Let dxdG(x)=xetanx,x∈(0,2π) Now, I=∫1/41/2x2etanπx2⋅dx=∫1/41/2πx22πxetanπx2⋅dx Let πx2=t⇒2πxdx=dt When x=21,t=4π and x=41,t=16π ∴I=∫π/16π/4tetantdt=g(t)∣4π16π=G(4π)−G(16π)
If d d x G(x)= e^tan x x, x ∈(0, π / 2), then ∫_1 / 4^1 / 2 2 x · e^tan (π x^2 ) d x is equal to
JEE Main 2012 — Mathematics Calculus
If dxdG(x)=xetanx,x∈(0,π/2), then ∫1/41/2x2⋅etan(πx2)dx is equal to
Held on 12 May 2012 · Verified 6 Jul 2026.
Options
- A
G(π/4)−G(π/16)
- B
2[G(π/4)−G(π/16)]
- C
π[G(1/2)−G(1/4)]
- D
G(1/2)−G(1/2)
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