The value of the integral 6 3 ( 4-cosec2 x 4 x )dx is :
- 11 3
- 16 3
- 323 3
- 643 3
Answer
(3) 323 3
Solution
Put t= x, dt= 2x dx; 2x=1+t2, cosec2x= 1+t2t2
4-cosec2x 4xdx= (4-cosec2x ) 2x 2x dx= [4- 1+t2t2 ] (1+t2 )dt
= [4+4t2- (1+t2 )2t2 ]dt= (2+3t2- 1t2 )dt
Limits: x= 6 t= 1 3, x= 3 t= 3
I= [2t+t3+ 1t ]1/ 3 3
= (2 3+3 3+ 1 3 )- ( 2 3+ 13 3+ 3 )
= 16 3- 163 3= 323 3
Answer: (3)