Question

evaluate the integral.
int 2sin^{3}4xcos^{3}4x dx
int 2sin^{3}4xcos^{3}4x dx=square

Explanation:

Step1: Rewrite using trig identity

First, use the identity $\sin A\cos A=\frac{1}{2}\sin2A$. Here $A = 4x$, so $\sin4x\cos4x=\frac{1}{2}\sin8x$. Then the integrand $2\sin^{3}4x\cos^{3}4x = 2(\sin4x\cos4x)^{3}=2(\frac{1}{2}\sin8x)^{3}=\frac{1}{4}\sin^{3}8x$.
So the integral becomes $\int\frac{1}{4}\sin^{3}8x dx=\frac{1}{4}\int\sin^{3}8x dx$.

Step2: Rewrite $\sin^{3}8x$

We know that $\sin^{3}\theta=\sin\theta(1 - \cos^{2}\theta)$. Let $\theta = 8x$, then $\sin^{3}8x=\sin8x(1 - \cos^{2}8x)$.
The integral is $\frac{1}{4}\int\sin8x(1 - \cos^{2}8x)dx$.

Step3: Use substitution

Let $u=\cos8x$, then $du=- 8\sin8x dx$, and $\sin8x dx=-\frac{1}{8}du$.
The integral $\frac{1}{4}\int\sin8x(1 - \cos^{2}8x)dx=\frac{1}{4}\int(1 - u^{2})(-\frac{1}{8})du=-\frac{1}{32}\int(1 - u^{2})du$.

Step4: Integrate term - by - term

Integrate $-\frac{1}{32}\int(1 - u^{2})du=-\frac{1}{32}(u-\frac{u^{3}}{3})+C$.

Step5: Substitute back

Substitute $u = \cos8x$ back into the result: $-\frac{1}{32}(\cos8x-\frac{\cos^{3}8x}{3})+C$.

Answer:

$-\frac{1}{32}(\cos8x-\frac{\cos^{3}8x}{3})+C$