∫(2x−1)2+5(2x−1)cos(2x−1)2+5dx
(2x−1)2+5=t2
2(2x−1)2dx=2tdt
2t2−5dx=tdt
So ∫2t2−5t2−5costdt=21sint+c
=21sin(2x−1)2+5+c
JEE Main 2021 — Mathematics Calculus
The integral ∫4x2−4x+6(2x−1)cos(2x−1)2+5dx is equal to (where c is a constant of integration)
Held on 18 Mar 2021 · Verified 6 Jul 2026.
21sin(2x−1)2+5+c
21cos(2x+1)2+5+c
21cos(2x−1)2+5+c
21sin(2x+1)2+5+c
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