Question

${\int }_{-\pi /2}^{\pi /2}\sin x\sqrt{1-{\cos }^{2}x}dx$

Answer 1

Consider the following question.

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\sin x\sqrt{1-{\cos }^{2}x}dx$

$={\int }_{\frac{-\pi }{2}}^0\sin x\sqrt{1-{\cos }^{2}x}dx+{\int }_0^{\frac{\pi }{2}}\sin x\sqrt{1-{\cos }^{2}x}dx$

$case.1$

$t=-\cos x$

$x=-\frac{\pi }{2},0$

$t=(0,1)$

$case.2)x=(\frac{\pi }{2},0)$

$t=(1,0)$

$\frac{dt}{\sin x}=dx$

$I={\int }_{\frac{-\pi }{2}}^0(-cosx)\sqrt{1-t^2}\frac{dt}{\cos x}$

$=-({\int }_{\frac{-\pi }{2}}^0\sqrt{1-t^2}dt+{\int }_0^{\frac{\pi }{2}}\sqrt{1-t^2}dt)$

$=-({(\left(\frac{1}{2}\sqrt{1-t^2}\right)-\frac{1}{2}\ln (1+\sqrt{1-t^2}))}_{-1}^0+{(\left(\frac{1}{2}\sqrt{1-t^2}\right)-\frac{1}{2}\ln (1+\sqrt{1-t^2}))}_0^1)$

$=-(\frac{1}{2}{(\left(\sin x\right)-\ln (1+\sin x))}_{\frac{-\pi }{2}}^0+\frac{1}{2}{(\left(\sin x\right)-\ln (1+\sin x))}_0^{\frac{\pi }{2}})+C$

$=-(\frac{1}{2}(0-1)+\frac{1}{2}(1-\ln 2-0\left)\right)$

$=\frac{1}{2}\ln 2$

Hence, this is the required answer.

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