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Question Number 112271 by bemath last updated on 07/Sep/20

$$\sqrt{\mathrm{bemath}}$$$$\int \sin \mathrm{x}\sqrt{1-\sin \mathrm{x}}\mathrm{dx}?$$

Answered by maths mind last updated on 07/Sep/20

$$1-sin\left(x\right)={(sin\left(\frac{x}{2}\right)-cos\left(\frac{x}{2}\right))}^2$$

Commented by bemath last updated on 07/Sep/20

$$\sqrt{1-\sin \mathrm{x}}=\mid \sin \frac{\mathrm{x}}{2}-\cos \frac{\mathrm{x}}{2}\mid$$

Answered by MJS_new last updated on 07/Sep/20

$$\int \sin x\sqrt{1-\sin x}dx=$$$$[t=\sin x\to dx=\frac{dt}{\cos x}]$$$$=\int \frac{t\sqrt{1-t}}{\sqrt{1-t^2}}dt=\int \frac{t}{\sqrt{1+t}}dt=\frac{2}{3}(t-2)\sqrt{t+1}$$$$\mathrm{but}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{losing}\mathrm{sign}\mathrm{changes}...$$$$\mathrm{other}\mathrm{idea}$$$$\int \sin x\sqrt{1-\sin x}dx=$$$$[t=\frac{x}{2}-\frac{\pi }{4}\to dx=2dt]$$$$=2\int \cos 2t\sqrt{1-\cos 2t}dt=$$$$=2\int ({2\cos }^{2}t-1)\sqrt{1-({2\cos }^{2}t-1)}dt=$$$$=2\sqrt{2}\int ({2\cos }^{2}t-1)\sqrt{1-{\cos }^{2}t}dt=$$$$=2\sqrt{2}\int ({2\cos }^{2}t\sin t-\sin t)dt=$$$$=\sqrt{2}\int (\sin 3t-\sin t)dt=$$$$=\sqrt{2}\cos t-\frac{\sqrt{2}}{3}\cos 3t=...$$$$\mathrm{but}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{also}\mathrm{losing}\mathrm{sign}\mathrm{changes}$$

Commented by MJS_new last updated on 07/Sep/20

$$\mathrm{found}\mathrm{it}!\mathrm{we}\mathrm{must}\mathrm{somehow}\mathrm{keep}\mathrm{the}\mathrm{factor}$$$$\sqrt{1-\sin x}\mathrm{in}\mathrm{the}\mathrm{result}.$$$$$$$$\int \sin x\sqrt{1-\sin x}dx=$$$$[t=\sin x\to dx=\frac{dt}{\cos x}]$$$$=\int \frac{t\sqrt{1-t}}{\sqrt{1-t^2}}dt=$$$$[u=\frac{\sqrt{1-t}}{\sqrt{1-t^2}}\to dt=-\frac{2(t+1)\sqrt{1-t^2}}{\sqrt{1-t}}]$$$$=2\int (\frac{1}{u^2}-\frac{1}{u^4})du=\frac{2}{3u^3}-\frac{2}{u}=$$$$=\frac{2(t-2)\sqrt{1-t^2}}{3\sqrt{1-t}}=$$$$=\frac{2(\sin x-2)\cos x}{3\sqrt{1-\sin x}}+C$$

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