How do you find the antiderivative of #1−cos(4x)#?

Question

Emma Aldridge · Accepted Answer

Let's say that, as an illustration unrelated to the issue:

#y=1/k*sin(kx)#

This would imply that:

#ky=sin(kx)#

And in the end, that

#k*(dy)/(dx)=k*cos(kx)#

#(dy)/(dx)=cos(kx)#

due to the chain rule.

If so, the following steps should be taken when resolving the issue:

#int1-cos(4x)dx#

#=x-1/4sin(4x)+C#

Isaiah Allen · Answer

undefined To find the antiderivative of $1 - \cos(4x)$, we can use the following steps:

1. Recognize that $\int 1 - \cos(4x) \, dx$ involves two separate integrals: one for $1$ and one for $-\cos(4x)$.

2. The antiderivative of $1$ with respect to $x$ is simply $x$.

3. To find the antiderivative of $-\cos(4x)$, we use the substitution method. Let $u = 4x$, then $du = 4 \, dx$, or equivalently, $dx = \frac{1}{4} \, du$.

4. Now, we rewrite the integral in terms of $u$: $\int -\cos(u) \cdot \frac{1}{4} \, du$.

5. The antiderivative of $-\cos(u)$ is $\sin(u)$. So, the antiderivative of $-\cos(4x)$ with respect to $u$ is $\sin(u)$.

6. Finally, we substitute $u = 4x$ back into the expression, giving us $\frac{1}{4} \sin(4x)$.

7. Combining the antiderivatives of $1$ and $-\cos(4x)$, we get the antiderivative of $1 - \cos(4x)$ as $x + \frac{1}{4} \sin(4x) + C$, where $C$ is the constant of integration.