How do you find b such that #intdx/(1+x^2)=pi/3# from [0,b]?

1 Answer
Oct 30, 2016

# b=sqrt(3) #

Explanation:

You need to use the standard result # int 1/(1+x^2)dx = arctanx+C #

We have # int_0^b 1/(1+x^2)dx = pi/3 #

# :. [arctanx]_0^b = pi/3 #
# :. arctanb - arctan 0 = pi/3 #
# :. arctanb - 0 = pi/3 #
# :. arctanb = pi/3 #
# :. b = tan(pi/3) #
# :. b = sqrt(3) #