Question

Найти неопределенные интегралы
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Answer 1

$1)\; \; \int \sqrt{3x^2+4x}\cdot dx=\Big [\; 3x^2+4x=3\cdot (x^2+\frac{4}{3}x)=3\cdot ((x+\frac{2}{3})^2-\frac{4}{9})\; \Big ]=\\\\=\sqrt3\cdot \int \sqrt{(x+\frac{2}{3})^2-\frac{4}{9}}\cdot dx=\Big [\; t=x+\frac{2}{3}\; ,\; \; dt=dx\; \Big ]=\sqrt3\cdot \int \sqrt{t^2-\frac{4}{9}}\cdot dt=\\\\=\Big [\; t=\frac{2}{3\, cosz}\; ,\; \; dt=\frac{2\cdot sinz\cdot dz}{3\cdot cos^2z}\; ,\; \; t^2-\frac{4}{9}=\frac{4}{9\cdot cos^2z}-\frac{4}{9}=\frac{4}{9}\cdot (\frac{1}{cos^2z}-1)=$

$=\frac{4}{9}\cdot tg^2z\; ]=\sqrt3\cdot \int \sqrt{\frac{4}{9}\cdot tg^2z}\cdot \frac{2\, sinz\cdot dz}{cos^2z}=\sqrt3\cdot \frac{2}{3}\cdot 2\cdot \int \frac{tgz\cdot sinz\cdot dz}{cos^2z}=\\\\=\frac{4}{\sqrt3}\cdot \int \frac{sinz\cdot sinz\cdot dz}{cos^3z}=\Big [\; u=sinz\; ,\; du=cosz\, dz\; ,\; dv=\frac{sinz\, dz}{cos^3z}\; ,\; v=\frac{1}{2cos^2z}\, ]=\\\\=\frac{4}{\sqrt3}\cdot \Big (sinz\cdot \frac{1}{2\, cos^2z}- \frac{1}{2}\int \frac{cosz\, dz}{cos^2z}\Big )=\frac{4}{\sqrt3}\cdot \Big (\frac{sinz}{2cos^2z}-\frac{1}{2}\cdot \int \frac{cosz\, dz}{1-sin^2z}\Big )=$

$=\frac{4}{\sqrt3}\cdot \frac{sinz}{2\, cos^2z}-\frac{2}{\sqrt3}\cdot \int \frac{d(sinz)}{1-sin^2z}=\frac{2}{\sqrt3}\cdot \frac{sinz}{ cos^2z}+\frac{2}{\sqrt3}\cdot \frac{1}{2}\cdot ln\Big |\frac{sinz-1}{sinz+1}\Big |+C=\\\\=\Big [\; cosz=\frac{2}{3t}\; ,\; z=arccos\frac{2}{3t}\; ,\; t=x+\frac{2}{3}\; ,\; ,z=arccos\frac{2}{3(x+\frac{2}{3})}=arccos\frac{2}{3x+2}\; \Big ]=\\\\=\frac{2}{\sqrt3}\cdot \frac{sin(arccos\frac{2}{3x+2})}{\frac{4}{(3x+2)^2}}+\frac{1}{\sqrt3}\cdot ln\Big |\frac{sin(arccos\frac{2}{3x+2})-1}{sin(arccos\frac{2}{3x+2})+1}\Big |+C=$

$=\frac{2}{\sqrt3}\cdot \frac{\sqrt{3x^2+12x}\, \cdot (3x+2)^2}{4}+\frac{1}{\sqrt3}\cdot ln\Big |\frac{\sqrt{3x^2+12x}\, -1}{\sqrt{3x^2+12x}\, +1}\Big |+C$

$2)\; \; \int sin(5-3x)\, dx=\Big [\; t=5-3x\; ,\; dt=-3\, dx\; \Big ]=\\\\=-\frac{1}{3}\int sint\, dt=-\frac{1}{3}\cdot (-cost)+C=\frac{1}{3}\cdot cos(5-3x)+C$