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OpenStudy (anonymous):
Evaluate integral sin(x)/cos^2(x)
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OpenStudy (anonymous):
answer that is given is secx+C but kinda confused how to get that
OpenStudy (raden):
sinx/cos^2x = sinx/cosx * 1/cosx = secxtanx obvious, the its integral given secx
OpenStudy (raden):
i mean : |dw:1383265660598:dw|
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