Question Number 224861 by fantastic last updated on 08/Oct/25
$$\int\frac{\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{1}+\mathrm{sin}\:{x}}}\:{dx} \\ $$
Answered by Mathswiz last updated on 08/Oct/25
Commented by som(math1967) last updated on 08/Oct/25
$$\sqrt{\mathrm{1}+{sinx}}\neq\sqrt{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$
Commented by Mathswiz last updated on 11/Oct/25
Answered by Ghisom_ last updated on 08/Oct/25
![∫((sin x)/( (√(1+sin x))))dx= [t=(√(1−sin x)) → dx=−((2dt)/( (√(2−t^2 ))))] =−2∫((t^2 −1)/(t^2 −2))dt=−2∫(1+(1/(t^2 −2)))dt= =−2t+((√2)/2)ln ∣((t+(√2))/(t−(√2)))∣ =...](https://www.tinkutara.com/question/Q224871.png)
$$\int\frac{\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{1}+\mathrm{sin}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{1}−\mathrm{sin}\:{x}}\:\rightarrow\:{dx}=−\frac{\mathrm{2}{dt}}{\:\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }}\right] \\ $$$$=−\mathrm{2}\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{2}}{dt}=−\mathrm{2}\int\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{2}}\right){dt}= \\ $$$$=−\mathrm{2}{t}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{{t}+\sqrt{\mathrm{2}}}{{t}−\sqrt{\mathrm{2}}}\mid\:=… \\ $$