Integral de : 6y*cos(3y^2)dx
Usuário anônimo:
é dx ou dy?
dy, escrevi errado
Vou tentar! É que estou meio enferrujado...
vlw cara esperando
Feito!
vlw mesmo cara
acho que bateu com a que eu fiz
Soluções para a tarefa
Respondido por
1
Consideremos 3y² = k,
Derivando,
Vamos a integral,
![\int6y\cdot\cos(3y^2)\,dy=\\\\\int\cos(3y^2)\cdot\underbrace{6y\,dy}_{dk}=\\\\\int\cos\,k\,dk=\\\\\left[\sin\,k\right]=\\\\\left [\sin(3y^2)\right]=\\\\\boxed{\sin(3y^2)+C}](https://tex.z-dn.net/?f=%5Cint6y%5Ccdot%5Ccos%283y%5E2%29%5C%2Cdy%3D%5C%5C%5C%5C%5Cint%5Ccos%283y%5E2%29%5Ccdot%5Cunderbrace%7B6y%5C%2Cdy%7D_%7Bdk%7D%3D%5C%5C%5C%5C%5Cint%5Ccos%5C%2Ck%5C%2Cdk%3D%5C%5C%5C%5C%5Cleft%5B%5Csin%5C%2Ck%5Cright%5D%3D%5C%5C%5C%5C%5Cleft+%5B%5Csin%283y%5E2%29%5Cright%5D%3D%5C%5C%5C%5C%5Cboxed%7B%5Csin%283y%5E2%29%2BC%7D "\int6y\cdot\cos(3y^2)\,dy=\\\\\int\cos(3y^2)\cdot\underbrace{6y\,dy}_{dk}=\\\\\int\cos\,k\,dk=\\\\\left[\sin\,k\right]=\\\\\left [\sin(3y^2)\right]=\\\\\boxed{\sin(3y^2)+C}")
Derivando,
Vamos a integral,
Perguntas interessantes