Tích phân [Lần 2]

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

Viết lại: $I= \displaystyle \int^{3}_{1} \sqrt{\frac{2+(x-2)}{2-(x-2)}}dx$
Đặt $x-2=2\cos2t\Rightarrow dx=-4\sin2tdt$;   Khi $\ x=1\Rightarrow t=\dfrac{\pi}{3}; \ x=3 \Rightarrow=\dfrac{\pi}{6}$
 $I=\displaystyle \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3} } \sqrt{\dfrac{2+2\cos2t}{2-2\cos2t}}.4\sin2tdt=4\displaystyle \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}2\cos^{2}tdt=2\displaystyle \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} (1+\cos2t)dt=(2t+\sin2t)\Bigg|_{\dfrac{\pi}{6}}^ {\dfrac{\pi}{3}}$

Bài 2. Tính tích phân:  $$I=\displaystyle \int_1^{e } \dfrac{1-\ln x}{x(x+\ln x)}dx$$

Ta có: \[\dfrac{{1 - \ln x}}{{x\left( {x + \ln x} \right)}} = \dfrac{{x + 1 - \left( {x + \ln x} \right)}}{{x\left( {x + \ln x} \right)}} = \dfrac{{x + 1}}{{x\left( {x + \ln x} \right)}} - \dfrac{1}{x}\]
\[ = \dfrac{{\left( {x + \ln x} \right)'}}{{x + \ln x}} - \left( {\ln x} \right)' = \left( {\ln \left( {x + \ln x} \right)} \right)' - \left( {\ln x} \right)' = \left( {\ln \dfrac{{x + \ln x}}{{\ln x}}} \right)'\]
Khi đó: \[I = \int\limits_1^e {\dfrac{{1 - \ln x}}{{x\left( {x + \ln x} \right)}}} dx = \left( {\ln \dfrac{{x + \ln x}}{{\ln x}}} \right)_1^e\]

Bài 3. Tính tích phân: $$I=\displaystyle \int_0^1 x^2 \ln (1+x) \mbox{d}x$$

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

Ta có: $I=\displaystyle \int_0^1 \ln {(x^2+x+1)} d\left(\dfrac{x^2}{2}\right)=\dfrac{x^2. \ln {(x^2+x+1)}}{2}\Bigg|_{0}^{1}- \dfrac{1}{2}\displaystyle \int_0^1 \dfrac{x^2(2x+1)dx}{x^2+x+1}=\dfrac{\ln 3}{2}-\dfrac{1}{2}J$

Xét : $J=\displaystyle \int_0^1 \dfrac{x^2(2x+1)dx}{x^2+x+1}=\displaystyle \int_0^1 \dfrac{2x(x^2+x+1)-(x^2+x+1)-\dfrac{1}{2}(2x+1)+\dfrac{3}{2}}{x^2+x+1}dx$

$J= \displaystyle \int_0^1 (2x-1)dx- \dfrac{1}{2}\displaystyle \int_0^1 \dfrac{d(x^2+x+1)}{x^2+x+1}+ 6 \displaystyle \int_0^1 \dfrac{dx}{(2x+1)^2+3}=(x^2-x)\Bigg|_{0}^{1}- \dfrac{1}{2} \ln {(x^2+x+1)}\Bigg|_{0}^{1}+6K=-\dfrac{\ln3}{2}+6K $

Xét:  $K=\displaystyle \int_0^1 \dfrac{dx}{(2x+1)^2+3}$

Đặt:  $2x+1=\sqrt{3}\tan t \Rightarrow dx=\dfrac{\sqrt{3}}{2}(\tan^2 t+1)dt$ . Khi $x=0 \Rightarrow  t=\dfrac{\pi}{6}; \ x=1 \Rightarrow  t=\dfrac{\pi}{3}$

Ta có: $K=\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}  \dfrac{\dfrac{\sqrt{3}}{2}(\tan^2 t+1)dt}{3(\tan^2 t+1)}=\dfrac{\sqrt{3}}{6}\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} dt=\dfrac{\sqrt{3}}{6}\Bigg|_{\dfrac{\pi}{6}}^ {\dfrac{\pi}{3}}  = \dfrac{\sqrt{3}\pi}{36}$

Bài 5. Tính:$$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{(1+\cos x)\{1-\tan ^ 2 \frac{x}{2}\tan (x+\sin x)\tan (x-\sin x)\}}{\tan (x+\sin x)}\ dx $$

$$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{(1+\cos x)\{1-\tan ^ 2 \frac{x}{2}\tan (x+\sin x)\tan (x-\sin x)\}}{\tan (x+\sin x)}\ dx $$
$$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{(1+\cos x)\{1-\left(\frac{1-\cos x }{1+\cos x}\right).\tan (x+\sin x)\tan (x-\sin x)\}}{\tan (x+\sin x)}\ dx $$

$$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1+ \cos x}{\tan (x+\sin x)}dx- \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \tan (x-\sin x).(1-\cos x)dx$$
Đặt $$x+\sin x=t\Leftrightarrow (1+\cos x)dx=dt$$
Và đặt $$x-\sin x=u\Leftrightarrow (1-\cos x)dx=du$$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{ \tan t}dt-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\tan udu$$
$$=\left[\ln \mid \sin x \mid \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}-\left[\ln \mid \sec x\mid \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$
$$=\ln \mid \sin \frac{\pi}{2}\mid-\ln \mid\sin \frac{\pi}{4}\mid-\ln \mid \sec \frac{\pi}{2}\mid+\ln \mid\sec \frac{\pi}{4}\mid$$