Tích phân

Question

$\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}\dfrac{sinx}{\sqrt{x^{2}+1}+x }dx$

$\int\limits_{-1}^{0}\frac{xln(x+2)}{\sqrt{4-x^{2}} }dx$

score 2 · Answer 1

$B = \int\limits_{ - 1}^0 {\frac{{xln(x + 2)}}{{\sqrt {4 - {x^2}} }}} dx$

Lưu ý: $y = - \sqrt {4 - {x^2}} \Rightarrow y' = \frac{x}{{\sqrt {4 - {x^2}} }}$

Đặt:

$u = ln(x + 2) \Rightarrow du = \frac{1}{{x + 2}}dx$

$dv = \frac{x}{{\sqrt {4 - {x^2}} }} \Leftarrow v = - \sqrt {4 - {x^2}} $

$B = - \sqrt {4 - {x^2}} ln(x + 2)\begin{cases}0 \\ -1 \end{cases} + \int\limits_{ - 1}^0 {\frac{{\sqrt {4 - {x^2}} }}{{x + 2}}} dx = - 2\ln 2 + \int\limits_{ - 1}^0 {\frac{{\sqrt {4 - {x^2}} }}{{x + 2}}} dx$

Đặt $x = 2\sin t,t \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$

$dx = 2\cos tdt$

$B = - 2\ln 2 + \int\limits_{ - \frac{\pi }{6}}^0 {\frac{{\sqrt {4 - 4{{\sin }^2}t} }}{{2\sin t + 2}}.2} \cos tdt = - 2\ln 2 + \int\limits_{ - \frac{\pi }{6}}^0 {\frac{{2{{\cos }^2}t}}{{\sin t + 1}}} dt$

$= - 2\ln 2 + 2\int\limits_{ - \frac{\pi }{6}}^0 {(1 - \sin t)} dt = - 2\ln 2 + 2 - \sqrt 3 + \frac{\pi }{3}$

score 3 · Answer 2

1/ $A = \int\limits_{\frac{{ - \pi }}{4}}^{\frac{\pi }{4}} {\frac{{\sin x}}{{\sqrt {{x^2} + 1} + x}}} dx = \int\limits_{\frac{{ - \pi }}{4}}^{\frac{\pi }{4}} {\left( {\sqrt {{x^2} + 1} - x} \right)\sin x} dx$

Đặt $t = - x \Rightarrow - dt = dx$

$A = - \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left( {\sqrt {{t^2} + 1} + t} \right)\sin t} dt = - \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left( {\sqrt {{t^2} + 1} - t + 2t} \right)\sin t} dt$

$\Leftrightarrow A = - \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left( {\sqrt {{t^2} + 1} - t} \right)\sin t} dt - 2\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {t\sin t} dt$

$\Leftrightarrow A = - A - 2\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {t\sin t} dt \Leftrightarrow A = - \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {t\sin t} dt$

Tới đây đặt $u=t, dv=sintdt$ là xong.