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Question Number 205432 by hardmath last updated on 21/Mar/24

$$\mathrm{Find}:\mathrm{\Omega }={\int }_0^{2\mathit{\pi }}\ln (\mathrm{sinx}+\sqrt{1+{\sin }^2\mathrm{x}})\mathrm{dx}$$

Answered by MathedUp last updated on 21/Mar/24

$$0$$$$\mathrm{cauz}-f\left(z\right)=f(-z),f(z+\pi )=-f\left(z\right),f(z+2\pi )=f\left(z\right)$$$$\mathrm{So},f\left(z\right)\mathrm{is}\mathrm{periodic}\mathrm{function}\mathrm{that}\mathrm{having}\mathrm{period}\pi$$$${\int }_0^{2\pi }f\left(z\right)\mathrm{d}z=0$$

Answered by mr W last updated on 21/Mar/24

$$\mathrm{\Omega }={\int }_0^{2\pi }\ln (\sin x+\sqrt{1+{\sin }^{2}x})dx$$$$={\int }_0^{\pi }\ln (\sin x+\sqrt{1+{\sin }^{2}x})dx+{\int }_{\pi }^{2\pi }\ln (\sin x+\sqrt{1+{\sin }^{2}x})dx$$$$={\int }_0^{\pi }\ln (\sin x+\sqrt{1+{\sin }^{2}x})dx+{\int }_0^{\pi }\ln (\sin (x+\pi )+\sqrt{1+{\sin }^2(x+\pi )})d(x+\pi )$$$$={\int }_0^{\pi }\ln (\sin x+\sqrt{1+{\sin }^{2}x})dx+{\int }_0^{\pi }\ln (-\sin x+\sqrt{1+{\sin }^{2}x})dx$$$$={\int }_0^{\pi }\ln (\sin x+\sqrt{1+{\sin }^{2}x})dx+{\int }_0^{\pi }\ln \frac{1}{\sin x+\sqrt{1+{\sin }^{2}x}}dx$$$$={\int }_0^{\pi }\ln (\sin x+\sqrt{1+{\sin }^{2}x})dx-{\int }_0^{\pi }\ln (\sin x+\sqrt{1+{\sin }^{2}x})dx$$$$=0$$

Commented by hardmath last updated on 21/Mar/24

$$\mathrm{cool}\mathrm{dear}\mathrm{professor}\mathrm{thankyou}$$

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