Find the indefinite integral of cos(2x) with respect to x.

Step-by-step solution

1. 1

   Guess the shape, then fix the constant. The antiderivative of $\cos$ is $\sin$, so the answer must look like $A\sin2x$. Differentiating that guess: $\dfrac{d}{dx}\left(A\sin2x\right)=2A\cos2x$. To match $\cos2x$ we need $2A=1$, so $A=\tfrac12$. The $\tfrac12$ is not decoration — it undoes the $2$ the chain rule would introduce.

2. 2

   Do it properly with substitution. Let $u=2x$, so $du=2\,dx$ and $dx=\tfrac12\,du$:

   $\int\cos2x\,dx=\int\cos u\cdot\frac12\,du=\frac12\int\cos u\,du=\frac12\sin u+C$

3. 3

   Substitute back.

   $\int\cos2x\,dx=\frac12\sin2x+C$

4. 4

   State the general rule. For any nonzero constant $a$,

   $\int\cos ax\,dx=\frac{1}{a}\sin ax+C$

   The factor out front is always the reciprocal of the inner coefficient. The same pattern gives $\int\sin ax\,dx=-\tfrac1a\cos ax+C$.

5. 5

   Note the most common error. Writing $\sin2x+C$ (no $\tfrac12$) or $2\sin2x+C$ (reciprocal inverted) are the two usual mistakes. Differentiating the answer is a five-second check that catches both.

6. 6

   Verify by differentiating. $\dfrac{d}{dx}\left(\tfrac12\sin2x\right)=\tfrac12\cdot2\cos2x=\cos2x$ ✓. Numerically, symmetric difference quotients of $\tfrac12\sin2x$ at $x=0.3,\ 1.4,\ -0.9$ match $\cos2x$ to five decimals ✓.