Question

Cum se rezolva această integrală
?

Answer 1

$\displaystyle I = \int_{1}^e \dfrac{dx}{x\sqrt{4-\ln^2 x}}$

Facem schimbarea de variabilă:

$\displaystyle \ln x = t \Rightarrow (\ln x)'\, dx = t'\, dt \Rightarrow \dfrac{1}{x}\, dx = dt\\ x = 1 \Rightarrow t = \ln 1 = 0 \\ x = e \Rightarrow t = \ln e = 1 \\ \\ I = \int_{0}^1 \dfrac{dt}{\sqrt{4-t^2}} =\int_{0}^1 \dfrac{dt}{\sqrt{2^2-t^2}} = \arcsin \dfrac{t}{2}\Bigg|_{0}^1 =\\ \\=\arcsin \dfrac{1}{2} - \arcsin 0 = \boxed{\dfrac{\pi}{6}}$

Answer 2

Răspuns:

pi/6

Explicație pas cu pas:

(1/x)dx=d(lnx)

$I=\int\limits^e_1 {\frac{dx}{x\sqrt{4-ln^{2} x} }} \, dx=\int\limits^e_1 {\frac{d(lnx)}{\sqrt{4-ln^{2} } x} \ =|t=lnx, (x=1; t=0), (x=e; t=1)| \\ = \int\limits^1_0 {\frac{1}{\sqrt{2^{2}-t^{2}}}} \, dt =$

$=arcsin\frac{t}{2}|_{0}^{1} =arcsin\frac{1}{2} -arcsin0=\frac{\pi}{6}.$