calculate-0-2pi-cos-2x-2cosx-sin-x-dx-

Question Number 62732 by mathmax by abdo last updated on 24/Jun/19

calculate ∫_0 ^(2π) ((cos(2x))/(2cosx −sin(x)))dx

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}{cosx}\:−{sin}\left({x}\right)}{dx}\: \\ $$

Answered by MJS last updated on 24/Jun/19

((cos (2(x+π)))/(2cos (x+π) −sin (x+π)))=−((cos 2x)/(2cos x −sin x)) ⇒ ⇒ ∫_0 ^π ((cos 2x)/(2cos x −sin x))dx=−∫_π ^(2π) ((cos 2x)/(2cos x −sin x))dx ⇒ ⇒ ∫_0 ^(2π) ((cos 2x)/(2cos x −sin x))dx=0

$$\frac{\mathrm{cos}\:\left(\mathrm{2}\left({x}+\pi\right)\right)}{\mathrm{2cos}\:\left({x}+\pi\right)\:−\mathrm{sin}\:\left({x}+\pi\right)}=−\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2cos}\:{x}\:−\mathrm{sin}\:{x}}\:\Rightarrow \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2cos}\:{x}\:−\mathrm{sin}\:{x}}{dx}=−\underset{\pi} {\overset{\mathrm{2}\pi} {\int}}\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2cos}\:{x}\:−\mathrm{sin}\:{x}}{dx}\:\Rightarrow \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2cos}\:{x}\:−\mathrm{sin}\:{x}}{dx}=\mathrm{0} \\ $$

Commented by mathmax by abdo last updated on 24/Jun/19

$${thank}\:{you}\:{sir}\:. \\ $$

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