How do you integrate int cos(3t)cos(4t)dt?

Apr 23, 2018

$\frac{1}{14} \sin 7 t + \frac{1}{2} \sin t + C$

Explanation:

Recall the cosine product identity

$\cos a \cos b = \frac{1}{2} \left[\cos \left(a + b\right) + \cos \left(a - b\right)\right]$

Using it, we can say

$\cos 3 t \cos 4 t = \frac{1}{2} \left[\cos \left(3 t + 4 t\right) + \cos \left(3 t - 4 t\right)\right]$

$= \frac{1}{2} \left(\cos 7 t + \cos \left(- t\right)\right)$

Cosine is an even function, so $\cos \left(- t\right) = \cos t$

$= \frac{1}{2} \left(\cos 7 t + \cos t\right)$

Thus, we get

$\frac{1}{2} \int \left(\cos 7 t + \cos t\right) \mathrm{dt} = \frac{1}{2} \int \cos 7 t \mathrm{dt} + \frac{1}{2} \int \cos t \mathrm{dt}$

$= \frac{1}{14} \sin 7 t + \frac{1}{2} \sin t + C$

Apr 23, 2018

Integrate: $\int \cos \left(3 t\right) \cos \left(4 t\right) \mathrm{dt}$

Use the identity $\cos \left(A\right) \cos \left(B\right) = \frac{1}{2} \left(\cos \left(A - B\right) + \cos \left(A + B\right)\right)$ where $A = 4 t$ and $B = 3 t$:

$\int \cos \left(3 t\right) \cos \left(4 t\right) \mathrm{dt} = \frac{1}{2} \int \cos \left(t\right) + \cos \left(7 t\right) \mathrm{dt}$

Separate into two integrals:

$\int \cos \left(3 t\right) \cos \left(4 t\right) \mathrm{dt} = \frac{1}{2} \int \cos \left(t\right) \mathrm{dt} + \frac{1}{2} \int \cos \left(7 t\right) \mathrm{dt}$

The two integrals are trivial:

$\int \cos \left(3 t\right) \cos \left(4 t\right) \mathrm{dt} = \frac{1}{2} \sin \left(t\right) + \frac{1}{14} \sin \left(7 t\right) + C$