Evaluate the integral: I = integral (sin 2x)/(sin² x + 3)

Evaluate the integral: $$ I = \int \frac{\sin 2x}{\sin^2 x + 3} \, dx. $$

Step 1 · Apply the double-angle identity

We begin by simplifying the integral. Notice that we can express the numerator using the double-angle identity for sine:

$$ \sin 2x = 2\sin x\cos x. $$

Substituting this into the integral, we get:

$$ I = \int \frac{2\sin x\cos x}{\sin^2x+3}\,dx. $$

Step 2 · Substitution

Next, we perform a substitution to simplify the integral. Let:

$$ u = \sin x \quad \text{so that} \quad du = \cos x\,dx. $$

Now, substitute $u$ and $du$ into the integral:

$$ I = \int \frac{2u}{u^2+3}\,du. $$

Step 3 · Apply the standard result

This is a standard integral of the form:

$$ \int \frac{u}{u^2+a^2}\,du = \frac{1}{2}\ln\left(u^2+a^2\right). $$

In our case, $a^2=3$, so:

$$ I = \int \frac{2u}{u^2+3}\,du = \ln\left(u^2+3\right). $$

Step 4 · Substitute back

Finally, substitute $u = \sin x$ back into the result:

$$ I = \ln\left(\sin^2x+3\right) + C, $$

where $C$ is the constant of integration.

Summary Table

Additional Notes

• The integrand is defined for all real $x$, since $\sin^2x+3 \geq 3 > 0$ always.
• Because the numerator is exactly $2u$ (matching the coefficient in $\int \frac{u}{u^2+a^2}du=\frac12\ln(u^2+a^2)$ scaled by $2$), the factor of $\frac12$ from the standard formula cancels the factor of $2$ from the numerator, leaving a clean $\ln(u^2+3)$ with no leftover constant.
• The result can be checked by differentiating $\ln(\sin^2x+3)$ with the chain rule and confirming it returns $\dfrac{\sin 2x}{\sin^2x+3}$.