$$\int \frac{\sin\frac{5x}{2}}{\sin\frac{x}{2}} dx$$ is equal to:

We have to evaluate the integral

$$I=\int \frac{\sin \dfrac{5x}{2}}{\sin \dfrac{x}{2}}\;dx.$$

To simplify the ratio of the two sine terms, we put

$$\theta=\frac{x}{2}\qquad\Longrightarrow\qquad x=2\theta.$$

With this substitution the integrand becomes

$$\frac{\sin\dfrac{5x}{2}}{\sin\dfrac{x}{2}} =\frac{\sin(5\theta)}{\sin\theta}.$$

Now we recall and state the standard formula obtained from Euler’s theorem (or from the geometric progression of the complex exponentials):

$$\frac{\sin n\theta}{\sin\theta}=t^{\,n-1}+t^{\,n-3}+t^{\,n-5}+\dots+t^{-(n-3)}+t^{-(n-1)}, \quad\text{where }t=e^{i\theta}.$$

For an odd integer $$n$$ the right-hand side contains every second power of $$t$$ in symmetrical pairs and one middle term $$1$$. Converting those paired terms with Euler’s formula $$t^k+t^{-k}=2\cos(k\theta)$$ gives

$$\frac{\sin n\theta}{\sin\theta}=1+2\cos(2\theta)+2\cos(4\theta)+\dots+2\cos\bigl((n-1)\theta\bigr).$$

In our question we have $$n=5$$, so

$$\frac{\sin(5\theta)}{\sin\theta}=1+2\cos(2\theta)+2\cos(4\theta).$$

Re-expressing the multiples of $$\theta$$ in terms of $$x$$ (remember $$\theta=\dfrac{x}{2}$$) we get

$$2\theta=x,\qquad 4\theta=2x,$$

and hence

$$\frac{\sin\dfrac{5x}{2}}{\sin\dfrac{x}{2}} =1+2\cos x+2\cos 2x.$$

So our integral becomes

$$I=\int\Bigl(1+2\cos x+2\cos 2x\Bigr)\;dx.$$

We now integrate term by term, using the elementary results

$$\int 1\,dx = x,\qquad \int \cos x\,dx = \sin x,\qquad \int \cos 2x\,dx = \frac{\sin 2x}{2}.$$

Substituting these gives

$$\begin{aligned} I&=\int 1\,dx+\int 2\cos x\,dx+\int 2\cos 2x\,dx\\[4pt] &=x+2\sin x+2\cdot\frac{\sin 2x}{2}+C\\[4pt] &=x+2\sin x+\sin 2x+C. \end{aligned}$$

Hence, the correct answer is Option A.