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Question Number 183865 by Michaelfaraday last updated on 31/Dec/22

Answered by MJS_new last updated on 31/Dec/22

=∫_0 ^π ∣1−2sin (x/2)∣dx=  =∫_0 ^(π/3) (2sin (x/2) −1)dx+∫_(π/3) ^π (1−2sin (x/2))dx=  =4((√3)−1)−(π/3)

$$=\underset{\mathrm{0}} {\overset{\pi} {\int}}\mid\mathrm{1}−\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\mid{dx}= \\ $$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{3}} {\int}}\left(\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\:−\mathrm{1}\right){dx}+\underset{\pi/\mathrm{3}} {\overset{\pi} {\int}}\left(\mathrm{1}−\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\right){dx}= \\ $$$$=\mathrm{4}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)−\frac{\pi}{\mathrm{3}} \\ $$

Commented by Michaelfaraday last updated on 31/Dec/22

thanks sir

$${thanks}\:{sir} \\ $$

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