Question

⚠️Gente alguém me ajuda pfvr?? calcular as primitivas a seguir:

obs: aplique a regra de integração por partes

Answer 1

Resposta:

$\left(x+1\right)\cos \left(2x\right)dx\\\\=\cos \left(2x\right)\left(x+1\right)dx\\\\=\cos \left(2x\right)dxx+\cos \left(2x\right)dx\cdot \:1\\\\=xcos \left(2x\right)dx+1\cdot \cos \left(2x\right)dx\\\\=xcos \left(2x\right)dx+\cos \left(2x\right)dx$

$\frac{d}{dx}\left(e^{2x}\sin \left(x\right)\right)\\\\=\frac{d}{dx}\left(e^{2x}\right)\sin \left(x\right)+\frac{d}{dx}\left(\sin \left(x\right)\right)e^{2x}\\\\=e^{2x}\cdot \:2\sin \left(x\right)+\cos \left(x\right)e^{2x}$

Answer 2

a) ∫ (x+1) cos(2x) dx

u=x+1 ==>du=dx

dv = cos(2x) dx  ==>  ∫dv =  ∫cos(2x) dx  ==>v=(1/2) * sen(2x)

∫ (x+1) cos(2x) dx = (x+1)/2 * sen(2x) -(1/2)* ∫ sen (2x) dx

∫ (x+1) cos(2x) dx = (x+1)/2 * sen(2x) -(1/2)* ((-1/2) * cos(2x) ) + c

∫ (x+1) cos(2x) dx = (x+1)/2 * sen(2x) +(1/4)* cos(2x)  + c

b) ∫e^(2x) * sen(x) dx

u=sen(x)  ==>du=cos(x) dx

dv = e^(2x)  dx ==> v=∫ e^(2x)  dx ==>  v = (1/2) *e^(2x)

∫e^(2x) * sen(x) dx  =(sen(x))/2 * e^(2x) - (1/2)* ∫ e^(2x)  cos(x) dx

______________________________________________

Resolvendo ∫ e^(2x)  cos(x) dx

u=cos(x) ==>du=-sen(x) dx

dv=e^(2x)  dx  ==>v=∫ e^(2x) dx ==>  v = (1/2) *e^(2x)

∫ e^(2x)  cos(x) dx =(cos(x))/2 *e^(2x) + (1/2)  ∫ e^(2x) *sen(x) dx

_____________________________________________

∫e^(2x) * sen(x) dx  =(sen(x))/2 * e^(2x) - (1/2)* [ (cos(x))/2 *e^(2x) + (1/2)  ∫ e^(2x) *sen(x) dx]

∫e^(2x) * sen(x) dx  =(sen(x))/2 * e^(2x) - [ (cos(x))/4 *e^(2x) + (1/4)  ∫ e^(2x) *sen(x) dx]

∫e^(2x) * sen(x) dx  =(sen(x))/2 * e^(2x) - (cos(x))/4 *e^(2x) - (1/4)  ∫ e^(2x) *sen(x) dx

∫e^(2x) * sen(x) dx +1/4 ∫ e^(2x) *sen(x) dx = (sen(x))/2 * e^(2x) - (cos(x))/4 *e^(2x)

(5/4)*  ∫e^(2x) * sen(x) dx = (sen(x))/2 * e^(2x) - (cos(x))/4 *e^(2x)

∫e^(2x) * sen(x) dx =(4/5) [ (sen(x))/2 * e^(2x) - (cos(x))/4 *e^(2x)]  + c