[Toán 12] Tính tích phân

G

giahung341_14

I=∫xx2+1dxI = \int {x\sqrt {{x^2} + 1} dx}
x=tan⁡t⇒x2+1=tan⁡2t+1=1cos⁡2t⇒dx=dtcos⁡2tx = \tan t \Rightarrow {x^2} + 1 = {\tan ^2}t + 1 = \frac{1}{{{{\cos }^2}t}} \Rightarrow dx = \frac{{dt}}{{{{\cos }^2}t}}
I=∫tan⁡t1cos⁡2tdt=∫tan⁡tcos⁡tdt=∫sin⁡t.1cos⁡3tdtI = \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
u=sin⁡t⇒du=cos⁡tdtu = \sin t \Rightarrow du = \cos tdt
dv=1cos⁡2t⇒v=tan⁡tdv = \frac{1}{{{{\cos }^2}t}} \Rightarrow v = \tan t
I=sin⁡2tcos⁡t−∫tan⁡tcos⁡tdt=sin⁡2tcos⁡t−∫sin⁡tdt=sin⁡2tcos⁡t+cos⁡t=1cos⁡t=1cos⁡(arctan⁡x)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)}}
 
T

truongduong9083

Gợi ý:
1. ∫01xx2+1dx=12∫01x2+1d(x2+1)=(x2+1)323∣01\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=x2+1t = x^2+1 là xong
 
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