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Question

Evaluate π0xsinx1+sinxdx.

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Solution

LetI=π0xsinx1+sinx(1)=π0(πx)sin(πx)1+sin(πx)=π0(πx)sinx1+sinxdx(2)

On addingboth(1)+(2),weget
2I=π0πsinxdx1+sinxI=π2π0sinx1+sinxdxI=π2π0(1sinx)sinx(1+sinx)(1sinx)dx=π2π0sinxsin2xcos2xdx=π2π0sinxcos2xdxπ2π0sin2xcos2xdx=π2π0tanxsecxdxπ2π01cos2xcos2xdx=π2[sec]π0π2π0sec2xdx+π2π0dx=π2[secπsec0]π2[tanx]π0+π2[x]π0=π2[1]π2[tanπtan0]+π2[π0]=ππ2[00]+π2[π]=π[π21]Thusπ0xsinx1+sinx=π[π21]

Hence, this is the answer.

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