Question

∫cos3xcos5xdx

Explanation:

Step1: Use the product - to - sum formula

We know that \(\cos A\cos B=\frac{1}{2}[\cos(A + B)+\cos(A - B)]\). Here \(A = 3x\) and \(B=5x\), so \(\cos3x\cos5x=\frac{1}{2}(\cos8x+\cos(- 2x))\). Since \(\cos(-\alpha)=\cos\alpha\), then \(\cos3x\cos5x=\frac{1}{2}(\cos8x+\cos2x)\)

Step2: Integrate term - by - term

\(\int\cos3x\cos5x dx=\frac{1}{2}\int(\cos8x+\cos2x)dx\)
Using the integral formula \(\int\cos(ax)dx=\frac{1}{a}\sin(ax)+C\) (\(a
eq0\))
\(\frac{1}{2}\int(\cos8x+\cos2x)dx=\frac{1}{2}(\frac{1}{8}\sin8x+\frac{1}{2}\sin2x)+C\)

Answer:

\(\frac{1}{16}\sin8x+\frac{1}{4}\sin2x + C\)