Question Number 165731 by mkam last updated on 07/Feb/22
$$\int\:\sqrt{{sinx}}\:{dx} \\ $$
Commented by puissant last updated on 07/Feb/22
Commented by Tawa11 last updated on 07/Feb/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by MJS_new last updated on 08/Feb/22
![∫(√(sin x)) dx=∫(√(1−sin^2 ((2x−π)/4))) dx= [t=((2x−π)/4) → dx=2dt] =2∫(√(1−sin^2 t)) dt=2E (t∣2) = =2E (((2x−π)/4)∣2) +C](https://www.tinkutara.com/question/Q165760.png)
$$\int\sqrt{\mathrm{sin}\:{x}}\:{dx}=\int\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}\:\rightarrow\:{dx}=\mathrm{2}{dt}\right] \\ $$$$=\mathrm{2}\int\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{t}}\:{dt}=\mathrm{2E}\:\left({t}\mid\mathrm{2}\right)\:= \\ $$$$=\mathrm{2E}\:\left(\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}\mid\mathrm{2}\right)\:+{C} \\ $$
Commented by peter frank last updated on 09/Feb/22
$$\mathrm{thank}\:\mathrm{you} \\ $$