question 1 - 1 point evaluate the indefinite integral given below. ∫4cos(7x)sin³(7x)dx provide your answer…

Explanation:

Step1: Use substitution

Let $u = \sin(7x)$, then $du=7\cos(7x)dx$, and $\cos(7x)dx=\frac{1}{7}du$. The integral $\int4\cos(7x)\sin^{3}(7x)dx$ becomes $\int4u^{3}\times\frac{1}{7}du$.

Step2: Simplify the integral

$\int4u^{3}\times\frac{1}{7}du=\frac{4}{7}\int u^{3}du$.

Step3: Integrate $u^{3}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n = 3$, we have $\frac{4}{7}\times\frac{u^{4}}{4}+C$.

Step4: Substitute back $u=\sin(7x)$

$\frac{4}{7}\times\frac{\sin^{4}(7x)}{4}+C=\frac{1}{7}\sin^{4}(7x)+C$.

Answer:

$\frac{1}{7}\sin^{4}(7x)+C$