\[\begin{align*} \boxed{\int \sin(2x) \, dx = -\dfrac{1}{2} \cos(2x) + C} \end{align*}\]
where \( C \) is the constant of integration.
Step-by-Step Solution:
To understand why the integral of \( \sin(2x) \) is \( -\dfrac{1}{2} \cos(2x) + C \), let’s use u-substitution.
\[\begin{align*} \int \sin(2x) \, dx \end{align*}\]
Since we know the common integral $\sin x$, and we’re dealing with \( 2x \) inside the sine function, it’s helpful to use substitution. Let:
\[\begin{align*} u = 2x \end{align*}\]
Now, differentiate \( u \) with respect to \( x \):
\[\begin{align*} \dfrac{du}{dx} = 2 \Rightarrow dx = \dfrac{du}{2} \end{align*}\]
Substituting \( u \) and $du$ into the integral:
\[\begin{align*} \int \sin(2x) \, dx = \int \sin(u) \cdot \dfrac{du}{2} \end{align*}\]
This simplifies to:
\[\begin{align*} \dfrac{1}{2} \int \sin(u) \, du \end{align*}\]
The integral of \( \sin(u) \) is \( -\cos(u) \), so we have:
\[\begin{align*} \dfrac{1}{2} \cdot (-\cos(u)) = -\dfrac{1}{2} \cos(u) \end{align*}\]
Finally, replace \( u \) with \( 2x \):
\[\begin{align*} -\dfrac{1}{2} \cos(2x) + C \end{align*}\]
So, the integral of \( \sin(2x) \) is:
\[\begin{align*} \boxed{\int \sin(2x) \, dx = -\dfrac{1}{2} \cos(2x) + C} \end{align*}\]
Don’t forget to add $+C$ as this is an indefinite integral!
Why This Works:
When we used substitution, we transformed \( \sin(2x) \) into \( \sin(u) \), which made the integral easier to solve as $\sin u$ is a standard integral. Substitution is a common method for integrals where a function’s argument (like \( 2x \)) has a multiplier.
Final Answer:
To recap:
\[\begin{align*} \boxed{\int \sin(2x) \, dx = -\dfrac{1}{2} \cos(2x) + C} \end{align*}\]
By understanding how to solve this integral, the method of $u$-substitution should come in mind for future problems where you spot common integrals!
