2 dx...

August 2022 1 9 Report

Вычислить интегралы: а) методом замены переменной; б) методом интегрирования по частям.
A) S 7x^3dx/5+2x^4
Б) S (2x-3)sin x/2 dx

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nafanya2014

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$\int{ \frac{7x^3dx}{5+2x^4} } \,=[5+2x^4=u;du=8x^3dx]= \\ \\= \frac{7}{8} \int{ \frac{8x^3dx}{5+2x^4} } \, = \frac{7}{8} \int{ \frac{du}{u} } \, =\frac{7}{8} ln|u|+C=\frac{7}{8} ln|5+2x^4|+C
$

$u=2x-3 \Rightarrow du=2dx \\ \\ dv=sin \frac{x}{2} dx\Rightarrow v=-2cos \frac{x}{2}$

$\int{(2x-3)sin \frac{x}{2} } \, dx =-2(2x-3)cos \frac{x}{2}- \int{(-cos \frac{x}{2}) } \, 2dx= \\ \\ =(6-4x)cos \frac{x}{2}+2\int{cos \frac{x}{2} } \,dx=(6-4x)cos \frac{x}{2}+4sin\frac{x}{2}+C$

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dx + 9y=0 3)y''+9y=0...

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