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Question Number 78198 by john santu last updated on 15/Jan/20

Commented by john santu last updated on 15/Jan/20

$$dearMjssir.$$$$iaskforyouropiniononthisanswer?$$$$rightornot.$$

Commented by MJS last updated on 15/Jan/20

$$\mathrm{right}!$$

Commented by john santu last updated on 15/Jan/20

$$thankssir$$

Commented by mathmax by abdo last updated on 15/Jan/20

$$letI={\int }_0^{\frac{\pi }{2}}\sqrt{1-sin\left(x\right)}dx\Rightarrow I={\int }_0^{\frac{\pi }{2}}\sqrt{1-cos(\frac{\pi }{2}-x)}dx$$$${=}_{\frac{\pi }{2}-x=t}-{\int }_0^{\frac{\pi }{2}}\sqrt{1-cost}(-dt)={\int }_0^{\frac{\pi }{2}}\sqrt{1-cost}dt$$$$={\int }_0^{\frac{\pi }{2}}\sqrt{2{sin}^2\left(\frac{t}{2}\right)}dt=\sqrt{2}{\int }_0^{\frac{\pi }{2}}sin\left(\frac{t}{2}\right)dt$$$$=-2\sqrt{2}{\left[cos\left(\frac{t}{2}\right)\right]}_0^{\frac{\pi }{2}}=-2\sqrt{2}(\frac{\sqrt{2}}{2}-1)=-2+2\sqrt{2}\Rightarrow$$$$I=2\sqrt{2}-2$$

Answered by MJS last updated on 15/Jan/20

$$\mathrm{two}\mathrm{other}\mathrm{possibilities}$$$$\left(1\right)$$$$\int \sqrt{1-\sin x}dx=$$$$[t=\sqrt{1-\sin x}\to dx=-\frac{2\sqrt{1-\sin x}}{\cos x}dx]$$$$=\int \frac{-2t}{\sqrt{2-t^2}}dt=2\sqrt{2-t^2}=2\sqrt{1+\sin x}+C$$$$$$$$\left(2\right)$$$$\int \sqrt{1-\sin x}dx=$$$$[t=\sin x\to dx=\frac{dt}{\cos x}]$$$$=\int \frac{dt}{\sqrt{t+1}}=2\sqrt{t+1}=2\sqrt{1+\sin x}+C$$

Commented by jagoll last updated on 15/Jan/20

$$waw...amazingsir$$

Commented by MJS last updated on 15/Jan/20

$$...\mathrm{sorry}\mathrm{there}\mathrm{had}\mathrm{been}\mathrm{a}\mathrm{typo},\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{correct}$$

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