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Question Number 46841 by maxmathsup by imad last updated on 01/Nov/18
calculate∫π4π3dxcosxsinx
Commented by maxmathsup by imad last updated on 01/Nov/18
wehaveI=∫π4π3dxcosxsinx=2∫π4π3dxsin(2x)=2x=t2∫π22π31sin(t)dt2=∫π22π3dtsint=tan(t2)=u∫tan(π4)tan(π3)12u1+u22du1+u2=∫13duu=[ln∣u∣]13=ln(3)=ln(3)2.
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18
∫π4π3dxcos2x×tanx∫π4π3d(tanx)tanx∣ln(tanx)∣π4π3=ln(3)−ln(1)=12ln3
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