Tích phân [Lần 1]

Bài 1. Tính tích phân: $$I=\displaystyle \int_0^1 x^2 \ln (1+x) \mbox{d}x$$

Dùng tích phân từng phần.
Đặt : $\begin{cases} u = \ln (x+1) \\ \mbox{d}v= x^2 \mbox{d}x \end{cases} \Rightarrow \begin{cases} \mbox{d}u = \dfrac{1}{x+1} \\\\ v= \dfrac{x^3}{3} \mbox{d}x \end{cases}$
Khi đó ta có : $$\begin{aligned}I= \left. \dfrac{x^3}{3} \ln (x+1) \right|_{0}^{1} - \dfrac{1}{3}\displaystyle \int_{0}^{1} \dfrac{x^3}{x+1} \mbox{d}x & = \dfrac{1}{3} \ln 2 - \dfrac{1}{3}\displaystyle \int_{0}^{1} \left(x^2 -x +1 - \dfrac{1}{x+1}\right) \mbox{d}x \\& = \dfrac{1}{3} \ln 2 - \left. \dfrac{1}{3} \left( \dfrac{x^3}{3}  - \dfrac{x^2}{2} +x - \ln (x+1) \right) \right|_{0}^{1}\\& = \dfrac{1}{3} \ln 2 - \dfrac{5}{18} +\dfrac{1}{3} \ln 2 \\& = \dfrac{2}{3} \ln 2 - \dfrac{5}{18} \end{aligned}$$

Bài 2. Tính tích phân : $$I= \displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x . (\sin x -1)\left(\dfrac{1}{\sin^3 x} -1 \right)\mbox{d}x$$

Ta có: $I=-\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos x(\sin x-1)dx + \displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\dfrac{\cos x}{\sin^2x}dx-\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\dfrac{\cos x}{\sin ^3 x}dx$

Hay $I=-\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-1)d(\sin x-1)+\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\dfrac{d(\sin x)}{\sin^2x}-\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\dfrac{d(\sin x)}{\sin^3x}$
$=-\dfrac{(\sin x-1)^2}{2}\Bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}}-\dfrac{1}{\sin x}\Bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}}+\dfrac{1}{2\sin^2x}\Bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}}= \dfrac{2\sqrt{2}-3}{4}$

Bài 3. Tính tích phân : $$\displaystyle \int _{1}^{e} \dfrac{\ln xdx}{x(1+\ln x+\sqrt{\ln^2x+1})}$$

Đặt $t=\sqrt{\ln^2x+1}\Rightarrow   2tdt=\dfrac{2\ln xdx}{x}$

                                      $x=0\Rightarrow t=1,x=e \Rightarrow t=\sqrt2$

$I=\displaystyle\int_{1}^{\sqrt2}\dfrac{t}{\sqrt{t^2+1}+t+1}dt$

Đến đây ta nhân liên hợp cho tích phân phía dưới ta có:

$I=\displaystyle\int_{1}^{\sqrt2}\dfrac{t(\sqrt{t^2+1}-t-1}{-2t})dt$

$\Leftrightarrow I=\displaystyle\int_{1}^{\sqrt2}\left(\dfrac{\sqrt{t^2+1}}{-2}+\dfrac{t}{2}+\dfrac{t+1}{2t}\right)dt$

$\Leftrightarrow I=\left(-\dfrac{x}{4}\sqrt{x^2+1}-\dfrac{1}{4}\ln (x+\sqrt{x^2+1})+t^2+\dfrac{1}{2}t+\dfrac{1}{2}\ln x\right)\bigg|^\sqrt2_1$

Bài 4. Tính tích phân : $$I = \displaystyle \int_{0}^{\frac{\pi}{4}} \left(1 + \tan x \tan \dfrac{x}{2} \right)\sin x  \mbox{d}x$$

Ta có:
$(1+  \tan x.\tan\dfrac{x}{2})\sin x=\sin x+ \tan x \tan\dfrac{x}{2}.2.\sin\dfrac{x}{2}.\cos\dfrac{x}{2}$
$\Rightarrow (1+  \tan x.\tan\dfrac{x}{2})\sin x=\sin x+ \tan x . 2\sin^{2}(\dfrac{x}{2})$
$\Rightarrow (1+  \tan x.\tan\dfrac{x}{2})\sin x=\sin x+ \tan x . (1-\cos x)$
$\Rightarrow (1+  \tan x.\tan\dfrac{x}{2})\sin x=\sin x+ \tan x -\sin x$
$\Rightarrow (1+  \tan x.\tan\dfrac{x}{2})\sin x=\tan x$
Vậy :
$I = \displaystyle \int_{0}^{\frac{\pi}{4}} \left(1 + \tan x \tan \dfrac{x}{2} \right)\sin x  \mbox{d}x =\displaystyle \int_{0}^{\frac{\pi}{4}} \tan x   \mbox{d}x = -\displaystyle \int_{0}^{\frac{\pi}{4}} \dfrac{1}{\cos x}   \mbox{d}\cos x=- \ln(\cos x) \bigg|^{\frac{\pi}{4}}_0$
$=-\ln(\dfrac{\sqrt{2}}{2})$

Bài 5. Tính tích phân: $$I=\displaystyle \int_0^{\dfrac{\pi}{2}} \dfrac{e^x.\sin x}{1+\sin 2x} dx$$

Xét : $J=\displaystyle \int_0^{\dfrac{\pi}{2}} \dfrac{e^x.\cos x}{1+\sin 2x} dx$

Ta có: $I+J=\displaystyle \int_0^{\dfrac{\pi}{2}}\dfrac{e^xdx}{\sin x+\cos x} \ (1)$

$I-J=\displaystyle \int_0^{\dfrac{\pi}{2}}\dfrac{e^x(\sin x- \cos x)dx}{(\sin x+\cos x)^2}=\displaystyle \int_0^{\dfrac{\pi}{2}}e^xd \left ( \dfrac{1}{\sin x+\cos x}\right)$
$=\dfrac{e^x}{\sin x+\cos x}\Bigg|_{0}^{\dfrac{\pi}{2}}-\displaystyle \int_0^{\dfrac{\pi}{2}}\dfrac{e^xdx}{\sin x+\cos x}$

Hay $I-J=e^{\dfrac{\pi}{2}}-1-\displaystyle \int_0^{\dfrac{\pi}{2}}\dfrac{e^xdx}{\sin x+\cos x} \ (2)$

Lấy $(1)+(2)$ ta có : $2I=e^{\dfrac{\pi}{2}}-1\Rightarrow I=\dfrac{1}{2}(e^{\dfrac{\pi}{2}}-1)$