Question

evaluate $int_{0}^{\frac{pi}{18}}5cos^{5}9x dx$. rewrite the given integral in a form that does not involve a power of cosine higher than 1. choose the correct answer below. a. $int_{0}^{\frac{pi}{18}}5(1 + sin9x)^{4}cos9x dx$ b. $int_{0}^{\frac{pi}{18}}5(1 + cos9x)^{4}sin9x dx$ c. $int_{0}^{\frac{pi}{18}}5(1+sin^{2}9x)^{2}cos9x dx$ d. $int_{0}^{\frac{pi}{18}}5(1 - cos9x)^{4}sin9x dx$ e. $int_{0}^{\frac{pi}{18}}5(1 - sin9x)^{4}cos9x dx$ f. $int_{0}^{\frac{pi}{18}}5(1-sin^{2}9x)^{2}cos9x dx$

Explanation:

Step1: Use the identity $\cos^{2}\theta = 1-\sin^{2}\theta$

We know that $\cos^{5}9x=\cos^{4}9x\cdot\cos9x = (\cos^{2}9x)^{2}\cdot\cos9x=(1 - \sin^{2}9x)^{2}\cdot\cos9x$.
So the integral $\int_{0}^{\pi/18}5\cos^{5}9x dx=\int_{0}^{\pi/18}5(1 - \sin^{2}9x)^{2}\cos9x dx$.

Answer:

F. $\int_{0}^{\pi/18}5(1 - \sin^{2}9x)^{2}\cos9x dx$