Question

Помогите, пожалуйста, с высшей математикой. Тема
Интегральное исчисление функции одной переменной

Answer 1

Объяснение:

7.

$\int {\frac{dx}{x^3\sqrt[3]{2-x^3} } } \, =\int {x^{-3}(2-x^3)^{-\frac{1}{3} }} \, dx =\\\\\begin{Vmatrix} Chebyshev:\; m=-3, \; n=3, \; p=-\frac{1}{3}\\\frac{m+1}{n}+p =\frac{-3+1}{3}-\frac{1}{3}=-1 \in \mathbb {Z}\\2-x^3=t^3x^3, \; t^3x^3+x^3=2,\;x^3(t^3+1)=2\\x^3 =2(t^3+1)^{-1}, \;x=\sqrt[3]{2}(t^3+1)^{-\frac{1}{3}}\\dx= -\frac{\sqrt[3]{2} }{3} (t^3+1)^{-\frac{4}{3} }(t^3+1)'dt=-\sqrt[3]{2}t^2(t^3+1)^{-\frac{4}{3} }dt \end{Vmatrix}=$

$\int {\underbrace{\frac{1}{2}(t^3+1) }_{x^{-3}}*\underbrace{t^{-1}*\frac{1}{\sqrt[3]{2} }(t^3+1)^\frac{1}{3} }_{(t^3x^3)^{-\frac{1}{3} }=t^{-1}x^{-1}}}* \, \underbrace{(-\sqrt[3]{2})t^2(t^3+1)^{-\frac{4}{3} }dt }_{dx}=\\-\frac{1}{2}\int {t} \, dt=-\frac{1}{2}*\frac{t^2}{2} +C= \begin{Vmatrix} 2-x^3=t^3x^3\\t^3=\frac{2-x^3}{x^3}\\t=\sqrt[3]{\frac{2-x^3}{x^3} } \end{Vmatrix}=-\frac{1}{4}\sqrt[3]{\big(\frac{2-x^3}{x^3} \big)^2}+C$

8.

$\int {\frac{dx}{5+4\sin x} } \,= \begin{Vmatrix} t=\rm tg \; \frac{x}{2} \\\sin x=\frac{2t}{1+t^2}\\x=2\rm arctg \; t\\dx=\frac{2dt}{1+t^2} \end{Vmatrix}=\int {\frac{\frac{2dt}{1+t^2} }{5+4\frac{2t}{1+t^2} } } \,=\int {\frac{\frac{2dt}{1+t^2} }{\frac{5+5t^2+8t}{1+t^2} } } \,=\\\int {\frac{2(1+t^2)}{(1+t^2)(5t^2+8t+5)} } \, dt =2\int {\frac{dt}{5t^2+8t+5} } \,=2\int {\frac{dt}{\big ((t\sqrt{5})^2+2*t\sqrt{5}*\frac{4}{\sqrt{5} } +(\frac{4}{\sqrt{5} })^2 \big)+1.8 } } \,=$

$2\int {\frac{dt}{\big(t\sqrt{5}+\frac{4}{\sqrt{5} } \big)^2+\big(\sqrt{1.8}\big)^2 } } \, =\frac{2}{\sqrt{5} } \int {\frac{dt\sqrt{5} }{\big(t\sqrt{5}+\frac{4}{\sqrt{5} } \big)^2+\big(\sqrt{1.8}\big)^2 } } \, =\\\frac{2}{\sqrt{5} } \int {\frac{d\big(t\sqrt{5}+\frac{4}{\sqrt{5} }\big) }{\big(t\sqrt{5}+\frac{4}{\sqrt{5} } \big)^2+\big(\sqrt{1.8}\big)^2 } } \, =$

$\frac{2}{\sqrt{5} }\frac{1}{\sqrt{1.8} }\rm arctg \frac{t\sqrt{5}+\frac{4}{\sqrt{5} }}{\sqrt{1.8} } +C=\frac{2}{3} \rm arctg\big(\frac{5}{3}t+\frac{4}{3}\big)+C=\frac{2}{3} \rm arctg\big(\frac{5}{3}\rm tg\frac{x}{2} +\frac{4}{3}\big)+C$

9.

$\int {x\rm tg^2x} \, dx\\\int {\rm tg^2x} \, dx=\begin{Vmatrix} \rm tg \; x=t\\x=\rm arctg \; t\\dx=\frac{dt}{1+t^2} \end{Vmatrix} =\int {\frac{t^2}{1+t^2} } \, dt=\int {\big(1-\frac{1}{1+t^2}\big) } \, dt=\\t-\rm arctg \; t+C=\rm tg \; x-\rm arctg(\rm tg \; x)+C=\rm tg \; x-x+C\\\\\int {x\rm tg^2x} \, dx =\begin{Vmatrix} u=x&du=dx\\dv=\rm tg^2xdx&v=\rm tg \; x-x\end{Vmatrix}=uv-\int {v} \, du=$

$x(\rm tg \; x-x)-\int {(\rm tg \; x-x)} \, dx =x(\rm tg \; x-x)-\underbrace{\int {\rm tg \; x} \, dx }_{I}+\int {x} \, dx \\I=\begin{Vmatrix} \rm tg \; x=t\\x=\rm arctg \; t\\dx=\frac{dt}{1+t^2} \end{Vmatrix}=\int {\frac{t}{1+t^2} } \, dt=\frac{1}{2} \int {\frac{2tdt}{1+t^2} } \, =\frac{1}{2} \int {\frac{d(1+t^2)}{1+t^2} } \,\\=\frac{1}{2}\ln|1+t^2|+C=\frac{1}{2}\ln|1+\rm tg^2x|+C=\frac{1}{2}\ln|\frac{1}{\cos^2x} |+C=\\\frac{1}{2}\ln|\cos^{-2}x|+C=-\ln|\cos x| +C$

$\int {x\rm tg^2x} \, dx =x(\rm tg \; x-x)+\ln|\cos x|+\frac{x^2}{2} +C=\\x\rm tg \;x+\ln|\cos x|-\frac{x^2}{2}+C$