Предмет: Алгебра, автор: Reideen

Найти неопределенные интегралы

Приложения:

Ответы

Автор ответа: mathkot

Ответ:

1) $\boxed{ \boldsymbol{ \displaystyle \int {\cos x \sin 2x \cos 4x} \, dx =\frac{\cos x}{4}+ \frac{ \cos 3x}{12} -\frac{ \cos 5x }{20} - \frac{ \cos 7x}{28} + C } }$

2) $\boxed{ \boldsymbol{\displaystyle \int {\cos^{2} 3x} \, dx =\frac{x}{2} + \frac{\sin 6x}{12} + C} }$

3) $\boxed{ \boldsymbol{\displaystyle \int {\cos^{4} 3x} \, dx = \frac{3x}{8} +\frac{\sin 6x}{12} + \frac{\sin 12x}{96} + C } }$

Примечание:

Формула понижения степени:

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

Преобразование произведения в сумму:

$\boxed{\sin x \cos y = \frac{1}{2} \bigg( \sin (x - y) + \sin (x + y) \bigg) }$

По таблице интегралов:

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

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

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

По свойствам интегралов:

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

Объяснение:

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

$\displaystyle = \frac{1}{2}\int { \cos x \bigg( \sin (-2x)+\sin (6x) \bigg)} \, dx = \frac{1}{2} \int { \cos x \bigg( \sin 6x - \sin 2x \bigg)} \, dx=$

$\displaystyle = \frac{1}{2} \int { \bigg(\cos x \sin 6x - \cos x\sin 2x \bigg)} \, dx= \frac{1}{2} \Bigg( \int { \cos x \sin 6x } \, dx - \int { \cos x\sin 2x} \, dx \Bigg)=$

$\displaystyle = \frac{1}{2} \Bigg( \int { \frac{1}{2}\bigg( \sin(6x - x) + \sin(6x + x) \bigg) } \, dx - \int { \frac{1}{2} \bigg( \sin (2x-x) + \sin (2x+x) \bigg) } \, dx \Bigg)=$

$\displaystyle = \frac{1}{2} \Bigg(\frac{1}{2} \int { \bigg( \sin 5x + \sin7x \bigg) } \, dx - \frac{1}{2}\int { \bigg( \sin x + \sin 3x \bigg) } \, dx \Bigg)=$

$\displaystyle = \frac{1}{2} \cdot \frac{1}{2} \Bigg( \int { \sin 5x } \, dx + \int { \sin7x } \, dx - \bigg( \int { \sin x } \, dx + \int { \sin 3x } \, dx \bigg) \Bigg)=$

$\displaystyle = \frac{1}{4} \Bigg( \frac{1}{5} \int { \sin 5x } \, d(5x) + \frac{1}{7} \int { \sin7x } \, d(7x) - \bigg( \int { \sin x } \, dx +\frac{1}{3} \int { \sin 3x } \, d(3x) \bigg) \Bigg)=$

$\displaystyle = \frac{1}{4} \Bigg( -\frac{ \cos 5x }{5} - \frac{ \cos 7x}{7} - \bigg( -\cos x -\frac{ \cos 3x}{3}\bigg) \Bigg) + C=$

$\displaystyle = \frac{1}{4} \Bigg( -\frac{ \cos 5x }{5} - \frac{ \cos 7x}{7} +\cos x +\frac{ \cos 3x}{3}\bigg) \Bigg) + C=$

$\displaystyle = \frac{\cos x}{4}+ \frac{ \cos 3x}{12} -\frac{ \cos 5x }{20} - \frac{ \cos 7x}{28} + C$

$\displaystyle \int {\cos^{2} 3x} \, dx = \int {\frac{1 + \cos 6x}{2} } \, dx = \frac{1}{2} \int {\bigg(1 + \cos 6x \bigg)} \, dx =$

$= \displaystyle \frac{1}{2} \Bigg(\int {1 } \, dx+\int { \cos 6x} \, dx \Bigg) = \frac{1}{2} \Bigg(x+ \frac{1}{6} \int { \cos 6x} \, d(6x) \Bigg) =$

$\displaystyle= \frac{1}{2} \Bigg(x+ \frac{\sin 6x}{6} \Bigg) + C = \frac{x}{2} + \frac{\sin 6x}{12} + C$

$\displaystyle \int {\cos^{4} 3x} \, dx = \int { (\cos^{2} 3x)^{2}} \, dx = \int {\bigg( \frac{1 + \cos 6x}{2} \bigg)^{2} } \, dx =$

$\displaystyle = \int {\frac{1 + 2\cos 6x + \cos^{2} 6x}{4} } \, dx = \frac{1}{4} \int {\bigg(1 + 2\cos 6x + \cos^{2} 6x \bigg) } \, dx =$

$\displaystyle = \frac{1}{4} \Bigg( \int {1 } \, dx + \int {2\cos 6x } \, dx + \int {\cos^{2} 6x } \, dx \Bigg) =$

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

а)

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

б)

$\displaystyle \int {2\cos 6x } \, dx = \frac{2}{6} \int {\cos 6x } \, d(6x) = \frac{\sin 6x}{3} + C_{2}$

в)

$\displaystyle \int {\cos^{2} 6x } \, dx = \int {\frac{1 +\cos 12x}{2} } \, dx = \frac{1}{2} \int {\bigg(1 + \cos 12x \bigg)} \, dx =$

$= \displaystyle \frac{1}{2} \Bigg(\int {1 } \, dx+\int { \cos 12x} \, dx \Bigg) = \frac{1}{2} \Bigg(x + \frac{1}{12} \int { \cos 12x} \, d(12x) \Bigg) =$

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

$\displaystyle= \frac{1}{2} \Bigg(x + \frac{\sin 12x}{12} \Bigg) + C_{3} = \frac{x}{2} + \frac{\sin 12x}{24} + C_{3}$

$\displaystyle = \frac{1}{4} \Bigg( x + C_{1} + \frac{\sin 6x}{3} + C_{2} + \frac{x}{2} - \frac{\sin 12x}{24} + C_{3} \Bigg) =$

$\displaystyle = \frac{1}{4} \Bigg( x + \frac{\sin 6x}{3} + \frac{x}{2} + \frac{\sin 12x}{24} \Bigg) + C = \frac{x}{4} + \frac{\sin 6x}{12} + \frac{x}{8} + \frac{\sin 12x}{96} + C =$