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2-0-sinx-cosx-dx-2cos-2-x-3sin-2-x-




Question Number 17033 by arnabpapu550@gmail.com last updated on 30/Jun/17
∫_((Π )/2) ^( 0)  ((sinx cosx dx)/(2cos^2 x+3sin^2 x))
Π20sinxcosxdx2cos2x+3sin2x
Answered by sma3l2996 last updated on 30/Jun/17
t=sinx⇒dt=cosxdx  I=∫_1 ^0 ((tdt)/(2+t^2 ))=(1/2)[ln∣2+t^2 ∣]_1 ^0 =(1/2)(ln2−ln3)=(1/2)ln(2/3)
t=sinxdt=cosxdxI=10tdt2+t2=12[ln2+t2]10=12(ln2ln3)=12ln23
Commented by arnabpapu550@gmail.com last updated on 30/Jun/17
Thank you very much.
Thankyouverymuch.

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