Question

Дифрівняння Вища математика
Знайти невизначений інтеграл​

Answer 1

Ответ:

a)

$\boldsymbol{\boxed{ \int {\sin ^{2} 4x} \, dx =\frac{x}{2} - \frac{\sin 8x}{16} + C}}$

b)

$\boldsymbol{\boxed{ \int {( 1 - 2x)\sin x} \, dx =2x \cos x - 2 \sin x-\cos x + C }}$

c)

$\boldsymbol{\boxed{ \int {\frac{x + 2}{\sqrt{x - 1} } } \, dx =\dfrac{2(\sqrt{x - 1} )^{3}}{3} + 6\sqrt{x-1} + C}}$

d)

$\boldsymbol{\boxed{\int {\frac{2x - 1}{x^{2} + 4x - 5} } \, dx =\ln|x^{2} + 4x - 5| - \dfrac{5}{6} \ln \bigg |\dfrac{x -1}{x + 5} \bigg | + C}}$

Примечание:

Формула понижения степени:

$\boxed{\sin ^{2} \alpha =\frac{1 - \cos 2\alpha }{2} }$

По свойствам интегралов:

$\boxed{ \displaystyle \int \sum\limits_{i=1}^n {C_{i}f_{i}(x)} \, dx = \sum\limits_{i=1}^nC_{i} \int {f_{i}(x)} \, dx}$

По таблице интегралов:

$\boxed{\displaystyle \int x^{n} \ dx = \frac{x^{n + 1}}{n + 1} + C; n \neq -1, x > 0}$

$\boxed{\int \frac{dx}{x^{2} - a^{2}} =\frac{1}{2a} \ln \bigg | \frac{x - a}{x + a} \bigg | + C }$

$\boxed{\displaystyle \int \sin x \ dx = -\cos x + C}$

$\boxed{\displaystyle \int \cos x \ dx = \sin x + C}$

$\boxed{\int \frac{dx}{x} = \ln|x| + C }$

Пошаговое объяснение:

а)

$\displaystyle \int {\sin ^{2} 4x} \, dx = \int {\frac{1 - \cos 8x}{2} } \, dx = \frac{1}{2} \int {(1 - \cos 8x)} \, dx =$

$\displaystyle = \frac{1}{2} \Bigg ( \int {1} \, dx - \int {\cos 8x} \, dx \Bigg) = \frac{1}{2} \Bigg ( x + C_{1} - \frac{1}{8} \int {\cos 8x} \, d(8x) \Bigg) =$

$\displaystyle = \frac{1}{2} \bigg ( x + C_{1} - \frac{\sin 8x}{8} + C_{2} \bigg) = \frac{x}{2} + \frac{C_{1}}{2} - \frac{\sin 8x}{16}+ \frac{C_{2}}{2} =\frac{x}{2} - \frac{\sin 8x}{16} + C$

d)

$\displaystyle \int {\frac{2x - 1}{x^{2} + 4x - 5} } \, dx = \int {\frac{2x - 1 + 4 - 4}{x^{2} + 4x - 5} } \, dx = \int {\frac{2x + 4 - 5}{x^{2} + 4x - 5} } \, dx =$

$\displaystyle = \int \bigg ( {\frac{2x + 4 }{x^{2} + 4x - 5} - \frac{ 5}{x^{2} + 4x - 5} } \bigg ) \, dx = \int {\frac{2x + 4 }{x^{2} + 4x - 5} } \, dx - \int {\frac{5 }{x^{2} + 4x - 5} } \, dx =$

$\displaystyle = \int {\frac{d(x^{2} + 4x - 5) }{(x^{2} + 4x - 5)} } - 5\int {\frac{dx }{x^{2} + 4x + 4 - 4 - 5} } = \ln|x^{2} + 4x - 5| + C_{1} -$

$\displaystyle -5\int {\frac{dx }{(x+ 2)^{2} - 9} } = \ln|x^{2} + 4x - 5| + C_{1} -5\int {\frac{d(x + 2) }{(x+ 2)^{2} - 3^{2}} } =$

$= \ln|x^{2} + 4x - 5| + C_{1} - \dfrac{5}{6} \ln \bigg |\dfrac{x + 2 - 3}{x + 2 + 3} \bigg | + C_{2} =$

$=\ln|x^{2} + 4x - 5| - \dfrac{5}{6} \ln \bigg |\dfrac{x -1}{x + 5} \bigg | + C$

b)

$\displaystyle \int {( 1 - 2x)\sin x} \, dx = \int {( \sin x - 2x\sin x)} \, dx = \int {\sin x} \, dx - \int {2x\sin x} \, dx =$

1)

$\displaystyle \int {\sin x} \, dx = -\cos x + C_{1}$

2)

$\displaystyle \int {2x\sin x} \, dx =$

-------------------------------------------------------------

$u = 2x \Longrightarrow du = (2x)' \ dx = 2 \ dx$

$dv = \sin x \ dx \Longrightarrow v = \displaystyle \int {\sin x} \, dx = -\cos x$

-------------------------------------------------------------

$= - 2x \cos x -\displaystyle \int {- 2 \cos x} \, dx = - 2x \cos x +\displaystyle2 \int { \cos x} \, dx = 2 \sin x - 2x \cos x +C_{2}$

3)

$\displaystyle \int {( 1 - 2x)\sin x} \, dx = \int {\sin x} \, dx - \int {2x\sin x} \, dx =$

$= -\cos x + C_{1} - 2 \sin x + 2x \cos x -C_{2} = 2x \cos x - 2 \sin x-\cos x + C$

c)

$\displaystyle \int {\frac{x + 2}{\sqrt{x - 1} } } \, dx =$

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Замена: $\sqrt{x - 1} = t \Longrightarrow t^{2} = x - 1 \Longrightarrow x = t^{2} + 1$

$dx = (t^{2} + 1)' \ dt = 2t \ dt$

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$\displaystyle = \int {\frac{2t( t^{2} + 1 + 2)}{t} } \, dt = 2\int{(t^{2} + 3)} \, dt = 2 \Bigg ( \int{t^{2} } \, dt +\int{3 } \, dt\Bigg)=$

$\displaystyle = 2 \Bigg ( \frac{t^{3}}{3} + C_{1}+ 3\int{ } \, dt\Bigg)= 2 \bigg (\frac{t^{3}}{3} + C_{1}+ 3t + C_{2} \bigg ) = \frac{2t^{3}}{3} + 6t + 2C_{1} + 2C_{2} =$

$= \dfrac{2t^{3}}{3} + 6t + C = \dfrac{2(\sqrt{x - 1} )^{3}}{3} + 6\sqrt{x-1} + C$