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Question Number 91479 by Zainal Arifin last updated on 01/May/20

$$\underset{-\pi /2}{\stackrel{\pi /2}{\int }}\sqrt{\cos x-{\cos }^{3}x}\mathrm{dx}=...$$

Commented by jagoll last updated on 01/May/20

$$\sqrt{\cos x(1-\cos {}^{2}x)}=\sqrt{\cos x}\sqrt{\sin {}^{2}x}$$$$\underset{-\frac{\pi }{2}}{\stackrel{0}{\int }}\sqrt{\cos x}(-\sin x)dx+\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\sqrt{\cos x}\sin xdx$$$$noweasytosolve$$

Commented by Zainal Arifin last updated on 01/May/20

$$\mathrm{Ok}.\mathrm{thanks}$$

Commented by Prithwish Sen 1 last updated on 01/May/20

$$\int \sqrt{\mathrm{cosx}}\mathrm{sinx}\mathrm{dx}\mathrm{put}\mathrm{cosx}={\mathrm{t}}^2\Rightarrow -\mathrm{sinxdx}=2\mathrm{tdt}$$$$=-\int \mathrm{t}.2\mathrm{t}.\mathrm{dt}=-\frac{2}{3}.{\left(\mathrm{cosx}\right)}^{\frac{3}{2}}$$$${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{\mathrm{cosx}}.\mathrm{sinxdx}=2{\int }_0^{\frac{\pi }{2}}\sqrt{\mathrm{cosx}}.\mathrm{sinxdx}=2{[-\frac{2}{3}{\left(\mathrm{cosx}\right)}^{\frac{2}{3}}]}_0^{\frac{\pi }{2}}$$$$=2.[-(0-\frac{2}{3})]=\frac{4}{3}\mathbf{please}\mathbf{check}.$$

Commented by jagoll last updated on 01/May/20

$$\sqrt{\sin {}^{2}x}\ne \sin x$$$$\sqrt{\sin {}^{2}x}=\mid \sin x\mid$$

Answered by john santu last updated on 01/May/20

Commented by Prithwish Sen 1 last updated on 01/May/20

$$\mathrm{I}\mathrm{think}\mathrm{in}0⩽\mathrm{x}⩽\frac{\pi }{2}\mathrm{sinx}\mathrm{and}\mathrm{cosx}⩾0$$$$\mathrm{sorry}\mathrm{fix}\mathrm{it}.$$

Commented by jagoll last updated on 01/May/20

$$0⩽x⩾\frac{\pi }{2}=0⩽x⩽\frac{\pi }{2}?$$

Commented by Prithwish Sen 1 last updated on 01/May/20

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.$$

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