I=∫02π1+ex96x2cos2xdx2I=2∫02π96x2cos2xdxI=96∫02πx2cos2xdx=48∫02πx2(1+cos2x)dx=2π2+48(0−0)−48∫02πxsin2xdx=2π2−12π+[0−0]=π(2π2−12)=π(απ2+β)⇒(α+β)2=100
If ∫−2π2π(1+ex)96x2cos2xdx=π(απ2+β),α,β∈Z, then (α+β)2 equals
Held on 28 Jan 2025 · Verified 6 Jul 2026.
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