Question

integral cos(3x)cos(4x) dx

Answer 1

i dont even no where to begin

Answer 2

the only way i think of to handle this is to use Euler's Formula's.\[\cos(3x)=\frac{e^{3ix}+e^{-3ix}}{2}\]
\[\cos(4x)=\frac{e^{4ix}+e^{-4ix}}{2}\] Multiplying and simplifying should make the problem easier to handle.

Answer 3

i havnet even come across that yet

Answer 4

then maybe using sum and difference formulas? cos of two angles and whatnot? it would be really easy with Eulers though =/

Answer 5

can you show me the first step with the sum and diffrence formulas

Answer 6

equation1 : \[\cos(u+v)=\cos(u)\cos(v)-\sin(u)\sin(v)\]
equation 2 : \[\cos(u-v)=\cos(u)\cos(v)+\sin(u)\sin(v)\]

adding equation1 to equation2 we get
\[\cos(u+v)+\cos(u-v)=2\cos(u)\cos(v)\]
now multiply both sides by 1/2
\[\cos(u)\cos(v)=\frac{1}{2}(\cos(u+v)+\cos(u-v))\]

Answer 7

\[\int\limits_{}^{}\cos(3x)\cos(4x)dx=\frac{1}{2}\int\limits_{}^{}(\cos(3x+4x)+\cos(3x-4x))dx\]
\[=\frac{1}{2}\int\limits_{}^{}(\cos(7x)+\cos(-x)) dx=\frac{1}{2}\int\limits_{}^{}(\cos(7x)+\cos(x)) dx\]

Answer 8

\[=\frac{1}{2}(\frac{1}{7}\sin(7x)+\sin(x))+C\]