Question

Помогите пожалуйста найти определенные интегралы.

Answer 1

$1)\; \; \int \frac{x^2\, dx}{5-x^6}=\int \frac{x^2\, dx}{5-(x^3)^2}=[\; t=x^3\; ,\; dt=3x^2\, dx\; ]=\frac{1}{3}\int \frac{dt}{5-t^2}=\\\\=\frac{1}{3}\cdot \frac{1}{2\sqrt5}\cdot ln\Big |\frac{\sqrt5+t}{\sqrt5-t}\Big |+C=\frac{1}{6\sqrt5}\cdot ln\Big |\frac{\sqrt5+x^3}{\sqrt5-x^3}\Big |+C\\\\2)\; \; \int \frac{dx}{\sqrt{x(3x+5)}}=\int \frac{dx}{\sqrt{3x^2+5x}}=\int \frac{dx}{\sqrt{3}\cdot \sqrt{x^2+\frac{5}{3}x}}=\frac{1}{\sqrt3}\int \frac{dx}{\sqrt{(x+\frac{5}{6})^2-\frac{25}{36}}}=$

$\Big [\; t=x+\frac{5}{6}\; ,\; dt=dx\; \Big ]=\frac{1}{\sqrt3}\int \frac{dt}{t^2-\frac{25}{36}}=\frac{1}{\sqrt3}\cdot \frac{1}{2\cdot \frac{5}{6}}\cdot ln\Big |\frac{t-\frac{5}{6}}{t+\frac{5}{6}}\Big |+C=\\\\=\frac{\sqrt3}{5}\cdot ln\Big |\frac{3x}{3x+5}\Big |+C\\\\3)\; \; \int x\cdot arctg\sqrt{x^2-1}\, dx=[\; t^2=x^2-1\; ,\; 2t\, dt=2x\, dx\; ,\; x\, dx=t\, dt]=\\\\=\int arctgt\cdot t\, dt=[u=arctgt\; ,\; du=\frac{dt}{1+t^2}\; ,\; dv=t\, dt\; ,\; v=\frac{t^2}{2}\; ]=$

$=uv-\int v\, du=\frac{t^2}{2}\cdot arctgt-\frac{1}{2}\int \frac{t^2\, dt}{1+t^2}=\frac{t^2}{2}\cdot arctgt-\frac{1}{2}\int (1-\frac{1}{1+t^2})dt=\\\\=\frac{t^2}{2}\cdot arctgt-\frac{1}{2}\cdot (t-arctgt)+C=\\\\=\frac{x^2-1}{2}\cdot arctg\sqrt{x^2-1}-\frac{1}{2} \cdot (\, \sqrt{x^2-1}-arctg\sqrt{x^2-1}\, )+C$

$4)\; \; \int\limits^1_0\frac{\sqrt[4]{x}-\sqrt[8]{x^3}}{\sqrt{x}}\, dx=[\; t=x^8\; ,\; x=\sqrt[8]{t}\; ,\; dx=\frac{t^{-7/8}}{8}\, dt\; ,\; t_1=0\; ,\; t_2=1\; ]=\\\\=\int\limits^1_0\frac{t^2-t^3}{t^4}\cdot \frac{dt}{8t^{7/8}}=\frac{1}{8} \int\limits^1_0\, \frac{1-t}{t^2\; \cdot \; t^{7/8}}\, dt=\frac{1}{8}\int\limits^1_0\frac{1-t}{t^{23/8}}\, dx=\frac{1}{8}\int\limits^1_0\, (t^{-23/8}-t^{-15/8})\, dt=\\\\=\frac{1}{8}\cdot \Big (\frac{t^{-15/8}}{-15/8}-\frac{t^{-7/8}}{-7/8}\Big )\Big |_0^1=\frac{1}{8}\cdot (-\frac{8}{15}+\frac{8}{7})=\frac{8}{105}$

$5)\; \; \int\limits^{\frac{\pi}{4}}_0\, \frac{tg^7x\, dx}{cos^2x}=\frac{tg^8x}{8}\Big |_0^{\frac{\pi}{4}}=\frac{1}{8}\cdot (1-0)=\frac{1}{8}$