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# Question-69027

Question Number 69027 by rajesh4661kumar@gmail.com last updated on 18/Sep/19
Commented by mathmax by abdo last updated on 18/Sep/19
$${I}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {sinxdx}\:−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid{sinx}\mid{dx}\:\:{but} \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{sinx}\:{dx}\:=_{{x}=\pi+{t}} \:\:\int_{−\pi} ^{\pi} {sin}\left(\pi+{t}\right){dt}\:=−\int_{−\pi} ^{\pi} \:{sint}\:{dt}\:=\mathrm{0}\left({odd}\:{function}\right) \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid{sinx}\mid{dx}\:=_{{x}=\pi+{t}} \:\:\int_{−\pi} ^{\pi} \mid{sint}\mid{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} {sint}\:{dt}\:=\mathrm{2}\left[−{cost}\right]_{\mathrm{0}} ^{\pi} \\$$$$=\mathrm{2}\left\{\mathrm{1}−\left(−\mathrm{1}\right)\right\}\:=\mathrm{4} \\$$
Commented by mathmax by abdo last updated on 18/Sep/19
$$\Rightarrow\:{I}\:=\mathrm{0}−\mathrm{4}\:=−\mathrm{4}\:. \\$$
Answered by MJS last updated on 18/Sep/19
$$\mathrm{sin}\:{x}\:−\mid\mathrm{sin}\:{x}\mid=\begin{cases}{\mathrm{0};\:\mathrm{0}\leqslant{x}<\pi}\\{\mathrm{2sin}\:{x};\:\pi\leqslant{x}\leqslant\mathrm{2}\pi}\end{cases} \\$$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\mathrm{sin}\:{x}\:−\mid\mathrm{sin}\:{x}\mid\:{dx}=\mathrm{2}\underset{\pi} {\overset{\mathrm{2}\pi} {\int}}\mathrm{sin}\:{x}\:{dx}=−\mathrm{2}\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{sin}\:{x}\:{dx}= \\$$$$=\mathrm{2}\left[\mathrm{cos}\:{x}\right]_{\mathrm{0}} ^{\pi} =−\mathrm{4} \\$$