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Question Number 52197 by maxmathsup by imad last updated on 04/Jan/19
calculate∫π4π3cosx−sinx2+sin(2x)dx
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19
2+sin2x1+sin2x+cos2x+2sinxcosx1+(sinx+cosx)2∫π4π3d(sinx+cosx)1+(sinx+cosx)2∣tan−1(sinx+cosx)∣π4π3tan−1(32+12)−tan−1(12+12)tan−1(1+32)−tan−1(2)tan−1(1+32−21+1+32)=tan−1(1+3−2222+1+32)=tan−1(1+3−222(2+1+3)tan−1(1+3−222+2+6)
Commented by Abdo msup. last updated on 05/Jan/19
thankyousirTanmay.
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