1) explicit f(a) =∫_1 ^(√3) arctan((a/x))dx with a>0 2) calculate ∫_1 ^(√3) arctan((2/x))dx and ∫


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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
	$1) f(a) =∫_1 ^(√3) arctan((a/x))dx by parts we get f(a) =[xarctan((a/x))]_1 ^(√3) −∫_1 ^(√3) x(−(a/x^2 ))×(1/(1+(a^2 /x^2 ))) dx =(√3)arctan((a/(√3)))−arctan(a) +∫_1 ^(√3) ((ax)/(x^2 +a^2 )) dx but ∫_1 ^(√3) ((ax)/(x^2 +a^2 )) dx =_(x=au) ∫_(1/a) ^((√3)/a) ((a(au))/(a^2 (1+u^2 ))) adu =a ∫_(1/a) ^((√3)/a) ((udu)/(1+u^2 )) =(a/2)[ln(1+u^2 )]_(1/a) ^((√3)/a) =(a/2){ln(1+(3/a^2 ))−ln(1+(1/a^2 ))} =(a/2)ln(((a^2 +3)/(a^2 +1))) ⇒ f(a) =(√3)arctan((a/(√3)))−arctan(a)+(a/2)ln(((3+a^2 )/(1+a^2 ))) 2)∫_1 ^(√3) artan((2/x))dx =f(2) =(√3) arctan((2/(√3)))−arctan(2)+ln((7/5)) ∫_1 ^(√3) arctan((3/x))dx =f(3) =(√3) arctan((√3))−arctan(3)+(3/2)ln((6/5))$
	$1) f (a) = \int_{1}^{\sqrt{3}} \arctan (\frac{a}{x}) dx by parts we get$ $f (a) = {[xarctan (\frac{a}{x})]}_{1}^{\sqrt{3}} - \int_{1}^{\sqrt{3}} x (- \frac{a}{x^{2}}) \times \frac{1}{1 + \frac{a^{2}}{x^{2}}} dx$ $= \sqrt{3} \arctan (\frac{a}{\sqrt{3}}) - \arctan (a) + \int_{1}^{\sqrt{3}} \frac{ax}{x^{2} + a^{2}} dx but$ $\int_{1}^{\sqrt{3}} \frac{ax}{x^{2} + a^{2}} dx =_{x = au} \int_{\frac{1}{a}}^{\frac{\sqrt{3}}{a}} \frac{a (au)}{a^{2} (1 + u^{2})} adu = a \int_{\frac{1}{a}}^{\frac{\sqrt{3}}{a}} \frac{udu}{1 + u^{2}}$ $= \frac{a}{2} {[\ln (1 + u^{2})]}_{\frac{1}{a}}^{\frac{\sqrt{3}}{a}} = \frac{a}{2} {\ln (1 + \frac{3}{a^{2}}) - \ln (1 + \frac{1}{a^{2}})} = \frac{a}{2} \ln (\frac{a^{2} + 3}{a^{2} + 1}) \Rightarrow$ $f (a) = \sqrt{3} \arctan (\frac{a}{\sqrt{3}}) - \arctan (a) + \frac{a}{2} \ln (\frac{3 + a^{2}}{1 + a^{2}})$ $2) \int_{1}^{\sqrt{3}} artan (\frac{2}{x}) dx = f (2) = \sqrt{3} \arctan (\frac{2}{\sqrt{3}}) - \arctan (2) + \ln (\frac{7}{5})$ $\int_{1}^{\sqrt{3}} \arctan (\frac{3}{x}) dx = f (3) = \sqrt{3} \arctan (\sqrt{3}) - \arctan (3) + \frac{3}{2} \ln (\frac{6}{5})$
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