Question

I need help solving the following integral

Answer 1

\[\int\limits_{-1}^{0} \cos(4 \pi x)\cos(n \pi x) dx\]

Answer 2

and i already know the trig identity that will allow me to do this, but i'd appreciate if someone could show me how to arrive to \[1/2 [ (\sin(n+4) \pi x)/(n+4) \pi ]+ 1/2 [ (\sin(n-4) \pi x)/(n-4) \pi]\] (from -1 to 0)

Answer 3

I'm assuming the identity you're referring to is
\[\cos mx~\cos nx=\frac{1}{2}\left(\cos(m+n)x+\cos(m-n)x\right)\]
So, you have
\[\cos4\pi x~\cos n\pi x=\frac{1}{2}\big(\cos[(4+n)\pi x]+\cos[(4-n)\pi x]\big)\]
Now, integrating term by term gives you what you need. One thing to note: cosine is an even function, which means \(\cos(-x)=\cos x\). Using this fact, you can rewrite
\[\begin{align*}\cos[(4-n)\pi x]&=\cos[-(n-4)\pi x]\\
&=\cos[(n-4)\pi x]
\end{align*}\]
\[\begin{align*}\int_{-1}^0\cos4\pi x~\cos n\pi x~dx&=\frac{1}{2}\int_{-1}^0\Big(\cos[(n+4)\pi x]+\cos[(n-4)\pi x]\Big)~dx\\
&=\frac{1}{2}\left[\frac{\sin[(n+4)\pi x]}{(n+4)\pi}+\frac{\sin[(n-4)\pi x]}{(n-4)\pi}\right]_{-1}^0
\end{align*}\]

Answer 4

thank you so much.