Question

intergal of (cos3xcos4x)dx

Answer 1

The equation... 1/2(cos(m-n)x+cos(m+n)x)

Answer 2

My answer.. sin7x/14-sinx/2

Answer 3

but in the answer book it's a +

Answer 4

Right. The integral of cosine is positive sine. Also, \(\cos(-x)=\cos x\) because cosine is an even function.

Answer 5

Sooo am I right lol or wrong?

Answer 6

The book is right.
\[\begin{align*}\frac{\cos(m-n)x+\cos(m+n)x}{2}&=\frac{1}{2}(\cos mx\cos nx+\sin mx\sin nx\\&\quad\quad\quad+\cos mx \cos nx-\sin mx \sin nx)\\\\
&=\cos mx\cos nx\end{align*}\]
Setting \(m=3\) and \(n=4\) and integrating the LHS gives
\[\begin{align*}\frac{1}{2}\int\cos(-1)x\,dx+\frac{1}{2}\int\cos(7)x\,dx&=\frac{1}{2}\int\cos x\,dx+\frac{1}{2}\int\cos7x\,dx\\\\
&=\frac{\sin x}{2}+\frac{\sin7x}{14}+C\end{align*}\]

Answer 7

Oh okay, so cos(-x)=cos(x) and sin(-x)=-sin(x)

Answer 8

?

Answer 9

Right. If you left it as \(\cos(-x)\), integrating would give
\[\int\cos(-x)\,dx=-\sin(-x)+C=\sin x+C\]
since \(\sin(-x)=-\sin x\).