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Question

Evaluate π0x1+sinxdx

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Solution

Let I=π0x1+sinxdx ...(1)
Substituting xπx
I=π0πx1+sin(πx)dx
[baf(x)dx=baf(a+bx)dx]
I=π0πx1+sinxdx ....(2)
Adding equations (1) and (2), we get
2I=π0π1+sinxdx
I=π2π011+sinxdx
I=π2π011+sinx×1sinx1sinxdx
I=π2π01sinx1sin2xdx=π2π01sinxcos2xdx
I=π2π0[1cos2xsinxcos2x]dx
I=π2π0[sec2xtanxsecx]dx
I=π2[tanxsecx]π0
I=π2[(tanπsecπ)(tan0sec0)]
I=π2[(0+1)(01)]=π
Hence, the value of required integration is π

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