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9pi-2-7pi-1-5-dx-1-sinx-




Question Number 187528 by sciencestudentW last updated on 18/Feb/23
∫_((9π)/2) ^((7π)/(1.5)) (dx/( (√(1−sinx))))=?
$$\underset{\frac{\mathrm{9}\pi}{\mathrm{2}}} {\overset{\frac{\mathrm{7}\pi}{\mathrm{1}.\mathrm{5}}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{1}−{sinx}}}=? \\ $$
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
$${I}=\int\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{sin}{x}}} \\ $$$$\:\:=\int\frac{{dx}}{\mid\mathrm{cos}\frac{{x}}{\mathrm{2}}−\mathrm{sin}\frac{{x}}{\mathrm{2}}\mid}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dx}}{\mathrm{cos}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{sec}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid+{C} \\ $$

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