Let I=∫π/245π/241+3tan2xdx.
Apply King's rule (x→π/4−x):
I=∫π/245π/241+3cot2xdx=∫π/245π/241+3tan2x3tan2xdx.
Adding: 2I=∫π/245π/24dx=245π−24π=6π.
I=12π.
JEE Main 2026 — Mathematics Calculus
The value of the integral ∫24π245π1+3tan2x dx is :
Held on 23 Jan 2026 · Verified 6 Jul 2026.
3π
12π
18π
6π
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