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Question Number 126749 by bemath last updated on 24/Dec/20

Commented by MJS_new last updated on 24/Dec/20

$$t=\tan \frac{x}{2}\mathrm{leads}\mathrm{to}$$$$2\int \frac{{(t+1)}^2}{t^4+10t^2+1}dt$$$$\mathrm{now}\mathrm{find}\mathrm{the}2\mathrm{square}\mathrm{factors}\mathrm{of}\mathrm{the}\mathrm{denominator}$$$$t^4+10t^2+1=(t^2+5-2\sqrt{6})(t^2+5+2\sqrt{6})$$$$\mathrm{and}\mathrm{decompose}...$$

Answered by liberty last updated on 24/Dec/20

$$I=\underset{0}{\stackrel{\pi }{\int }}\frac{1+\sin x}{2-(2\cos {}^{2}x-1)}dx=\underset{0}{\stackrel{\pi }{\int }}\frac{1+\sin x}{3-2\cos {}^{2}x}dx$$$$I_1=\underset{0}{\stackrel{\pi }{\int }}\frac{1}{3-2\cos {}^{2}x}dx$$$$I_2=\underset{0}{\stackrel{\pi }{\int }}\frac{\sin x}{3-2\cos {}^{2}x}dx=-\underset{0}{\stackrel{\pi }{\int }}\frac{d\left(\cos x\right)}{3-2\cos {}^{2}x}$$$$I_2=\underset{-1}{\stackrel{1}{\int }}\frac{dt}{3-2t^2}=2\underset{0}{\stackrel{1}{\int }}\frac{dt}{(\sqrt{3}-t\sqrt{2})(\sqrt{3}+t\sqrt{2})}$$$$I=I_1+I_2$$

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