evaluate $int_{0}^{\frac{pi}{18}}5cos^{5}9x dx$.\n\ne. $u = \frac{1}{9}sin9x$\n\nrewrite the integral found…

evaluate $int_{0}^{\frac{pi}{18}}5cos^{5}9x dx$.\n\ne. $u = \frac{1}{9}sin9x$\n\nrewrite the integral found above using this change of variables.\n\n$int_{0}^{1}left(\frac{5}{9}(1 - 2u^{2}+u^{4})\right)du$\n\nevaluate the given definite integral.\n\n$int_{0}^{\frac{pi}{18}}5cos^{5}9x dx=square$\n\n(type an exact answer.)

evaluate $int_{0}^{\frac{pi}{18}}5cos^{5}9x dx$.\n\ne. $u = \frac{1}{9}sin9x$\n\nrewrite the integral found above using this change of variables.\n\n$int_{0}^{1}left(\frac{5}{9}(1 - 2u^{2}+u^{4})\right)du$\n\nevaluate the given definite integral.\n\n$int_{0}^{\frac{pi}{18}}5cos^{5}9x dx=square$\n\n(type an exact answer.)

Answer

Explanation:

Step1: Expand the integrand

We have the integral $\int_{0}^{\frac{\pi}{18}}\frac{5}{9}(1 - 2u^{2}+u^{4})du$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we integrate term - by - term.

Step2: Integrate each term

$\int_{0}^{\frac{\pi}{18}}\frac{5}{9}(1 - 2u^{2}+u^{4})du=\frac{5}{9}\left[\int_{0}^{1}1du-2\int_{0}^{1}u^{2}du+\int_{0}^{1}u^{4}du\right]$. For $\int_{0}^{1}1du=u\big|{0}^{1}=1 - 0 = 1$. For $2\int{0}^{1}u^{2}du=2\times\frac{u^{3}}{3}\big|{0}^{1}=\frac{2}{3}(1 - 0)=\frac{2}{3}$. For $\int{0}^{1}u^{4}du=\frac{u^{5}}{5}\big|_{0}^{1}=\frac{1}{5}(1 - 0)=\frac{1}{5}$.

Step3: Calculate the result

$\frac{5}{9}\left(1-\frac{2}{3}+\frac{1}{5}\right)=\frac{5}{9}\left(\frac{15 - 10+3}{15}\right)=\frac{5}{9}\times\frac{8}{15}=\frac{8}{27}$.

Answer:

$\frac{8}{27}$