Question Number 168556 by mokys last updated on 13/Apr/22
$$\int\sqrt{{sinx}}\:{dx} \\ $$
Answered by MJS_new last updated on 13/Apr/22
![this question keeps reappearing; I′ve given the answer before... ∫(√(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/Q168599.png)
$$\mathrm{this}\:\mathrm{question}\:\mathrm{keeps}\:\mathrm{reappearing};\:\mathrm{I}'\mathrm{ve}\:\mathrm{given} \\ $$$$\mathrm{the}\:\mathrm{answer}\:\mathrm{before}… \\ $$$$\int\sqrt{\mathrm{sin}\:{x}}\:{dx}=\int\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}\:\rightarrow\:{dx}=\mathrm{2}{dt}\right] \\ $$$$=\mathrm{2}\int\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{t}}\:{dt}=\mathrm{2E}\:\left({t}\mid\mathrm{2}\right)\:= \\ $$$$=\mathrm{2E}\:\left(\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}\mid\mathrm{2}\right)\:+{C} \\ $$
Commented by peter frank last updated on 14/Apr/22
$$\mathrm{thanks} \\ $$