Find the integral of 1/(9x^2+4) dx with the limits [-∞...∞] using the substitution 2/3tanu=x?
I was wondering if someone could walk me through the actual substitution step by step since I wind up with int((6+6tan^2u)/(4+4tan^2u))dx for the limits [1/2pi...1/2pi], which doesn't give me the right answer.
More Questions From This User See All
Answers & Comments
Verified answer
x = (2/3)tanu
dx = (2/3)sec^2 u du
integral of 1/(9x^2+4) dx
= integral of (2/3)sec^2 u/(4tan^2 u + 4) du
= 1/6 integral of du from -(1/2)pi to (1/2)i
= pi/6
x = 2tanu/3
dx/du = (2/3) sec^2 u = (2/3)(1+tan^2u)
1/(9x^2+4) dx = 1/[4(1+tan^2u)] *(2/3)(1+tan^2u)du
= Integral (1/6)du = u/6 = (1/6) * tan^-1(3x/2)