Find-0-2-ln-sinx-1-sin-2-x-dx-

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

Find: Ω = ∫_0 ^( 2𝛑) ln (sinx + (√(1 + sin^2 x))) dx

$$\mathrm{Find}:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{2}\boldsymbol{\pi}} \:\mathrm{ln}\:\left(\mathrm{sinx}\:+\:\sqrt{\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{x}}\right)\:\mathrm{dx} \\ $$

Answered by MathedUp last updated on 21/Mar/24

0 cauz −f(z)=f(−z) , f(z+π)=−f(z) , f(z+2π)=f(z) So , f(z) is periodic function that having period π ∫_0 ^(2π) f(z)dz=0

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

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

Ω=∫_0 ^(2π) ln (sin x+(√(1+sin^2 x)))dx =∫_0 ^π ln (sin x+(√(1+sin^2 x)))dx+∫_π ^(2π) ln (sin x+(√(1+sin^2 x)))dx =∫_0 ^π ln (sin x+(√(1+sin^2 x)))dx+∫_0 ^π ln (sin (x+π)+(√(1+sin^2 (x+π))))d(x+π) =∫_0 ^π ln (sin x+(√(1+sin^2 x)))dx+∫_0 ^π ln (−sin x+(√(1+sin^2 x)))dx =∫_0 ^π ln (sin x+(√(1+sin^2 x)))dx+∫_0 ^π ln (1/(sin x+(√(1+sin^2 x))))dx =∫_0 ^π ln (sin x+(√(1+sin^2 x)))dx−∫_0 ^π ln (sin x+(√(1+sin^2 x)))dx =0

$$\Omega=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{ln}\:\left(\mathrm{sin}\:{x}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\:\left(\mathrm{sin}\:{x}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx}+\int_{\pi} ^{\mathrm{2}\pi} \mathrm{ln}\:\left(\mathrm{sin}\:{x}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\:\left(\mathrm{sin}\:{x}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx}+\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\:\left(\mathrm{sin}\:\left({x}+\pi\right)+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\left({x}+\pi\right)}\right){d}\left({x}+\pi\right) \\ $$$$\:\:=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\:\left(\mathrm{sin}\:{x}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx}+\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\:\left(−\mathrm{sin}\:{x}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\:\left(\mathrm{sin}\:{x}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx}+\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{sin}\:{x}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}}{dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\:\left(\mathrm{sin}\:{x}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx}−\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\:\left(\mathrm{sin}\:{x}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx} \\ $$$$\:\:=\mathrm{0} \\ $$

Commented by hardmath last updated on 21/Mar/24

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

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