[haunted2]

how do you integrate 1/(9x^2 + 1) with upper limit as 1/ (root3) and lower limit as 0
Thanks in advance. I have been trying for a very long time and just couldnt get the answer which is pi/9 while I keep getting pi/3.

[PECTOPAH]

Originally Posted by haunted2

how do you integrate 1/(9x^2 + 1) with upper limit as 1/ (root3) and lower limit as 0
Thanks in advance. I have been trying for a very long time and just couldnt get the answer which is pi/9 while I keep getting pi/3.

1/(9x^2 + 1) = 1/((3x)^2 + 1)

after integration (use formula)
(1/3)tan^-1 (3x) i guess u miss out the 1/3 here.. u have to integrate the 3x inside

apply limit
then u get pi/9

[haunted2]

Oh thanks for your help