## Calculus: Early Transcendentals 8th Edition

$\frac{1}{2}ln(2) - \frac{1}{2}$
Evaluate the integral: $\int sin(x)ln(cos(x))dx$ with an area of x from 0 to $\pi/3$ (You can't add the limits to the integral sign, yet.) Use the following rule: $\int u \times dv = u\times v-\int v\times du$ where: $u = ln(cos(x))$ $du = \frac{-sin(x)}{cos(x)}$ $dv = sin(x) dx$ $v = -cos(x)$ $\int sin(x)ln(cos(x))dx = [ln(cos(x)) \times - cos(x)] - \int -cos(x) \times \frac{-sin(x)}{cos(x)} dx = [-cos(x)ln(cos(x))] - \int sin(x) dx = [-cos(x)ln(cos(x))] - [-cos(x)] = [cos(x) - cos(x)ln(cos(x))]$ Now adding in the limits 0 to $\frac{\pi}{3}$ gives us: $(cos(\frac{\pi}{3}) - cos(\frac{\pi}{3})ln(cos(\frac{\pi}{3})) - (cos(0) - cos(0)ln(cos(0)) = \frac{1}{2} - \frac{1}{2}ln(\frac{1}{2}) - (1 - ln(1)) = \frac{1}{2} - \frac{1}{2}ln(\frac{1}{2}) -1 + ln(1) = -\frac{1}{2} + \frac{1}{2}ln(2) = \frac{1}{2}ln(2) - \frac{1}{2}$