int_0^($\frac{2}{3}$) dx/(4+9x^2) equals:

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KollegeApply · Accepted Answer

pi/4 Explanation: Let `I = int_0^($\frac{2}{3}$) dx/(4 + 9x^2) = $\frac{1}{9}$ int_0^($\frac{2}{3}$) dx/($\frac{4}{9}$ + x^2) = $\frac{1}{9}$ * $\frac{1}{2}$ [tan^-1 (x/($\frac{2}{3}$))]_0^($\frac{2}{3}$) = $\frac{1}{6}$ [tan^-1 (3x)/2]_0^($\frac{2}{3}$) = $\frac{1}{6}$ [tan^-1 1 - tan^-1 0] = $\frac{1}{6}$ [pi/4 - 0] = pi/24`

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