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Question Number 211886 by Spillover last updated on 23/Sep/24

Answered by Ghisom last updated on 23/Sep/24

$$\int \sqrt{{\cos }^{2}x+\cos x}dx=\int \sqrt{2\cos x}\cos \frac{x}{2}dx=$$$$[t=\frac{\sqrt{2}\sin \frac{x}{2}}{\sqrt{\cos x}}\iff x=2\arctan \frac{t}{\sqrt{t^2+2}}\to dx=\frac{\sqrt{{2\cos }^{3}x}}{\cos \frac{x}{2}}dt]$$$$=2\int \frac{dt}{{(t^2+1)}^2}=\frac{t}{t^2+1}+\arctan t=$$$$=\sqrt{2\cos x}\sin \frac{x}{2}+\arctan \frac{\sqrt{2}\sin \frac{x}{2}}{\sqrt{\cos x}}+C$$

Answered by Spillover last updated on 23/Sep/24

$$\sqrt{\cos {}^{2}x+\cos x}dx$$$$Let\cos x=\sin {}^{2}tdx=\frac{-2\sin t\cos t}{\sin x}dt=-\frac{2\sin t\cos t}{\sqrt{1-\cos {}^{2}t}}dt$$$$\frac{2\sin t\cos t}{\sqrt{1-\cos {}^{2}t}}=-\frac{2\sin t\cos t}{\sqrt{1-\sin {}^{4}t}}dt=-\frac{2\sin t\cos t}{\sqrt{(1-\sin {}^{2}t)(1+p\sin {}^{2}t)}}dt$$$$dx=-\frac{2\sin t\cos t}{\sqrt{(1-\sin {}^{2}t)(1+\sin {}^{2}t)}}dt$$$$\int \sqrt{\cos {}^{2}x+\cos x}dx=\int \sqrt{\sin {}^{4}t+\sin {}^{2}t}dx$$$$=\int (\sqrt{\sin {}^{4}t+\sin {}^{2}t}\times -\frac{2\sin t\cos t}{\sqrt{(1-\sin {}^{2}t)(1+\sin {}^{2}t)}}dt)$$$$-2\int \sin {}^{2}tdt=-2\left(\frac{-\sin t\cos t+t}{2}\right)$$$$\sin t\cos t-t$$$$\sqrt{\cos x(1-\cos x)}-{\sin }^{-1}\sqrt{\cos x}+C$$$$$$

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