Rearranging, we get,
∫π/6π/3(2dxd(tan4x)⋅sin43x+tan4x⋅2dxd(sin43x))dx
=21∫π/6π/3(dxd(tan4x⋅sin43x))dx
=21[tan4x⋅sin43x]π/6π/3
=21[(3)4×0−(31)4×1]
=−21×91=−181
JEE Main 2020 — Mathematics Calculus
The integral ∫6π3πtan3x⋅sin23x(2sec2x⋅sin23x+3tanx⋅sin6x)dx is equal to:
Held on 4 Sept 2020 · Verified 6 Jul 2026.
187
−91
−181
29
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