1-explicit-f-a-1-3-arctan-a-x-dx-with-a-gt-0-2-calculate-1-3-arctan-2-x-dx-and-1-3-arctan-3-x-dx-

Question Number 99146 by mathmax by abdo last updated on 18/Jun/20

1) explicit f (a) = \int_{1}^{\sqrt{3}} \arctan (\frac{a}{x}) dx with a > 0

2) calculate \int_{1}^{\sqrt{3}} \arctan (\frac{2}{x}) dx and \int_{1}^{\sqrt{3}} \arctan (\frac{3}{x}) dx

Answered by mathmax by abdo last updated on 19/Jun/20

1) f (a) = \int_{1}^{\sqrt{3}} \arctan (\frac{a}{x}) dx by parts we get

f (a) = {[xarctan (\frac{a}{x})]}_{1}^{\sqrt{3}} - \int_{1}^{\sqrt{3}} x (- \frac{a}{x^{2}}) \times \frac{1}{1 + \frac{a^{2}}{x^{2}}} dx

= \sqrt{3} \arctan (\frac{a}{\sqrt{3}}) - \arctan (a) + \int_{1}^{\sqrt{3}} \frac{ax}{x^{2} + a^{2}} dx but

\int_{1}^{\sqrt{3}} \frac{ax}{x^{2} + a^{2}} dx =_{x = au} \int_{\frac{1}{a}}^{\frac{\sqrt{3}}{a}} \frac{a (au)}{a^{2} (1 + u^{2})} adu = a \int_{\frac{1}{a}}^{\frac{\sqrt{3}}{a}} \frac{udu}{1 + u^{2}}

= \frac{a}{2} {[\ln (1 + u^{2})]}_{\frac{1}{a}}^{\frac{\sqrt{3}}{a}} = \frac{a}{2} {\ln (1 + \frac{3}{a^{2}}) - \ln (1 + \frac{1}{a^{2}})} = \frac{a}{2} \ln (\frac{a^{2} + 3}{a^{2} + 1}) \Rightarrow

f (a) = \sqrt{3} \arctan (\frac{a}{\sqrt{3}}) - \arctan (a) + \frac{a}{2} \ln (\frac{3 + a^{2}}{1 + a^{2}})

2) \int_{1}^{\sqrt{3}} artan (\frac{2}{x}) dx = f (2) = \sqrt{3} \arctan (\frac{2}{\sqrt{3}}) - \arctan (2) + \ln (\frac{7}{5})

\int_{1}^{\sqrt{3}} \arctan (\frac{3}{x}) dx = f (3) = \sqrt{3} \arctan (\sqrt{3}) - \arctan (3) + \frac{3}{2} \ln (\frac{6}{5})