Integral $\int_0^{\pi/2} \sin(ax)\cos(x)\,dx$ I have to evaluate an integral $I(a) = \sin(ax)\cos(x)$ from $0$ to $\pi/2$.The variable of $a$ is not is greater than $1$:
$$\int_0^{\pi/2} \sin(ax)\cos(x)\,dx$$
I attempted to change the function to $[\sin(ax+x)+\sin(ax-x)]/2$ and then integrate, but I am left with  (-)cosines with a zero in the denominator. How do I remedy this? Am I missing something conceptual.

Your approach is good. Integrate the following identity term-wise:
$$
   \sin(a x) \cos(x) = \frac{1}{2} \sin\left(\left(1+a\right)x\right) - \frac{1}{2} \sin\left(\left(1-a\right) x\right)
$$
getting
$$
   \int_0^{\pi/2} \sin(a x) \cos(x) \mathrm{d}x = \frac{1}{2} \int_0^{\pi/2} \sin\left(\left(1+a\right)x\right)\mathrm{d}x - \frac{1}{2} \int_0^{\pi/2} \sin\left(\left(1-a\right)x\right)\mathrm{d}x
$$
And using $\int \sin(\omega x) \mathrm{d}x = -\frac{1}{\omega} \cos(\omega x)$, hence
$$
  \frac{1}{2} \int_0^{\pi/2} \sin\left(\left(1\pm a\right)x\right)\mathrm{d}x = \left. -\frac{1}{2(1 \pm a)} \cos\left(\left(1 \pm a\right) x \right)  \right|_{x = 0}^{x = \pi/2} = \frac{1}{2 \left(1\pm a\right)} \left(1- \cos\left(\frac{\pi}{2} \left(1 \pm a\right)\right)\right) = \frac{\sin^2\left(\frac{\pi}{2} \frac{1 \pm a}{2}\right)}{ 1\pm a} 
$$
Further simplifying the difference of $+$ and $-$ parts we get
$$
  \int_0^{\pi/2} \sin(a x) \cos(x) \mathrm{d}x = \frac{\sin\left(\frac{\pi}{2} a \right) - a}{1-a^2}
$$
