Question

How do you integrate $\int{\cos 6xdx}$?

Answer 1

Hint: We first describe and explain the relation between derivative and anti-derivative. We find the integral value of the integration $\int{\cos mxdx}$. We express the verification as the derivative form of $\dfrac{\sin mx}{m}+c$. We then put the value of $m=6$ in the equation of $\int{\cos mxdx}=\dfrac{\sin mx}{m}+c$ to find the integral solution of $\int{\cos 6xdx}$.

Complete step by step solution:
The given integration is for trigonometric identity.
We know that $\int{\cos mxdx}=\dfrac{\sin mx}{m}+c$.
We first show the integral process then we put the values for $m$.
We take the differentiation of the integral value. The integral being $\dfrac{\sin mx}{m}+c$.
We take $\dfrac{d}{dx}\left( \dfrac{\sin mx}{m}+c \right)$. We know that differentiation of $\sin mx$ is $m\cos mx$.
We take the fraction $\dfrac{1}{m}$ as constant.
Therefore, \[\dfrac{d}{dx}\left( \dfrac{\sin mx}{m}+c \right)=\dfrac{1}{m}\times \dfrac{d}{dx}\left( \sin mx \right)+\dfrac{dc}{dx}\].
Now differentiation of constant is 0.
Therefore, \[\dfrac{d}{dx}\left( \dfrac{\sin mx}{m}+c \right)=\dfrac{1}{m}\times \left( m\cos mx \right)=\cos mx\].
Now by definition of integration or anti-derivative we get $\int{\cos mxdx}=\dfrac{\sin mx}{m}+c$.
The condition for this integration is that $m\ne 0$ is a constant independent of $x$.
Now we place the value of $m=6$ in the equation of $\int{\cos mxdx}=\dfrac{\sin mx}{m}+c$.
We get $\int{\cos 6xdx}=\dfrac{\sin 6x}{6}+c$.

Therefore, the integration of $\int{\cos 6xdx}$ is $\dfrac{\sin 6x}{6}+c$.

Note: We need to remember the concept of integral and anti-derivative comes from the same concept. We can never differentiate them. Also, the condition for the integral where $m\ne 0$, if the value of m is 0 then the integral becomes integration of a constant.