Evaluate the integral: I = integral sin(2x+3) dx

Evaluate the integral: $$ I = \int \sin(2x+3)\,dx. $$

Explanation of Method

A sine of a linear expression $2x+3$ is integrated with the substitution $u=2x+3$, converting $dx$ into a constant multiple of $du$ before applying the standard sine antiderivative.

Substitute $u$ for the linear argument of the sine.
Convert $dx$ into $du$.
Integrate using the standard sine antiderivative.
Substitute back and add the constant of integration.

Step 1 · Substituting a new variable

Let:

$$ u = 2x+3. $$

Then, differentiate both sides:

$$ \frac{du}{dx} = 2 \quad \Rightarrow \quad du = 2\,dx \quad \Rightarrow \quad dx = \frac{du}{2}. $$

Step 2 · Transforming the integral

Rewriting the integral in terms of $u$:

$$ I = \int \sin u \cdot \frac{du}{2}. $$

Step 3 · Evaluating the integral

Since $\int \sin u\,du = -\cos u$, we obtain:

$$ I = \frac{1}{2}\int \sin u\,du = \frac{1}{2}(-\cos u) = -\frac{1}{2}\cos u. $$

Step 4 · Substituting back $u$

Replacing $u = 2x+3$:

$$ I = -\frac{1}{2}\cos(2x+3) + C. $$

Summary Table

Quantity	Value
Substitution	$u = 2x+3$
$dx$ in terms of $du$	$\tfrac12 du$
Antiderivative	$-\tfrac12\cos u$

Additional Notes

The integrand is defined for all real $x$, since $\sin$ is defined everywhere.
This follows the general rule $\int \sin(kx+b)\,dx = -\tfrac1k\cos(kx+b)+C$; only the coefficient of $x$ (here $k=2$) affects the constant out front — the additive constant $3$ inside the sine has no effect on it.
The result can be checked by differentiating $-\tfrac12\cos(2x+3)$ with the chain rule and confirming it returns $\sin(2x+3)$.

Result

The solution to the integral is:

$$ \int \sin(2x+3)\,dx = -\frac{1}{2}\cos(2x+3) + C. $$