Q.

The value of the integral ∫π245π24dx1+tan2x3 is:                 [2026]

1 π3  
2 π18  
3 π12  
4 π6  

Ans.

(3)

I=∫π245π24dx1+tan2x3    ...(1)

Apply King

I=∫π245π24dx1+tan2(π4-x)3

I=∫π245π24dx1+cot2x3    ...(2)

Add (1) + (2) ⇒2I=∫π245π24(1) dx

I=12(π6)=π12