Find the indefinite integral of 4e^(2x) - sin(2x) with respect to x.

Step-by-step solution

1. 1

   Split the integral by linearity. Integration distributes over sums and pulls out constants, so the problem becomes two independent standard forms:

   $\int\left(4e^{2x}-\sin 2x\right)dx=4\int e^{2x}dx-\int\sin 2x\,dx$

2. 2

   Integrate the exponential, dividing by the inner derivative. Because $\frac{d}{dx}e^{2x}=2e^{2x}$, undoing the chain rule means dividing by $2$:

   $\int e^{2x}dx=\frac{e^{2x}}{2}\;\Longrightarrow\; 4\int e^{2x}dx=4\cdot\frac{e^{2x}}{2}=2e^{2x}$

   Skipping that division is the single most common mistake with $e^{kx}$.

3. 3

   Integrate the sine, and track the two sign flips. The antiderivative of $\sin 2x$ is $-\tfrac{\cos 2x}{2}$, and it enters with a minus sign in front:

   $-\int\sin 2x\,dx=-\left(-\frac{\cos 2x}{2}\right)=+\frac{\cos 2x}{2}$

   Two negatives make the cosine term positive — this is the detail worth pausing on.

4. 4

   Assemble the answer with a single constant.

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

   One $C$ suffices; the arbitrary constants from the two pieces merge into it.

5. 5

   Differentiate to verify.

   $\frac{d}{dx}\left(2e^{2x}+\frac{1}{2}\cos 2x+C\right)=2\cdot 2e^{2x}+\frac{1}{2}\left(-2\sin 2x\right)=4e^{2x}-\sin 2x\;\checkmark$

   Differentiating the result is always available as a check for an indefinite integral, and it takes seconds compared with the integration itself.