The integral ∫π/6π/4sin2x(tan5x+cot5x)dx equals:
(A)201tan−1(931)
(B)101(4π−tan−1(931))✓ Correct
(C)40π
(D)51(4π−tan−1(331))
Answer: (B)
Solution
Set t=tanx (dt=sec2xdx), so I=∫1/312t(t5+t−5)dt=∫1/312(t10+1)t4dt. Now put u=t5, du=5t4dt: I=101∫3−5/21u2+1du=101[tan−1u]3−5/21=101(4π−tan−1931).
JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option