Bài 2: Tích phân

Em biến đổi dx thành d(2x+1) sau đó dùng tích phân cơ bản là ok

Tính \(I=\int_0^{\dfrac{\pi}{2}}\dfrac{cos^{2017}x}{sin^{2017}x+cos^{2017}}dx\left(1\right)\)

Đặt \(t=cosx\Rightarrow sinx=\sqrt{1-cos^2x}\)

\(\Rightarrow dt=-sinx.dx\)

\(\Rightarrow I=\int_0^1\dfrac{t^{2017}.}{\sqrt{1-t^2}.\left(\left(\sqrt{1-t^2}\right)^{2017}+t^{2017}\right)}dt\)

Đặt: \(t=siny\Rightarrow\sqrt{1-t^2}=cosy\)

\(\Rightarrow dt=cosy.dy\)

\(\Rightarrow I=\int_0^{\dfrac{\pi}{2}}\dfrac{sin^{2017}y.cosy}{cosy\left(cos^{2017}y+sin^{2017}y\right)}dy=\int_0^{\dfrac{\pi}{2}}\dfrac{sin^{2017}y}{\left(cos^{2017}y+sin^{2017}y\right)}\)

\(\Rightarrow I=\int_0^{\dfrac{\pi}{2}}\dfrac{sin^{2017}x}{\left(cos^{2017}x+sin^{2017}x\right)}\left(2\right)\)

Cộng (1) và (2) ta được

\(2I=\int_0^{\dfrac{\pi}{2}}\dfrac{sin^{2017}x+cos^{2017}x}{sin^{2017}x+cos^{2017}x}dx=\int_0^{\dfrac{\pi}{2}}1dx\)

\(=x|^{\dfrac{\pi}{2}}_0=\dfrac{\pi}{2}\)

\(\Rightarrow I=\dfrac{\pi}{4}\)

Thế lại bài toán ta được

\(\dfrac{\pi}{4}+t^2-6t+9-\dfrac{\pi}{4}=0\)

\(\Leftrightarrow t^2-6t+9=0\)

\(\Leftrightarrow t=3\)

Chọn đáp án C

^{_{\(\left\{{}\begin{matrix}u=1+x\\dv=c\Leftrightarrow os\left(2x\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{2}sin\left(2x\right)\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}sin\left(2x\right)\left(1+x\right)|^{\dfrac{\pi}{4}}_0-\int\limits^{\dfrac{\pi}{4}}_0\dfrac{1}{2}sin\left(2x\right)dx\)}}

tới đây bạn tự giải được chứ

đáp án là:

\(\dfrac{3√2-ln(1+√2)}{8}\)

Nhưng cách tính thì em ko biết! Mong mọi người giúp đỡ ạ!

Bài 3.9:

a)

\(\int ^{1}_{0}(y^3+3y^2-2)dy=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{y^4}{4}+y^3-2y \right )=\frac{-3}{4}\)

b) \(\int ^{4}_{1}\left (t+\frac{1}{\sqrt{t}}-\frac{1}{t^2}\right)dt=\left.\begin{matrix} 4\\ 1\end{matrix}\right|\left ( \frac{t^2}{2}+2\sqrt{t}+\frac{1}{t} \right )=\frac{35}{4}\)

d) Ta có:

\(\int ^{1}_{0}(3^s-2^s)^2ds=\int ^{1}_{0}(9^s+4^s-2.6^s)ds=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{9^s}{\ln 9}+\frac{4^s}{\ln 4}-\frac{2.6^s}{\ln 6} \right )\)

\(=\frac{8}{\ln 9}+\frac{3}{\ln 4}-\frac{10}{\ln 6}\)

h)

Ta có \(\int ^{\frac{5\pi}{4}}_{\pi}\frac{\sin x-\cos x}{\sqrt{1+\sin 2x}}dx=\int ^{\frac{5\pi}{4}}_{\pi}\frac{\sin x-\cos x}{\sqrt{\sin^2x+\cos^2x+2\sin x\cos x}}dx\)

\(=\int ^{\frac{5\pi}{4}}_{\pi}\frac{-d(\sin x+\cos x)}{|\sin x+\cos x|}=\int ^{\frac{5\pi}{4}}_{\pi}\frac{d(\sin x+\cos x)}{\sin x+\cos x}=\left.\begin{matrix} \frac{5\pi}{4}\\ \pi\end{matrix}\right|\ln |\sin x+\cos x|=\ln (\sqrt{2})\)

Bài 3.10:

a)

Đặt \(t=1-x\) thì:

\(\int ^{2}_{1}x(1-x)^5dx=\int ^{-1}_{0}t^5(1-t)d(1-t)=\int ^{0}_{-1}t^5(1-t)dt\)

\(=\left.\begin{matrix} 0\\ -1\end{matrix}\right|\left ( \frac{t^6}{6}-\frac{t^7}{7} \right )=\frac{-13}{42}\)

b) Đặt \(\sqrt{e^x-1}=t\) \(\Rightarrow x=\ln (t^2+1)\)

Khi đó

\(\int ^{\ln 2}_{0}\sqrt{e^x-1}dx=\int ^{1}_{0}td(\ln (t^2+1))=\int ^{1}_{0}t.\frac{2t}{t^2+1}dt\)

\(=\int ^{1}_{0}\frac{2t^2}{t^2+1}dt=\int ^{1}_{0}2dt-\int ^{1}_{0}\frac{2}{t^2+1}dt=\left.\begin{matrix} 1\\ 0\end{matrix}\right|2t-\int ^{1}_{0}\frac{2dt}{t^2+1}=2-\int ^{1}_{0}\frac{2dt}{t^2+1}\)

Với \(\int ^{1}_{0}\frac{2dt}{t^2+1}\), đặt \(t=\tan m\)

\(\Rightarrow \int ^{1}_{0}\frac{2dt}{t^2+1}=\int ^{\frac{\pi}{4}}_{0}\frac{2d(\tan m)}{\tan ^2m+1}=\int ^{\frac{\pi}{4}}_{0}2\cos ^2md(\tan m)\)

\(=\int ^{\frac{\pi}{4}}_{0}2dm=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|2m=\frac{\pi}{2}\)

Do đó \(\int ^{\ln 2}_{0}\sqrt{e^x-1}dx=2-\frac{\pi}{2}\)

Bài 3.10

c) Đặt \(t=\sqrt[3]{1-x}\Rightarrow x=1-t^3\)

\(\int ^{9}_{1}x\sqrt[3]{1-x}dx=\int ^{-2}_{0}(1-t^3)td(1-t^3)=\int ^{0}_{-2}3t^2.t(1-t^3)dt\)

\(\left.\begin{matrix} 0\\ -2\end{matrix}\right|\left ( \frac{3t^4}{4}-\frac{3t^7}{7} \right )=\frac{-468}{7}\)

e) Đặt \(t=\frac{1}{x}\) suy ra:

\(\int ^{2}_{1}\frac{\sqrt{x^2+1}}{x^4}dx=\int ^{\frac{1}{2}}_{1}t^4\sqrt{\frac{1}{t^2}+1}d\left(\frac{1}{t}\right)\)

\(=\int ^{1}_{\frac{1}{2}}\frac{t^4\sqrt{t^2+1}}{t}.\frac{1}{t^2}dt=\int ^{1}_{\frac{1}{2}}t\sqrt{t^2+1}dt=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\sqrt{t^2+1}d(t^2+1)\)

\(=\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left(\frac{1}{2}.\frac{2}{3}\sqrt{(t^2+1)^3}\right)=\frac{2\sqrt{2}}{3}-\frac{5\sqrt{5}}{24}\)

Lời giải:

a) Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos 2xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\frac{\sin 2x}{2}\end{matrix}\right.\)

\(\Rightarrow \int x\cos 2xdx=\frac{x\sin 2x}{2}-\int \frac{\sin 2x}{2}dx=\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}\)

\(\Rightarrow \int ^{\frac{\pi}{2}}_{0}x\cos 2xdx=\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\left ( \frac{x\sin 2x}{2}+\frac{\cos 2x}{4} \right )=\frac{-1}{2}\)

b) Đặt \(\left\{\begin{matrix} u=x\\ dv=e^{-2x}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\frac{-e^{-2x}}{2}\end{matrix}\right.\)

\(\Rightarrow \int xe^{-2x}dx=\frac{-xe^{-2x}}{2}+\int \frac{e^{-2x}}{2}dx=\frac{-xe^{-2x}}{2}-\frac{e^{-2x}}{4}\)

\(\Rightarrow \int ^{\ln 2}_{0}xe^{-2x}dx=\left.\begin{matrix} \ln 2\\ 0\end{matrix}\right|\left ( \frac{-xe^{-2x}}{2}-\frac{e^{2x}}{4} \right )=\frac{3}{16}-\frac{\ln 2}{8}\)

c)

\(\int ^{1}_{0}\ln (2x+1)dx=\frac{1}{2}\int ^{1}_{0}\ln (2x+1)d(2x+1)=\frac{1}{2}\int ^{3}_{1}\ln tdt\)

Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t\)

Do đó \(\frac{1}{2}\int ^{3}_{1}\ln tdt=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\left(\frac{t\ln t-t}{2}\right)=\frac{3\ln 3}{2}-1\)

d)

Ta có \(\int ^{3}_{2}(\ln (x-1)-\ln (x+1))dx=\int ^{3}_{2}\ln (x-1)d(x-1)-\int ^{3}_{2}\ln (x+1)d(x+1)\)

\(=\int ^{2}_{1}\ln tdt-\int ^{4}_{3}\ln tdt\)

Theo phần c, ta đã chỉ ra được \(\int \ln tdt=t\ln t-t\), do đó:

\(\int ^{2}_{1}\ln tdt-\int ^{4}_{3}\ln tdt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|(t\ln t-t)-\left.\begin{matrix} 4\\ 3\end{matrix}\right|(t\ln t-t)=\ln \left(\frac{27}{64}\right)\)

e)

Xét \(\int (x+1-\frac{1}{x})e^{x+\frac{1}{x}}dx=\int e^{x+\frac{1}{x}}dx+\int \left (x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx\)

Đặt \(\left\{\begin{matrix} u=e^{x+\frac{1}{x}}\\ dv=dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}dx\\ v=x\end{matrix}\right.\)

\(\Rightarrow \int e^{x+\frac{1}{x}}dx=xe^{x+\frac{1}{x}}-\int \left(x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx\)

Do đó \(\int \left(x+1-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx=xe^{x+\frac{1}{x}}\)

\(\int ^{2}_{\frac{1}{2}}\left(x+1-\frac{x}{x}\right)e^{x+\frac{1}{x}}dx=\left.\begin{matrix} 2\\ \frac{1}{2}\end{matrix}\right|xe^{x+\frac{1}{x}}=\frac{3e^{\frac{5}{2}}}{2}\)

g)

Có \(\int x\cos x\sin^2xdx=\frac{1}{2}\int x\sin 2x\sin xdx=\frac{1}{4}\int x(\cos x-\cos 3x)dx\)

Xét \( \int x\cos xdx\). Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\sin x\end{matrix}\right.\)

\(\Rightarrow \int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x\)

Tương tự, \(\int x\cos 3xdx=\frac{x\sin 3x}{3}+\frac{\cos 3x}{9}\)

Do đó, \(\int ^{\frac{\pi}{2}}_{0}x\cos x\sin^2xdx=\int ^{\frac{\pi}{2}}_{0}\frac{1}{4}x(\cos x-\cos 3x)dx\)

\(=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|(x\sin x+\cos x-\frac{x\sin 3x}{3}-\frac{\cos 3x}{9})=\frac{3\pi-4}{18}\)

h)

\(\left\{\begin{matrix} u=xe^x\\ dv=\frac{1}{(x+1)^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=e^x(x+1)dx\\ v=\frac{-1}{x+1}\end{matrix}\right.\)

\(\Rightarrow \int \frac{xe^x}{(x+1)^2}dx=\frac{-xe^x}{x+1}+\int e^xdx=\frac{-xe^x}{x+1}+e^x=\frac{e^x}{x+1}\)

Do đó \(\int ^{1}_{0}\frac{xe^x}{(x+1)^2}dx=\frac{e}{2}-1\)

i)

Có \(\int \frac{1+x\ln x}{x}e^xdx=\int \frac{e^x}{x}dx+\int \ln xe^xdx\)

Đặt \(\left\{\begin{matrix} u=\ln x\\ dv=e^xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dx}{x}\\ v=e^x\end{matrix}\right.\Rightarrow \int \ln xe^xdx=\ln xe^x-\int \frac{e^x}{x}dx\)

\(\Rightarrow \int ^{e}_{1}\frac{1+x\ln x}{x}e^xdx=\left.\begin{matrix} e\\ 1\end{matrix}\right|\ln xe^x=e^e\)

Lời giải:

Đặt \(I=\int \frac{\sqrt{x^2-1}dx}{x^3}\)

Nguyên hàm từng phần:

Đặt \(\left\{\begin{matrix} u=\sqrt{x^2-1}\\ dv=\frac{1}{x^3}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{x}{\sqrt{x^2-1}}dx\\ v=\frac{-1}{2x^2}\end{matrix}\right.\)

\(\Rightarrow I=\frac{-\sqrt{x^2-1}}{2x^2}+\int \frac{dx}{x\sqrt{x^2-1}}\)

Xét \(\int \frac{dx}{x\sqrt{x^2-1}}=\int \frac{d(x^2)}{2x^2\sqrt{x^2-1}}\). Đặt \(\sqrt{x^2-1}=t\rightarrow x^2=t^2+1\)

Khi đó, \(\int \frac{dx}{x\sqrt{x^2-1}}=\int \frac{d(t^2+1)}{2t(t^2+1)}=\int \frac{dt}{t^2+1}\)

Đặt \(t=\tan m\), đây là một dạng toán đặt quen thuộc, ta thu

được \(\int \frac{dx}{x\sqrt{x^2-1}}=\int \frac{dt}{t^2+1}=m=\tan ^{-1}t=\tan ^{-1}(\sqrt{x^2-1})\)

Do đó, \(\int \frac{\sqrt{x^2-1}dx}{x^3}=\frac{-\sqrt{x^2-1}}{2x^2}+\frac{1}{2}\tan ^{-1}(\sqrt{x^2-1})\)

\(\Rightarrow \int ^{\sqrt{2}}_{1}\frac{\sqrt{x^2-1}}{x^3}dx=\frac{\pi}{8}-\frac{1}{4}\)