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Question Number 53465 by maxmathsup by imad last updated on 22/Jan/19

$$find{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{cosx-{cos}^{3}x}dx$$

Commented by maxmathsup by imad last updated on 23/Jan/19

$$letI={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{cosx-{cos}^3}dx\Rightarrow I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{cosx}\sqrt{1-{cos}^{2}x}dx$$$$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{cosx}\mid sinx\mid dx=2{\int }_0^{\frac{\pi }{2}}\sqrt{cosx}sinxdx$$$$=-2{\int }_0^{\frac{\pi }{2}}{\left(cosx\right)}^{\frac{1}{2}}\left(d\left(cosx\right)\right)dx$$$$=-2{\left[\frac{1}{1+\frac{1}{2}}{cos}^{1+\frac{1}{2}}x\right]}_0^{\frac{\pi }{2}}=-2.\frac{2}{3}{\left[{cos}^{\frac{3}{2}}x\right]}_0^{\frac{\pi }{2}}=-\frac{4}{3}(-1)=\frac{4}{3}$$

Answered by ajfour last updated on 22/Jan/19

$${\int }_{-\pi /2}^{\pi /2}\sqrt{\cos x}\mid \sin x\mid dx$$$$=-2{\int }_0^{\pi /2}\sqrt{\cos x}\left(\sin x\right)dx$$$$=-2{\int }_1^0\sqrt{t}dt=\frac{4}{3}.$$

Commented by malwaan last updated on 23/Jan/19

$$\mathrm{t}=\mathrm{cosx}$$$$\Rightarrow \mathrm{dt}=-\mathrm{sinxdx}$$$$\mathrm{where}\mathrm{is}(-)?$$

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