin22x/2. = (1+cos22x)/2. Now let cos2x = t
 
 
 
 
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[Post New]8 Feb 2008 18:31:50 IST
    Subject: integration-NCERT problem-1
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priyesh (1586)

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thanks
"Imagination is more important than knowledge."
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[Post New]8 Feb 2008 18:46:38 IST
    Subject: Re:integration-NCERT problem-1
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nadeemoidu (1184)

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Divide by cos4x and put tanx =t . wouldn't this be easier?
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[Post New]8 Feb 2008 21:40:22 IST
    Subject: Re:integration-NCERT problem-1
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priyesh (1586)

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ya this method would be easier
 thanks nadeem
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