Evaluate the integral: I = integral sin 2/3 x dx
Evaluate the integral: $$ I = \int \sin \frac{2}{3}x \, dx. $$
Explanation of Method
Whenever the argument of $\sin$ or $\cos$ is a linear expression in $x$, the substitution $u=(\text{argument})$ converts $dx$ to a constant multiple of $du$, and the standard $\int \sin u\,du=-\cos u$ formula applies directly.
- Substitute $u$ for the argument of the sine function.
- Convert $dx$ into $du$ using the constant derivative.
- Integrate using the standard sine antiderivative.
- Substitute back and add the constant of integration.
Step 1 · Substitution
Let:
$$ u = \frac{2}{3}x. $$
Differentiating both sides with respect to $x$, we get:
$$ du = \frac{2}{3}\,dx. $$
Solving for $dx$:
$$ dx = \frac{3}{2}\,du. $$
Step 2 · Rewrite the integral
Substituting $u$ and $dx$ into the integral:
$$ I = \int \sin u \cdot \frac{3}{2}\,du. $$
Factor out the constant:
$$ I = \frac{3}{2}\int \sin u\,du. $$
Step 3 · Solve the integral
We use the standard integral formula:
$$ \int \sin u\,du = -\cos u. $$
Applying this:
$$ I = \frac{3}{2}(-\cos u) = -\frac{3}{2}\cos u. $$
Step 4 · Substitute back $u$
Recalling that $u = \dfrac{2}{3}x$, we substitute back:
$$ I = -\frac{3}{2}\cos\left(\frac{2}{3}x\right) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = \tfrac23 x$ |
| $dx$ in terms of $du$ | $\tfrac32 du$ |
| Antiderivative | $-\tfrac32\cos u$ |
Additional Notes
- The integrand is defined for all real $x$, since $\sin$ is defined everywhere.
- The constant $\tfrac32$ appears because $dx=\tfrac32 du$ is the reciprocal of the coefficient $\tfrac23$ inside the sine — in general, $\int \sin(kx)\,dx=-\tfrac1k\cos(kx)+C$ for any nonzero $k$.
- The result can be checked by differentiating $-\tfrac32\cos\left(\tfrac23x\right)$ with the chain rule and confirming it returns $\sin\left(\tfrac23x\right)$.
Result
The solution to the integral is: