Question

Знайти інтеграл методом інтегрування частинами. Интеграл x cos 7x dx.​

Answer 1

Ответ:

$∫x \cos(7x) dx$

$u = x \\ dv = \cos(7x) dx$

$u = x \\ du = x'dx \\ du = 1dx \\ du = dx$

$dv = \cos(7x) dx \\ 1dv = \cos(7x) dx \\ ∫1dv = ∫ \cos(7x) dx \\ v = \frac{ \sin(7x) }{7}$

$du = dx \\ v = \frac{ \sin(7x) }{7}$

$x \times \frac{ \sin(7x) }{7} - ∫ \frac{ \sin(7x) }{7} dx$

$x \times \frac{ \sin(7x) }{7} - \frac{1}{7} \times ∫ \sin(7x) dx$

$x \times \frac{ \sin(7x) }{7} - \frac{1}{7} \times ∫ \sin(7x) \times \frac{1}{7} dt$

$x \times \frac{ \sin(7x) }{7} - \frac{1}{7} \times ∫ \frac{ \sin(7x) }{7} dt$

$x \times \frac{ \sin(7x) }{7} - \frac{1}{7} \times ∫ \frac{ \sin(t) }{7} dt$

$x \times \frac{ \sin(7x) }{7} - \frac{1}{7} \times \frac{1}{7} \times ∫(t)dt$

$x \times \frac{ \sin(7x) }{7} - \frac{1 \times 1}{7 \times 7} \times ∫ \sin(t) dt$

$x \times \frac{ \sin(7x) }{7} - \frac{1}{49} \times ∫ \sin(t) dt$

$x \times \frac{ \sin(7x) }{7} - \frac{1}{49} \times ( - \cos(t) )$

$x \times \frac{ \sin(7x) }{7} - \frac{1}{49} \times ( - \cos(7x) )$

$\frac{x \times \sin(7x) }{7} + \frac{ \cos(7x) }{49}$

$\frac{x \times \sin(7x) }{7} + \frac{ \cos(7x) }{49} + C, C∈R$