evaluate the integral.\n intcos4\thetasin8\theta d\theta \n intcos4\thetasin8\theta d\theta=square

Explanation:

Step1: Use product - to - sum formula

We know that $\cos A\sin B=\frac{1}{2}[\sin(A + B)-\sin(A - B)]$. Here $A = 4\theta$ and $B=8\theta$, so $\cos4\theta\sin8\theta=\frac{1}{2}[\sin(4\theta + 8\theta)-\sin(4\theta-8\theta)]=\frac{1}{2}(\sin12\theta+\sin4\theta)$.

Step2: Integrate term - by - term

$\int\cos4\theta\sin8\theta d\theta=\frac{1}{2}\int(\sin12\theta+\sin4\theta)d\theta$. The integral of $\sin ax$ is $-\frac{1}{a}\cos ax + C$. So $\frac{1}{2}\int(\sin12\theta+\sin4\theta)d\theta=\frac{1}{2}\left(-\frac{1}{12}\cos12\theta-\frac{1}{4}\cos4\theta\right)+C=-\frac{1}{24}\cos12\theta-\frac{1}{8}\cos4\theta + C$. If this is a definite integral (the limits are not shown in the problem, assuming it's an indefinite integral), the antiderivative is $-\frac{1}{24}\cos12\theta-\frac{1}{8}\cos4\theta + C$.

Answer:

$-\frac{1}{24}\cos12\theta-\frac{1}{8}\cos4\theta + C$