Oblicz całki oznaczone: a) ſ xe^(x^2) dx, [2,0] b) ſ cos3xdx [π/3, -2] c) ſ sqrt{3x-5}/ x-2; [7,3].... Question from @Justys9630

January 2019 1 11 Report

Oblicz całki oznaczone:
a) ſ xe^(x^2) dx, [2,0]
b) ſ cos3xdx [π/3, -2]
c) ſ sqrt{3x-5}/ x-2; [7,3]

loitzl9006 $\int x\cdot e^{x^2}dx \ \ \ \ \ |x^2=t, \ 2xdx=dt, \ xdx=\frac12dt|=\\=\int \frac12e^t\cdot dt=\frac12e^t+C=\frac12e^{x^2}+C\\ \int_2^0x\cdot e^{x^2}dx=\frac12e^{x^2}\biggl|^0_2=\frac12e^0-\frac12e^4=1-\frac12e^4$

$\int \cos3xdx = \ \ \ \ \ \ |3x=t, \ \ 3dx=dt, \ \ dx=\frac13dt|\\=\int \frac13\cos tdt=\frac13\sin t+C=\frac13\sin3x+C\\ \int_{\frac{\pi}3}^{-2}\cos3xdx=\frac13\sin3x\biggl|^{-2}_{\frac{\pi}3}=\frac13\sin(-6)-\frac13\sin\pi=-\frac13\sin6-0=\\=-\frac13\sin6$

$\int\frac{\sqrt{3x-5}}{x-2}dx= \ \ \ \ \ |\sqrt{3x-5}=t, \ \ 3x-5=t^2\ \to \ x=\frac{t^2+5}3\ \to \ \\ \to\ dx=\frac{2t}3dt\\ \int\frac{\sqrt{3x-5}}{x-2}dx=\int\frac{t}{\frac{t^2+5}3-2}\cdot\frac{2t}3dt=\int\frac{2t^2}{3(\frac{t^2+5}3-\frac63)}dt=\int\frac{2t^2}{3\cdot\frac{t^2-1}3}dt=\\=\int\frac{2t^2}{t^2-1}dt=2\int\frac{t^2}{t^2-1}dt=2\int\frac{t^2-1+1}{t^2-1}dt=2\int\frac{t^2-1}{t^2-1}dt+2\cdot\int\frac1{t^2-1}dt=\\=2\cdot\int1dt+2\cdot\int\frac1{t^2-1}dt=2t+2\int\frac1{t^2-1}dt$

$\frac1{t^2-1}=\frac1{(t-1)(t+1)}=\frac A{t-1}+\frac B{t+1} \ \ \ \ \ |\cdot(t-1)(t+1)\\ 1=A(t+1)+B(t-1)\\ 1=At+A+Bt-B\\ 0t+1=(A+B)t+A-B\\ \left \{ {{0=A+B} \atop {1=A-B}} \right. \\ 0+1=2A\ \to \ A=\frac12\ \to \ B=-\frac12\\ \frac1{t^2-1}=\frac1{2(t-1)}-\frac1{2(t+1)}\\ \int\frac1{t^2-1}dt=\frac12\ln|t-1|-\frac12\ln|t+1|+C=\frac12\ln|\frac{t-1}{t+1}|+C\\ \int\frac{t^2}{t^2-1}dt=2t+\ln|\frac{t-1}{t+1}|+C=2\sqrt{3x-5}+\ln|\frac{t+1-2}{t+1}|+C=$
$=2\sqrt{3x-5}+\ln|1-\frac2{t+1}|+C=2\sqrt{3x-5}+\ln|1-\frac2{\sqrt{3x-5}+1}|+C\\ \int^7_3\frac{\sqrt{3x-5}}{x-2}dx=2\sqrt{3x-5}+\ln|1-\frac2{\sqrt{3x-5}+1}|\biggl|^7_3=\\=2\sqrt{3\cdot7-5}+\ln|1-\frac2{\sqrt{3\cdot7-5}+1}|-2\sqrt{3\cdot3-5}-\ln|1-\frac2{\sqrt{3\cdot3-5}+1}|=\\=2\sqrt{16}+\ln|1-\frac2{\sqrt{16}+1}|-2\sqrt4-\ln|1-\frac2{\sqrt4+1}|=\\=2\cdot4+\ln|1-\frac25|-2\cdot2-\ln|1-\frac23|=\\=8+\ln\frac35-4-\ln\frac13=4+\ln3-\ln5-\ln1+\ln3=4+2\ln3-\ln5$

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