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Question Number 192916 by Mingma last updated on 31/May/23

Answered by ARUNG_Brandon_MBU last updated on 31/May/23

$$I={\int }_0^{\frac{\pi }{2}}(\frac{1-\sin 2x}{1+\cos x}+\frac{1-\cos 2x}{1+\sin x})dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+\cos x}+2{\int }_0^{\frac{\pi }{2}}\frac{\cos x}{1+\cos x}\left(\sin xdx\right)+2{\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{1+\sin x}dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{1-\cos x}{{\sin }^{2}x}dx+2{\int }_0^1\frac{u}{1+u}du+2{\int }_0^{\frac{\pi }{2}}(\sin x-1+\frac{1}{1+\sin x})dx$$$$={[\cot x-\frac{1}{\sin x}]}_{\frac{\pi }{2}}^0+2{[u-\ln (1+u)]}_0^1+2{[\cos x+x-\tan x+\frac{1}{\cos x}]}_{\frac{\pi }{2}}^0$$$$=1+2(1-\ln 2)+2(2-\frac{\pi }{2})=7-\pi -2\ln 2$$

Commented by aba last updated on 31/May/23

$$\mathrm{Error}\mathrm{I}=3-\pi +2\ln 2$$

Commented by aba last updated on 31/May/23

$$$$$$I={\int }_0^{\frac{\pi }{2}}(\frac{1-\sin 2x}{1+\cos x}+\frac{1-\cos 2x}{1+\sin x})dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+\cos x}-2{\int }_0^{\frac{\pi }{2}}\frac{\cos x}{1+\cos x}\left(\sin xdx\right)+2{\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{1+\sin x}dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{1-\cos x}{{\sin }^{2}x}dx-2{\int }_0^1\frac{u}{1+u}du+2{\int }_0^{\frac{\pi }{2}}(\sin x-1+\frac{1}{1+\sin x})dx$$$$={[+\cot x-\frac{1}{\sin x}]}_{\frac{\pi }{2}}^0-2{[u-\ln (1+u)]}_0^1+2{[\cos x-x+\mathrm{tanx}-\frac{1}{\mathrm{cosx}}]}_{\frac{\pi }{2}}^0$$$$$$$$=1-2(1-\ln 2)+2(2-\frac{\pi }{2})=3-\pi +2\ln 2$$

Commented by Mingma last updated on 01/Jun/23

Perfect ��

Commented by ARUNG_Brandon_MBU last updated on 04/Jun/23

Yeah you're right. Thanks!

Answered by aba last updated on 31/May/23

$$\mathrm{a}=3\wedge \mathrm{b}=-1\wedge \mathrm{c}=2\Rightarrow \mathrm{a}+\mathrm{b}-\mathrm{c}=0$$

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