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Robert · Accepted Answer

For all x, | sinx| ≤ 1, Hence we have∫ ∞
0
[sinx]2
1 + x2
dx ≤
∫ ∞
0
1
1 + x2
dx
= [arctanx]∞0 = arctan(∞)− arctan 0 = (nπ +
π
2
)− (nπ + 0)
=
π
2
Hence we have ∫ ∞
0
[sinx]2
1 + x2
dx ≤ π
2
1

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