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Mathematics 13 Online
OpenStudy (anonymous):

Find the area under curve y= 1 / cos^2 t between t=0 and t=Pi/4

OpenStudy (anonymous):

\[\int\limits_{0}^{\pi/4}\frac{ 1 }{ \cos^2t}dt\]

OpenStudy (anonymous):

integration of \[\sec^2t\] is tan(t) then puting the limits u get tanpi/4 - tan0 =1-0=1 sq.unit

OpenStudy (anonymous):

Where did sec^2 t come from?

OpenStudy (anonymous):

Ok I got it - fundamental identities. Thanks!

OpenStudy (anonymous):

anytime .:) ..

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