Question Number 142687 by iloveisrael last updated on 04/Jun/21
![∫ (dx/( (√(1−sin x)) (√(1+cos x)))) =?](https://www.tinkutara.com/question/Q142687.png)
$$\:\:\:\:\:\:\int\:\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{sin}\:{x}}\:\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}}\:=? \\ $$
Answered by qaz last updated on 04/Jun/21
![∫(dx/( (√(1−sin x))(√(1+cos x)))) =(1/( (√2)))∫(dx/((sin (x/2)−cos (x/2))cos (x/2))) =(1/( (√2)))∫(dx/(2sin (x/2)cos (x/2)−cos^2 (x/2))) =(1/( (√2)))∫((sec^2 (x/2))/(2tan (x/2)−1))dx =(1/( (√2)))ln∣2tan (x/2)−1∣+C](https://www.tinkutara.com/question/Q142707.png)
$$\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}\sqrt{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{dx}}{\left(\mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}−\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{dx}}{\mathrm{2sin}\:\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}}−\mathrm{cos}\:^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{sec}\:^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}}{\mathrm{2tan}\:\frac{\mathrm{x}}{\mathrm{2}}−\mathrm{1}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\mid\mathrm{2tan}\:\frac{\mathrm{x}}{\mathrm{2}}−\mathrm{1}\mid+\mathrm{C} \\ $$
Commented by iloveisrael last updated on 05/Jun/21
![(√(1−sin x)) = ∣sin (x/2)−cos (x/2)∣](https://www.tinkutara.com/question/Q142762.png)
$$\sqrt{\mathrm{1}−\mathrm{sin}\:{x}}\:=\:\mid\mathrm{sin}\:\frac{{x}}{\mathrm{2}}−\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\mid \\ $$