Question

Qual a Integral de -3x² cos(4x) dx ?

Answer 1

Coloca a constate para fora
=$-3\int \:x^2\cos \left(4x\right)dx$

Aplica a integração parte por parte
=$-3\left(x^2\frac{\sin \left(4x\right)}{4}-\int \:2x\frac{\sin \left(4x\right)}{4}dx\right)$
=-$3\left(\frac{x^2\sin \left(4x\right)}{4}-\int \frac{x\sin \left(4x\right)}{2}dx\right)$

De novo coloca a constate pra fora
=$-3\left(\frac{x^2\sin \left(4x\right)}{4}-\frac{1}{2}\int \:x\sin \left(4x\right)dx\right)$
De novo integração por partes

=$-3\left(\frac{x^2\sin \left(4x\right)}{4}-\frac{1}{2}\left(x\left(-\frac{\cos \left(4x\right)}{4}\right)-\int \:1\left(-\frac{\cos \left(4x\right)}{4}\right)dx\right)\right)$

Coloca de novo a constante pra fora
=$-3\left(\frac{x^2\sin \left(4x\right)}{4}-\frac{1}{2}\left(-\frac{x\cos \left(4x\right)}{4}-\left(-\frac{1}{4}\int \cos \left(4x\right)dx\right)\right)\right)$

Aplica substituição de integral 
=$-3\left(\frac{x^2\sin \left(4x\right)}{4}-\frac{1}{2}\left(-\frac{x\cos \left(4x\right)}{4}-\left(-\frac{1}{4}\int \cos \left(u\right)\frac{1}{4}du\right)\right)\right)$


$u=4x:\quad \quad du=4dx,\:\quad \:dx=\frac{1}{4}du$

Coloca a constate pra fora novamente

=$=-3\left(\frac{x^2\sin \left(4x\right)}{4}-\frac{1}{2}\left(-\frac{x\cos \left(4x\right)}{4}-\left(-\frac{1}{4}\int \cos \left(4x\right)dx\right)\right)\right)$

Usa integração comum 

=$-3\left(\frac{x^2\sin \left(4x\right)}{4}-\frac{1}{2}\left(-\frac{x\cos \left(4x\right)}{4}-\left(-\frac{1}{4}\frac{1}{4}\sin \left(u\right)\right)\right)\right)$

Substitui ''U''

=$-3\left(\frac{x^2\sin \left(4x\right)}{4}-\frac{1}{2}\left(-\frac{x\cos \left(4x\right)}{4}-\left(-\frac{1}{4}\frac{1}{4}\sin \left(4x\right)\right)\right)\right)$

Simplifica...

=$-3\left(\frac{x^2\sin \left(4x\right)}{4}+\frac{\frac{x\cos \left(4x\right)}{4}-\frac{\sin \left(4x\right)}{16}}{2}\right)$

Adiciona a constante 

=$-3\left(\frac{x^2\sin \left(4x\right)}{4}+\frac{\frac{x\cos \left(4x\right)}{4}-\frac{\sin \left(4x\right)}{16}}{2}\right)+C$