Evaluate the integral: I = integral cos(1-3x) dx
Evaluate the integral: $$ I = \int \cos(1-3x)\,dx. $$
Explanation of Method
A cosine of a linear expression with negative slope, $1-3x$, is handled the same way as any linear-argument trig integral: substitute $u$ for the argument, convert $dx$ to $du$ (picking up a negative constant here), and apply the standard cosine antiderivative.
- Substitute $u$ for the linear argument of the cosine.
- Convert $dx$ into $du$, tracking the negative sign.
- Integrate using the standard cosine antiderivative.
- Substitute back and add the constant of integration.
Step 1 · Substituting a new variable
Let:
$$ u = 1 - 3x. $$
Then, differentiate both sides:
$$ \frac{du}{dx} = -3 \quad \Rightarrow \quad du = -3\,dx \quad \Rightarrow \quad dx = \frac{du}{-3}. $$
Step 2 · Transforming the integral
Rewriting the integral in terms of $u$:
$$ I = \int \cos u \cdot \frac{du}{-3}. $$
Step 3 · Evaluating the integral
Since $\int \cos u\,du = \sin u$, we obtain:
$$ I = \frac{1}{-3}\int \cos u\,du = -\frac{1}{3}\sin u. $$
Step 4 · Substituting back $u$
Replacing $u = 1-3x$:
$$ I = -\frac{1}{3}\sin(1-3x) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = 1-3x$ |
| $dx$ in terms of $du$ | $-\tfrac13 du$ |
| Antiderivative | $-\tfrac13\sin u$ |
Additional Notes
- The integrand is defined for all real $x$, since $\cos$ is defined everywhere.
- The negative slope of the argument ($-3$ instead of $3$) directly produces the overall negative coefficient in the answer — compare with a positive-slope cosine argument, where $\int \cos(3x)\,dx=\tfrac13\sin(3x)+C$ has no minus sign.
- The result can be checked by differentiating $-\tfrac13\sin(1-3x)$ with the chain rule and confirming it returns $\cos(1-3x)$.
Result
The solution to the integral is: