Saturday

September 20, 2014

September 20, 2014

Posted by **Arisa** on Friday, June 19, 2009 at 8:40am.

Evaluate the integrals:

limit 0 to pi/4 ∫ [sec^2x]/[5+tanx] dx

limit 0 to pi/6 ∫ [3cos3x]/[3+sin3x] dx

limit 0 to 3 ∫ [2x-1]/[x^2-x+1] dx

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By squaring sin^2 + cos^2, show that sin^4x + cos^4x = 1/4 (3+cos4x)

Hence, evaluate ∫(sin^4x+cos^4x) dx

(limit from 0 to pi/8)

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How do i express (2x+1)/(x+3) in the form of p + q/(x+3)???

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