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Câu a)
\(\int \frac{1}{\cos^4x}dx=\int \frac{\sin ^2x+\cos^2x}{\cos^4x}dx=\int \frac{\sin ^2x}{\cos^4x}dx+\int \frac{1}{\cos^2x}dx\)
Xét \(\int \frac{1}{\cos^2x}dx=\int d(\tan x)=\tan x+c\)
Xét \(\int \frac{\sin ^2x}{\cos^4x}dx=\int \frac{\tan ^2x}{\cos^2x}dx=\int \tan^2xd(\tan x)=\frac{\tan ^3x}{3}+c\)
Vậy :
\(\int \frac{1}{\cos ^4x}dx=\frac{\tan ^3x}{3}+\tan x+c\)
\(\Rightarrow \int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{dx}{\cos^4 x}=\)\(\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|\left ( \frac{\tan ^3 x}{3}+\tan x+c \right )=\frac{44}{9\sqrt{3}}\)
Câu b)
\(\int \frac{(x+1)^2}{x^2+1}dx=\int \frac{x^2+1+2x}{x^2+1}dx=\int dx+\int \frac{2xdx}{x^2+1}\)
\(=x+c+\int \frac{d(x^2+1)}{x^2+1}=x+\ln (x^2+1)+c\)
Do đó:
\(\int ^{1}_{0}\frac{(x+1)^2}{x^2+1}dx=\left.\begin{matrix} 1\\ 0\end{matrix}\right|(x+\ln (x^2+1)+c)=\ln 2+1\)
Câu c)
\(\int \frac{x^2+2\ln x}{x}dx=\int xdx+2\int \frac{2\ln x}{x}dx\)
\(=\frac{x^2}{2}+c+2\int \ln xd(\ln x)\)
\(=\frac{x^2}{2}+c+\ln ^2x\)
\(\Rightarrow \int ^{2}_{1}\frac{x^2+2\ln x}{x}dx=\left.\begin{matrix} 2\\ 1\end{matrix}\right|\left ( \frac{x^2}{2}+\ln ^2x +c \right )=\frac{3}{2}+\ln ^22\)
Câu d)
\(\int^{2}_{1} \frac{x^2+3x+1}{x^2+x}dx=\int ^{2}_{1}dx+\int ^{2}_{1}\frac{2x+1}{x^2+x}dx\)
\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|x+\int ^{2}_{1}\frac{d(x^2+x)}{x^2+x}=1+\left.\begin{matrix} 2\\ 1\end{matrix}\right|\ln |x^2+x|=1+\ln 6-\ln 2\)
\(=1+\ln 3\)
\(I_1=3\int_1^2x^2dx+\int_1^2\cos xdx+\int_1^2\frac{dx}{x}=x^3\)\(|^2 _1\)+\(\sin x\)\(|^2_1\) +\(\ln\left|x\right|\)\(|^2_1\)
\(=\left(8-1\right)+\left(\sin2-\sin1\right)+\left(\ln2-\ln1\right)\)
\(=7+\sin2-\sin1+\ln2\)
b) \(I_2=4\int_1^2\frac{dx}{x}-5\int_1^2x^4dx+2\int_1^2\sqrt{x}dx\)
\(=4\left(\ln2-\ln1\right)-\left(2^5-1^5\right)+\frac{4}{3}\left(2\sqrt{2}-1\sqrt{1}\right)\)
\(=4\ln2+\frac{8\sqrt{2}}{3}-32\frac{1}{3}\)
a.
\(\int\limits^{\sqrt{7}}_0\dfrac{x^3}{\sqrt[3]{x^2+1}}dx\)
Đặt \(\sqrt[3]{x^2+1}=u\Rightarrow x^2+1=u^3\Rightarrow x^2=u^3-1\Rightarrow x.dx=\dfrac{3}{2}u^2du\)
\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=\sqrt{7}\Rightarrow u=2\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^2_1\dfrac{\left(u^3-1\right).\dfrac{3}{2}u^2du}{u}=\int\limits^2_1\dfrac{3}{2}\left(u^4-u\right)du=\dfrac{3}{2}\left(\dfrac{1}{5}u^5-\dfrac{1}{2}u^2\right)|^2_1\)
\(=\dfrac{141}{20}\)
b.
Đặt \(\sqrt{x+3}=u\Rightarrow x=u^2-3\Rightarrow dx=2udu\)
\(\left\{{}\begin{matrix}x=1\Rightarrow u=2\\x=6\Rightarrow u=3\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^3_2\dfrac{u+1}{u^2-3+2}.2udu=\int\limits^3_2\dfrac{2udu}{u-1}=\int\limits^3_22\left(1+\dfrac{1}{u-1}\right)du\)
\(=2\left(u+ln\left|u-1\right|\right)|^3_2=2\left(1+ln2\right)\)
Nhìn đề dữ dội y hệt cr của tui z :( Để làm từ từ
Lập bảng xét dấu cho \(\left|x^2-1\right|\) trên đoạn \(\left[-2;2\right]\)
| x | -2 | -1 | 1 | 2 |
| \(x^2-1\) | 0 | 0 |
\(\left(-2;-1\right):+\)
\(\left(-1;1\right):-\)
\(\left(1;2\right):+\)
\(\Rightarrow I=\int\limits^{-1}_{-2}\left|x^2-1\right|dx+\int\limits^1_{-1}\left|x^2-1\right|dx+\int\limits^2_1\left|x^2-1\right|dx\)
\(=\int\limits^{-1}_{-2}\left(x^2-1\right)dx-\int\limits^1_{-1}\left(x^2-1\right)dx+\int\limits^2_1\left(x^2-1\right)dx\)
\(=\left(\dfrac{x^3}{3}-x\right)|^{-1}_{-2}-\left(\dfrac{x^3}{3}-x\right)|^1_{-1}+\left(\dfrac{x^3}{3}-x\right)|^2_1\)
Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính
2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)
\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)
\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)
Câu 1)
Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2\ln x}{x}\\ v=\frac{-1}{x}\end{matrix}\right.\)
\(\int \left ( \frac{\ln}{x} \right )^2dx=\frac{-\ln^2x}{x}+2\int \frac{\ln x}{x^2}dx\)
Đặt \(\left\{\begin{matrix} t=\ln x\\ dk=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dt=\frac{1}{x}dx\\ k=-\frac{1}{x}\end{matrix}\right.\Rightarrow \int \frac{\ln x}{x^2}dx=-\frac{\ln x}{x}+\int \frac{1}{x^2}dx=\frac{-\ln x}{x}-\frac{1}{x}\)
\(\Rightarrow I=\left.\begin{matrix} e\\ 1\end{matrix}\right|\left(\frac{-\ln^2 x}{x}-\frac{2\ln x}{x}-\frac{2}{x}\right)=2-\frac{5}{e}\)
Câu 2)
\(I=\int ^{\frac{\pi}{4}}_{0}\frac{x}{1+\cos 2x}dx=\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{x}{\cos^2x}dx\)
Đặt \(\left\{\begin{matrix} u=x\\ dv=\frac{dx}{\cos^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\tan x\end{matrix}\right.\Rightarrow I=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{x\tan x}{2}-\frac{1}{2}\int^{\frac{\pi}{4}}_{0} \tan xdx\)
\(=\frac{\pi}{8}+\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{d(\cos x)}{\cos x}=\frac{\pi}{8}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{\ln |\cos x|}{2}=\frac{\pi}{8}+\frac{\ln\frac{\sqrt{2}}{2}}{2}\)
Ở tất cả các dạng bài như thế này em chỉ cần ghi nhớ công thức:
\(d(u(x))=u'(x)dx\)
Câu 1)
Ta có \(I_1=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} e^{\sin x}\cos xdx=\int _{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{\sin x}d(\sin x)\)
Đặt \(\sin x=t\Rightarrow I_1=\int ^{1}_{\frac{\sqrt{2}}{2}}e^tdt=\left.\begin{matrix} 1\\ \frac{\sqrt{2}}{2}\end{matrix}\right|e^t=e-e^{\frac{\sqrt{2}}{2}}\)
Câu 2)
\(I_2=\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}}e^{2\cos x+1}\sin xdx=\frac{-1}{2}\int ^\frac{\pi}{2}_{\frac{\pi}{4}}e^{2\cos x+1}d(2\cos x+1)\)
Đặt \(2\cos x+1=t\Rightarrow I_2=\frac{-1}{2}\int ^{1}_{1+\sqrt{2}}e^tdt\)
\(=\frac{-1}{2}.\left.\begin{matrix} 1\\ 1+\sqrt{2}\end{matrix}\right|e^t=\frac{-1}{2}(e-e^{1+\sqrt{2}})\)
Câu 3:
Có \(I_3=\int ^{e}_{1}\frac{e^{2\ln x+1}}{x}dx=\int ^{e}_{1}e^{2\ln x+1}d(\ln x)\)
\(=\frac{1}{2}\int ^{e}_{1}e^{2\ln x+1}d(2\ln x+1)\)
Đặt \(2\ln x+1=t\Rightarrow I_3=\frac{1}{2}\int ^{3}_{1}e^tdt=\frac{1}{2}.\left.\begin{matrix} 3\\ 1\end{matrix}\right|e^t=\frac{1}{2}(e^3-e)\)
Câu 4:
\(I_4=\int ^{1}_{0}xe^{x^2+2}dx=\frac{1}{2}\int ^{1}_{0}e^{x^2+2}d(x^2+2)\)
Đặt \(x^2+2=t\Rightarrow I_4=\frac{1}{2}\int ^{3}_{2}e^tdt=\frac{1}{2}.\left.\begin{matrix} 3\\ 2\end{matrix}\right|e^t=\frac{1}{2}(e^3-e^2)\)
a/ \(I=\int\limits^1_0\dfrac{1}{\left(x^2+3\right)\left(x^2+1\right)}dx=\dfrac{1}{2}\int\limits^1_0\left(\dfrac{1}{x^2+1}-\dfrac{1}{x^2+3}\right)dx\)
\(=\dfrac{1}{2}\left(arctanx-\dfrac{1}{\sqrt{3}}arctan\dfrac{x}{\sqrt{3}}\right)|^1_0=\dfrac{\pi}{8}-\dfrac{\pi\sqrt{3}}{36}\)
b/ \(I=\int\dfrac{x^2-1}{x^4+1}dx=\int\dfrac{1-\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}dx\)
Đặt \(x+\dfrac{1}{x}=t\Rightarrow\left(1-\dfrac{1}{x^2}\right)dx=dt\) ; \(x^2+\dfrac{1}{x^2}=t^2-2\)
\(\Rightarrow I=\int\dfrac{dt}{t^2-2}=\int\dfrac{dt}{\left(t-\sqrt{2}\right)\left(t+\sqrt{2}\right)}=\dfrac{1}{2\sqrt{2}}\int\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)dt\)
\(\Rightarrow I=\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{t-\sqrt{2}}{t+\sqrt{2}}\right|+C=\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|+C\)
c/ \(I=\int\dfrac{dx}{x\left(x^3+1\right)}=\int\dfrac{x^2dx}{x^3\left(x^3+1\right)}\)
Đặt \(x^3+1=t\Rightarrow3x^2dx=dt\)
\(\Rightarrow I=\dfrac{1}{3}\int\dfrac{dt}{\left(t-1\right)t}=\dfrac{1}{3}\int\left(\dfrac{1}{t-1}-\dfrac{1}{t}\right)dt=\dfrac{1}{3}ln\left|\dfrac{t-1}{t}\right|+C\)
\(\Rightarrow I=\dfrac{1}{3}ln\left|\dfrac{x^3}{x^3+1}\right|+C\)
d/ \(I=\int\limits^1_0\dfrac{xdx}{x^4+x^2+1}\)
Đặt \(x^2=t\Rightarrow2xdx=dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=1\Rightarrow t=1\end{matrix}\right.\)
\(I=\dfrac{1}{2}\int\limits^1_0\dfrac{dt}{t^2+t+1}=\dfrac{1}{2}\int\limits^1_0\dfrac{dt}{\left(t+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}=\dfrac{2}{3}\int\limits^1_0\dfrac{dt}{\dfrac{4}{3}\left(t+\dfrac{1}{2}\right)^2+1}\)
Đặt \(t+\dfrac{1}{2}=\dfrac{\sqrt{3}}{2}tanu\Rightarrow dt=\dfrac{\sqrt{3}}{2}.\dfrac{du}{cos^2u}\); \(\left\{{}\begin{matrix}t=0\Rightarrow u=\dfrac{\pi}{6}\\t=1\Rightarrow u=\dfrac{\pi}{3}\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{2}{3}.\dfrac{\sqrt{3}}{2}\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{du}{cos^2u\left(tan^2u+1\right)}=\dfrac{\sqrt{3}}{3}\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}du=\dfrac{\pi\sqrt{3}}{18}\)
a.
Đặt \(\sqrt{1-x^2}=u\Rightarrow x^2=1-u^2\Rightarrow xdx=-udu\)
\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=0\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^0_1\left(1-u^2\right).u.\left(-udu\right)=\int\limits^1_0\left(u^2-u^4\right)du=\left(\dfrac{1}{3}u^3-\dfrac{1}{5}u^5\right)|^1_0\)
\(=\dfrac{2}{15}\)
b.
\(\int\limits^2_1\dfrac{dx}{x^2-2x+2}=\int\limits^2_1\dfrac{dx}{\left(x-1\right)^2+1}\)
Đặt \(x-1=tanu\Rightarrow dx=\dfrac{1}{cos^2u}du\)
\(\left\{{}\begin{matrix}x=1\Rightarrow u=0\\x=2\Rightarrow u=\dfrac{\pi}{4}\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{1}{tan^2u+1}.\dfrac{1}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{cos^2u}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0du\)
\(=u|^{\dfrac{\pi}{4}}_0=\dfrac{\pi}{4}\)