Question Number 78198 by john santu last updated on 15/Jan/20
Commented by john santu last updated on 15/Jan/20
$${dear}\:{Mjs}\:{sir}. \\ $$$${i}\:{ask}\:{for}\:{your}\:{opinion}\:{on}\:{this}\:{answer}? \\ $$$${right}\:{or}\:{not}. \\ $$
Commented by MJS last updated on 15/Jan/20
$$\mathrm{right}! \\ $$
Commented by john santu last updated on 15/Jan/20
$${thanks}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 15/Jan/20
![let I =∫_0 ^(π/2) (√(1−sin(x)))dx ⇒I =∫_0 ^(π/2) (√(1−cos((π/2)−x)))dx =_((π/2)−x =t) −∫_0 ^(π/2) (√(1−cost))(−dt) =∫_0 ^(π/2) (√(1−cost))dt =∫_0 ^(π/2) (√(2sin^2 ((t/2))))dt =(√2)∫_0 ^(π/2) sin((t/2))dt =−2(√2)[cos((t/2))]_0 ^(π/2) =−2(√2)(((√2)/2)−1) =−2 +2(√2) ⇒ I =2(√2)−2](Q78247.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−{sin}\left({x}\right)}{dx}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx} \\ $$$$=_{\frac{\pi}{\mathrm{2}}−{x}\:={t}} \:\:\:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−{cost}}\left(−{dt}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−{cost}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)}{dt}\:=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}\left(\frac{{t}}{\mathrm{2}}\right){dt} \\ $$$$=−\mathrm{2}\sqrt{\mathrm{2}}\left[{cos}\left(\frac{{t}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=−\mathrm{2}\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\mathrm{1}\right)\:=−\mathrm{2}\:+\mathrm{2}\sqrt{\mathrm{2}}\:\Rightarrow \\ $$$${I}\:=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2} \\ $$
Answered by MJS last updated on 15/Jan/20
![two other possibilities (1) ∫(√(1−sin x))dx= [t=(√(1−sin x)) → dx=−((2(√(1−sin x)))/(cos x))dx] =∫((−2t)/(√(2−t^2 )))dt=2(√(2−t^2 ))=2(√(1+sin x))+C (2) ∫(√(1−sin x))dx= [t=sin x → dx=(dt/(cos x))] =∫(dt/(√(t+1)))=2(√(t+1))=2(√(1+sin x))+C](Q78205.png)
$$\mathrm{two}\:\mathrm{other}\:\mathrm{possibilities} \\ $$$$\left(\mathrm{1}\right) \\ $$$$\int\sqrt{\mathrm{1}−\mathrm{sin}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{1}−\mathrm{sin}\:{x}}\:\rightarrow\:{dx}=−\frac{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{sin}\:{x}}}{\mathrm{cos}\:{x}}{dx}\right] \\ $$$$=\int\frac{−\mathrm{2}{t}}{\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }}{dt}=\mathrm{2}\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{1}+\mathrm{sin}\:{x}}+{C} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\int\sqrt{\mathrm{1}−\mathrm{sin}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sin}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{cos}\:{x}}\right] \\ $$$$=\int\frac{{dt}}{\sqrt{{t}+\mathrm{1}}}=\mathrm{2}\sqrt{{t}+\mathrm{1}}=\mathrm{2}\sqrt{\mathrm{1}+\mathrm{sin}\:{x}}+{C} \\ $$
Commented by jagoll last updated on 15/Jan/20
$${waw}...{amazing}\:{sir} \\ $$
Commented by MJS last updated on 15/Jan/20
$$...\mathrm{sorry}\:\mathrm{there}\:\mathrm{had}\:\mathrm{been}\:\mathrm{a}\:\mathrm{typo},\:\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct} \\ $$