Bài 2: Tích phân

Đặt \(ln\left(x^2+1\right)=t\Rightarrow\dfrac{x}{x^2+1}dx=\dfrac{1}{2}dt\)

\(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=\sqrt{e^{23}-1}\Rightarrow t=23\end{matrix}\right.\)

\(I=\dfrac{1}{2}\int\limits^{23}_0f\left(t\right)dt=\dfrac{1}{2}\int\limits^{23}_0f\left(x\right)dx=1\)

Đặt \(\sqrt{5-x}=t\Rightarrow5-x=t^2\Rightarrow dx=-2t.dt\)

\(\left\{{}\begin{matrix}x=-4\Rightarrow t=3\\x=4\Rightarrow t=1\end{matrix}\right.\)

\(I=\int\limits^1_3\dfrac{-2t.dt}{1+t}=\int\limits^3_1\left(2-\dfrac{2}{t+1}\right)dt=\left(2t-2ln\left|t+1\right|\right)|^3_1=4-2ln2\)

\(\Rightarrow ab=8\)

Giao điểm của 2 pt đường tròn \(\left(\dfrac{\sqrt{3}}{2};\dfrac{1}{2}\right)\)

\(\left\{{}\begin{matrix}x=r\cos\varphi\\y=r\sin\varphi\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}r\in\left[2\sin\varphi;1\right]\\\varphi\in\left[0;\dfrac{\pi}{6}\right]\\\left|J\right|=r\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^{\dfrac{\pi}{6}}_0d\varphi\int\limits^1_{2\sin\varphi}\sqrt{r^2}.rdr=\int\limits^{\dfrac{\pi}{6}}_0d\varphi\int\limits^1_{2\sin\varphi}r^2dr=\dfrac{1}{3}\int\limits^{\dfrac{\pi}{6}}_0\left(1-8\sin^3\varphi\right)d\varphi=...\)

Ta có: 2 ∫ 1 [2 + f(x)] dx = 2 ∫ 1 [2 + x^2] dx = 2 [2x + (1/3)x^3]1→1 = 2 [(2+1/3) - (2+1/3)] = 0 Vậy đáp án là D. 7/3.

\(\int\limits^3_1f\left(x\right)dx=-2+1=-1\)

Đặt \(2x+2=u\Rightarrow2xdx=du\Rightarrow dx=\dfrac{1}{2}du\)

\(\left\{{}\begin{matrix}x=0\Rightarrow u=2\\x=2\Rightarrow u=6\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^6_2f\left(u\right).\dfrac{1}{2}du=\dfrac{1}{2}\int\limits^6_2f\left(u\right)du=\dfrac{1}{2}\int\limits^6_2f\left(x\right)dx=\dfrac{1}{2}.6=3\)

\(I=\int\limits^e_1xlnxdx+\int\limits^e_1\dfrac{lnx}{x}dx=I_1+I_2\)

Xét \(I_1\) , đặt \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{x^2}{2}\end{matrix}\right.\)

\(\Rightarrow I_1=\dfrac{x^2}{2}lnx|^e_1-\int\limits^e_1\dfrac{x}{2}=\dfrac{e^2}{2}-\dfrac{e}{2}+\dfrac{1}{2}\)

Xét \(I_2=\int\limits^e_1\dfrac{lnx}{x}dx=\int\limits^e_1lnx.d\left(lnx\right)=\dfrac{ln^2x}{2}|^e_1=\dfrac{1}{2}\)

\(\Rightarrow I=\dfrac{e^2}{2}-\dfrac{e}{2}+1\)

Từ GT ta lấy tích phân 2 vế cận từ 0 đến 1 ; sẽ được : 

\(\int\limits^1_0f\left(x+1\right)dx+\int\limits^1_03f\left(3x+2\right)dx-\int\limits^1_04f\left(4x+1\right)dx-\int\limits^1_0f\left(2^x\right)dx=\int\limits^1_0\dfrac{3dx}{\sqrt{x+1}+\sqrt{x+2}}\left(1\right)\)

\(\int\limits^1_0\dfrac{3dx}{\sqrt{x+1}+\sqrt{x+2}}=\int\limits^1_03\left(\sqrt{x+2}-\sqrt{x+1}\right)dx\)  = 

\(2\left[\left(x+2\right)\sqrt{x+2}-\left(x+1\right)\sqrt{x+1}\right]\dfrac{1}{0}\)  = \(2+6\sqrt{3}-8\sqrt{2}\left(2\right)\)

Dễ thấy : \(\int\limits^1_0f\left(x+1\right)dx=\int\limits^2_1f\left(t\right)dt=\int\limits^2_1f\left(x\right)dx\)

\(\int\limits^1_03f\left(3x+2\right)dx=\int\limits^5_2f\left(t\right)dt=\int\limits^5_2f\left(x\right)dx\)  (3)

\(\int\limits^1_04f\left(4x+1\right)=\int\limits^5_1f\left(t\right)dt=\int\limits^5_1f\left(x\right)dx\left(4\right)\)

\(\int\limits^1_0f\left(2^x\right)dx=\int\limits^2_1\dfrac{f\left(t\right)dt}{tln2}=\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(t\right)dt}{t}=\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(x\right)dx}{x}\)  (5)

Thay (2) ; (3) ; (4) ; (5) vào (1) ta được : 

\(\int\limits^2_1f\left(x\right)dx+\int\limits^5_2f\left(x\right)dx-\int\limits^5_1f\left(x\right)dx-\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(x\right)dx}{x}=2+6\sqrt{3}-8\sqrt{2}\)

\(\Leftrightarrow\int\limits^2_1\dfrac{f\left(x\right)dx}{x}=\left(2+6\sqrt{3}-8\sqrt{2}\right)ln2\)

Lời giải:

\(\int ^{\frac{\pi}{2}}_{0}\frac{\sin 2x\cos x}{1+\cos x}dx=\int ^{\frac{\pi}{2}}_{0}\frac{2\sin x\cos ^2x}{\cos x+1}dx=2\int ^{\frac{\pi}{2}}_{0}\frac{\cos^2x\sin xdx}{\cos x+1}\)

\(=2\int ^{\frac{\pi}{2}}_{0}\frac{-\cos ^2xd(\cos x)}{\cos x+1}=2\int ^{0}_{1}\frac{-t^2dt}{t+1}=2\int ^{1}_{0}\frac{t^2}{t+1}dt\)

\(=2\int^1_0\frac{(t^2-1)+1}{t+1}dt=2\int ^1_0(t-1+\frac{1}{t+1})dt\)

\(=2(\frac{t^2}{2}-t+\ln|t+1|)|^{1}_0=2\ln 2-1\)