Question

question 7 evaluate ∫a²b³dx where a = sin(f(x)), b = cos(g(x)), and f(x)=g(x)=6x. you must show all steps. enter your final answer.

Explanation:

Step1: Substitute the functions

Since $A = \sin(6x)$, $B=\cos(6x)$, then $A^{2}B^{3}=\sin^{2}(6x)\cos^{3}(6x)$.

Step2: Rewrite $\cos^{3}(6x)$

We know that $\cos^{3}(6x)=\cos(6x)\cos^{2}(6x)=\cos(6x)(1 - \sin^{2}(6x))$. So the integral becomes $\int\sin^{2}(6x)\cos(6x)(1 - \sin^{2}(6x))dx$.

Step3: Use substitution

Let $u = \sin(6x)$, then $du=6\cos(6x)dx$, and $\cos(6x)dx=\frac{1}{6}du$.
The integral $\int\sin^{2}(6x)\cos(6x)(1 - \sin^{2}(6x))dx=\frac{1}{6}\int u^{2}(1 - u^{2})du$.

Step4: Expand the integrand

Expand $u^{2}(1 - u^{2})=u^{2}-u^{4}$. So we have $\frac{1}{6}\int(u^{2}-u^{4})du$.

Step5: Integrate term - by - term

$\frac{1}{6}\int(u^{2}-u^{4})du=\frac{1}{6}(\frac{u^{3}}{3}-\frac{u^{5}}{5})+C$.

Step6: Substitute back $u = \sin(6x)$

The result is $\frac{1}{6}(\frac{\sin^{3}(6x)}{3}-\frac{\sin^{5}(6x)}{5})+C=\frac{\sin^{3}(6x)}{18}-\frac{\sin^{5}(6x)}{30}+C$.

Answer:

$\frac{\sin^{3}(6x)}{18}-\frac{\sin^{5}(6x)}{30}+C$