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1-10-sin-x-x-dx-




Question Number 84036 by M±th+et£s last updated on 08/Mar/20
∫_(−1) ^(10) sin(x−∣x∣) dx
$$\int_{−\mathrm{1}} ^{\mathrm{10}} {sin}\left({x}−\mid{x}\mid\right)\:{dx} \\ $$
Answered by TANMAY PANACEA last updated on 09/Mar/20
∫_(−1) ^0 sin(x+x)dx+∫_0 ^(10) sin(x−x)dx  =((−∣cos2x∣_(−1) ^0 )/2)+0  =((−1)/2){(cos0−cos(−2)}=((cos(−2)−1)/2)=((−1+cos2)/2)
$$\int_{−\mathrm{1}} ^{\mathrm{0}} {sin}\left({x}+{x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{10}} {sin}\left({x}−{x}\right){dx} \\ $$$$=\frac{−\mid{cos}\mathrm{2}{x}\mid_{−\mathrm{1}} ^{\mathrm{0}} }{\mathrm{2}}+\mathrm{0} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\left\{\left({cos}\mathrm{0}−{cos}\left(−\mathrm{2}\right)\right\}=\frac{{cos}\left(−\mathrm{2}\right)−\mathrm{1}}{\mathrm{2}}=\frac{−\mathrm{1}+{cos}\mathrm{2}}{\mathrm{2}}\right. \\ $$
Commented by M±th+et£s last updated on 09/Mar/20
thaks sir nice solution
$${thaks}\:{sir}\:{nice}\:{solution}\: \\ $$
Commented by TANMAY PANACEA last updated on 09/Mar/20
most welcomesir
$${most}\:{welcomesir} \\ $$

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