[Toán 12] Tính tích phân

G

giahung341_14

[tex]I = \int {x\sqrt {{x^2} + 1} dx} [/tex]
[tex]x = \tan t \Rightarrow {x^2} + 1 = {\tan ^2}t + 1 = \frac{1}{{{{\cos }^2}t}} \Rightarrow dx = \frac{{dt}}{{{{\cos }^2}t}}[/tex]
[tex]I = \int {\tan t\sqrt {\frac{1}{{{{\cos }^2}t}}} } dt = \int {\frac{{\tan t}}{{\cos t}}} dt = \int {\sin t.\frac{1}{{{{\cos }^3}t}}} dt[/tex]
[tex]u = \sin t \Rightarrow du = \cos tdt[/tex]
[tex]dv = \frac{1}{{{{\cos }^2}t}} \Rightarrow v = \tan t[/tex]
[tex]I = \frac{{{{\sin }^2}t}}{{\cos t}} - \int {\tan t\cos tdt} = \frac{{{{\sin }^2}t}}{{\cos t}} - \int {\sin tdt} = \frac{{{{\sin }^2}t}}{{\cos t}} + \cos t = \frac{1}{{\cos t}} = \frac{1}{{\cos (\arctan x)}}[/tex]
 
T

truongduong9083

Gợi ý:
1. $\displaystyle \int_{0}^{1} x\sqrt{x^2 +1} dx=\dfrac{1}{2}\int_{0}^{1} \sqrt{x^2 +1} d(x^2+1) = \dfrac{( \sqrt{x^2 +1} )^{\dfrac{3}{2}}}{3}\bigg|^1_0$
2. Đặt $t = x^2+1$ là xong
 
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