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Bạn xem lại xem có type thiếu đề không? \((x+\frac{\pi}{6})\) có sin hay cos, tan ở phía trước không?
a)
Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)
\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)
\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)
b)
\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)
\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)
c)
Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).
Đặt \(x+1=t\)
\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
\(I=\int\dfrac{x^3dx}{\left(x^8-4\right)^2}\)
Đặt \(x^4=t\Rightarrow x^3dx=\dfrac{1}{4}dt\Rightarrow I=\dfrac{1}{4}\int\dfrac{dt}{\left(t^2-2\right)^2}=\dfrac{1}{4}\int\dfrac{dt}{\left(t-\sqrt{2}\right)^2\left(t+\sqrt{2}\right)^2}\)
\(=\dfrac{1}{32}\int\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)^2dt=\dfrac{1}{32}\int\left(\dfrac{1}{\left(t-\sqrt{2}\right)^2}+\dfrac{1}{\left(t+\sqrt{2}\right)^2}-\dfrac{2}{\left(t+\sqrt{2}\right)\left(t-\sqrt{2}\right)}\right)dt\)
\(=\dfrac{1}{32}\int\left(\dfrac{1}{\left(t-\sqrt{2}\right)^2}+\dfrac{1}{\left(t+\sqrt{2}\right)^2}-\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)\right)dt\)
\(=\dfrac{1}{32}\left(\dfrac{-1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}-\dfrac{1}{\sqrt{2}}ln\left|\dfrac{t-\sqrt{2}}{t+\sqrt{2}}\right|\right)+C\)
\(=\dfrac{1}{32}\left(\dfrac{-1}{x^4-\sqrt{2}}-\dfrac{1}{x^4+\sqrt{2}}-\dfrac{1}{\sqrt{2}}ln\left|\dfrac{x^4-\sqrt{2}}{x^4+\sqrt{2}}\right|\right)+C\)
2/ \(I=\int\dfrac{\left(2x+1\right)dx}{\left(x^2+x-1\right)\left(x^2+x+3\right)}=\dfrac{1}{4}\int\left(\dfrac{1}{x^2+x-1}-\dfrac{1}{x^2+x+3}\right)\left(2x+1\right)dx\)
\(=\dfrac{1}{4}\int\left(\dfrac{2x+1}{x^2+x-1}-\dfrac{2x+1}{x^2+x+3}\right)dx\)
\(=\dfrac{1}{4}\left(\int\dfrac{d\left(x^2+x-1\right)}{x^2+x-1}-\int\dfrac{d\left(x^2+x+3\right)}{x^2+x+3}\right)\)
\(=\dfrac{1}{4}ln\left|\dfrac{x^2+x-1}{x^2+x+3}\right|+C\)
3/ Đặt \(\sqrt[3]{x}=t\Rightarrow x=t^3\Rightarrow dx=3t^2dt\)
\(\Rightarrow I=\int\dfrac{3t^2.sint.dt}{t^2}=3\int sint.dt=-3cost+C=-3cos\left(\sqrt[3]{x}\right)+C\)
4/ \(I=\int\dfrac{dx}{1+cos^2x}=\int\dfrac{\dfrac{1}{cos^2x}dx}{\dfrac{1}{cos^2x}+1}\)
Đặt \(t=tanx\Rightarrow\left\{{}\begin{matrix}dt=\dfrac{1}{cos^2x}dx\\\dfrac{1}{cos^2x}=1+tan^2x=1+t^2\end{matrix}\right.\)
\(\Rightarrow I=\int\dfrac{dt}{1+t^2+1}=\int\dfrac{dt}{t^2+2}=\dfrac{1}{2}\int\dfrac{dt}{\left(\dfrac{t}{\sqrt{2}}\right)^2+1}\)
\(=\dfrac{1}{2}.\sqrt{2}.arctan\left(\dfrac{t}{\sqrt{2}}\right)+C=\dfrac{1}{\sqrt{2}}arctan\left(\dfrac{tanx}{\sqrt{2}}\right)+C\)
5/ \(I=\int\dfrac{sinx+cosx}{4+2sinx.cosx-sin^2x-cos^2x}dx=\int\dfrac{sinx+cosx}{4-\left(sinx-cosx\right)^2}dx\)
Đặt \(sinx-cosx=t\Rightarrow\left(cosx+sinx\right)dx=dt\)
\(\Rightarrow I=\int\dfrac{dt}{4-t^2}=-\int\dfrac{dt}{\left(t-2\right)\left(t+2\right)}=\dfrac{1}{4}\int\left(\dfrac{1}{t+2}-\dfrac{1}{t-2}\right)dt\)
\(=\dfrac{1}{4}ln\left|\dfrac{t+2}{t-2}\right|+C=\dfrac{1}{4}ln\left|\dfrac{sinx-cosx+2}{sinx-cosx-2}\right|+C\)
Ơ bài 1 nhầm số 4 thành số 2 rồi, bạn sửa lại 1 chút nhé :D
Còn 1 cách làm khác nữa là lượng giác hóa
Đặt \(x^4=2sint\Rightarrow x^3dx=\dfrac{1}{2}cost.dt\)
\(\Rightarrow I=\dfrac{1}{2}\int\dfrac{cost.dt}{\left(4sin^2t-4\right)^2}=\dfrac{1}{32}\int\dfrac{cost.dt}{cos^4t}=\dfrac{1}{32}\int\dfrac{dt}{cos^3t}\)
Đặt \(\left\{{}\begin{matrix}u=\dfrac{1}{cost}\\dv=\dfrac{dt}{cos^2t}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{sint.dt}{cos^2t}\\v=tant\end{matrix}\right.\)
\(\Rightarrow32I=\dfrac{tant}{cost}-\int\dfrac{tant.sint.dt}{cos^2t}=\dfrac{sint}{cos^2t}-\int\dfrac{sin^2t.dt}{cos^3t}\)
\(=\dfrac{sint}{1-sin^2t}-\int\dfrac{1-cos^2t}{cos^3t}dt=\dfrac{sint}{1-sin^2t}-\int\dfrac{dt}{cos^3t}+\int\dfrac{1}{cosx}dx\)
Chú ý rằng \(\int\dfrac{dt}{cos^3t}=32I\)
\(\Rightarrow32I=\dfrac{sint}{1-sin^2t}-32I+\int\dfrac{cost.dt}{cos^2t}\)
\(\Rightarrow64I=\dfrac{sint}{1-sin^2t}-\int\dfrac{d\left(sint\right)}{sin^2t-1}=\dfrac{sint}{1-sin^2t}-\dfrac{1}{2}ln\left|\dfrac{sint-1}{sint+1}\right|+C\)
\(\Rightarrow I=\dfrac{1}{64}\left(\dfrac{2x^4}{4-x^8}-\dfrac{1}{2}ln\left|\dfrac{x^4-2}{x^4+2}\right|\right)+C\)
\(a,\int sin2x.cosxdx=\int\dfrac{1}{2}\left[sin3x+sinx\right]dx=\dfrac{1}{2}\int sin3xdx+\dfrac{1}{2}\int sinxdx=\dfrac{-1}{6}cos3x-\dfrac{1}{2}cosx\)
\(I=\int\dfrac{x}{1-cos2x}dx=\int\dfrac{x}{2sin^2x}dx\)
Đặt \(\left\{{}\begin{matrix}u=\dfrac{x}{2}\\dv=\dfrac{1}{sin^2x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{2}\\v=-cotx\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int cotxdx=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int\dfrac{cosx.dx}{sinx}\)
\(=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int\dfrac{d\left(sinx\right)}{sinx}=\dfrac{-x.cotx}{2}+\dfrac{1}{2}ln\left|sinx\right|+C\)
2/ Câu 2 bữa trước làm rồi, bạn coi lại nhé
3/ \(I=\int\left(2x+1\right)ln^2xdx\)
Đặt \(\left\{{}\begin{matrix}u=ln^2x\\dv=\left(2x+1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{2lnx}{x}dx\\v=x^2+x\end{matrix}\right.\)
\(\Rightarrow I=\left(x^2+x\right)ln^2x-\int\left(2x+2\right)lnxdx=\left(x^2+x\right)ln^2x-I_1\)
\(I_1=\int\left(2x+2\right)lnx.dx\) \(\Rightarrow\left\{{}\begin{matrix}u=lnx\\dv=\left(2x+2\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x^2+2x\end{matrix}\right.\)
\(\Rightarrow I_1=\left(x^2+2x\right)lnx-\int\left(x+2\right)dx=\left(x^2+2x\right)ln-\dfrac{x^2}{2}+2x+C\)
\(\Rightarrow I=\left(x^2+x\right)ln^2x-\left(x^2+2x\right)lnx+\dfrac{x^2}{2}-2x+C\)
4/ \(I=\int\left(2x-1\right)cosx.dx\) \(\Rightarrow\left\{{}\begin{matrix}u=2x-1\\dv=cosx.dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=sinx\end{matrix}\right.\)
\(\Rightarrow I=\left(2x-1\right)sinx-2\int sinx.dx=\left(2x-1\right)sinx+2cosx+C\)
5/ \(I=\int\left(x^2+x+1\right)e^xdx\) \(\Rightarrow\left\{{}\begin{matrix}u=x^2+x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\left(2x+1\right)dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=\left(x^2+x+1\right)e^x-\int\left(2x+1\right)e^xdx\)
\(I_1=\int\left(2x+1\right)e^xdx\) \(\Rightarrow\left\{{}\begin{matrix}u=2x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I_1=\left(2x+1\right)e^x-2\int e^xdx=\left(2x+1\right)e^x-2e^x+C=\left(2x-1\right)e^x+C\)
\(\Rightarrow I=\left(x^2+x+1\right)e^x-\left(2x-1\right)e^x+C=\left(x^2-x+2\right)e^x+C\)
6/ \(I=\int\left(2x+1\right).ln\left(x+2\right)dx\)
\(\Rightarrow\left\{{}\begin{matrix}u=ln\left(x+2\right)\\dv=\left(2x+1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x+2}\\v=x^2+x\end{matrix}\right.\)
\(\Rightarrow I=\left(x^2+x\right)ln\left(x+2\right)-\int\dfrac{x^2+x}{x+2}dx\)
\(=\left(x^2+x\right)ln\left(x+2\right)-\int\left(x-1+\dfrac{2}{x+2}\right)dx\)
\(I=\left(x^2+x\right)ln\left(x+2\right)-\dfrac{x^2}{2}+x-2ln\left|x+2\right|+C\)
Nhớ quy tắc ưu tiên khi tính nguyên hàm từng phần:
- Đặt u sẽ ưu tiên các hàm ln, log đầu tiên (luôn luôn đặt các hàm này là u nếu có mặt), sau đó đến các hàm đa thức P(x), sau đó là lượng giác hoặc e^
- Đặt dv thì theo thứ tự ngược lại, ưu tiên đặt lượng giác (sin, cos) và e^
Câu a)
\(I=\int ^{1}_{0}\frac{x(e^x+1)+1}{e^x+1}dx=\int ^{1}_{0}xdx+\int ^{1}_{0}\frac{dx}{e^x+1}\)
\(=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2}{2}+\int ^{1}_{0}\frac{d(e^x)}{e^x(e^x+1)}=\frac{1}{2}+\left.\begin{matrix} 1\\ 0\end{matrix}\right|\ln\left | \frac{e^x}{e^x+1} \right |\)
\(\Leftrightarrow I=\frac{3}{2}+\ln 2-\ln (e+1)\)
Câu d)
\(I=\int ^{e}_{1}\ln(x+1)d(x)=\int ^{e}_{1}\ln (x+1)d(x+1)\)
Đặt \(\left\{\begin{matrix} u=\ln (x+1)\\ dv=d(x+1)\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{d(x+1)}{x+1}\\ v=x+1\end{matrix}\right.\)
\(\Rightarrow I=\left.\begin{matrix} e\\ 1\end{matrix}\right|(x+1)\ln (x+1)-\int ^{e}_{1}d(x+1)\)
\(=(e+1)\ln \left ( \frac{e+1}{e} \right )-2\ln \left (\frac{2}{e}\right )\)
Câu b)
Đặt \(\tan \frac{x}{2}=t\). Ta có:
\(\left\{\begin{matrix} dt=d\left ( \tan \frac{x}{2} \right )=\frac{1}{2\cos ^2\frac{x}{2}}dx=\frac{t^2+1}{2}dx\rightarrow dx=\frac{2dt}{t^2+1}\\\ \cos x=\frac{1-t^2}{t^2+1}\end{matrix}\right.\)
\( I=\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{1}{1+\cos x}dx}_{A}+\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{d(\cos x)}{\cos x+1}}_{B}\)
Có \(B=\int ^{\frac{\pi}{2}}_{0}\frac{d(\cos x+1)}{\cos x+1}=\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\ln |\cos x+1|=-\ln 2\)
\(A=\int ^{1}_{0}\frac{2dt}{(t^2+1)\frac{2}{t^2+1}}=\int ^{1}_{0}dt=1\)
\(\Rightarrow I=A+B=1-\ln 2\)
Hi
... ;-;