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Question Number 187528 by sciencestudentW last updated on 18/Feb/23

∫_((9π)/2) ^((7π)/(1.5)) (dx/( (√(1−sinx))))=?

7π1.59π2dx1sinx=?

Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23

I=∫(dx/( (√(1−sinx))))    =∫(dx/(∣cos(x/2)−sin(x/2)∣))=(1/( (√2)))∫(dx/(cos((x/2)+(π/4))))    =(1/2)ln∣sec((x/2)+(π/4))+tan((x/2)+(π/4))∣+C

I=dx1sinx=dxcosx2sinx2=12dxcos(x2+π4)=12lnsec(x2+π4)+tan(x2+π4)+C

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