Answer to Question #223046 in Calculus for Bawe Bilanyuy

Question

Evaluate ∫₀^π √(1+cos8xdx)

Answer 1

$\int_{0}^{\pi}\sqrt{1+cos8x}dx=\int_{0}^{\pi}\sqrt{2cos^24x}dx=\\ =\int_{0}^{\pi}\sqrt{2}|cos4x|dx=\\ =\int_{0}^{\frac{\pi}{8}}\sqrt{2}cos4xdx-\int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\sqrt{2}cos4xdx+\\ +\int_{\frac{3\pi}{8}}^{\frac{5\pi}{8}}\sqrt{2}cos4xdx-\int_{\frac{5\pi}{8}}^{\frac{7\pi}{8}}\sqrt{2}cos4xdx+\\ +\int_{\frac{7\pi}{8}}^{\pi}\sqrt{2}cos4xdx=\\ =\sqrt{2}\cdot \frac{1}{4}\cdot sin4x|^{\frac{\pi}{8}}_{0} - \sqrt{2}\cdot \frac{1}{4}\cdot sin4x|^{\frac{3\pi}{8}}_{\frac{\pi}{8}}+\\ +\sqrt{2}\cdot \frac{1}{4}\cdot sin4x|^{\frac{5\pi}{8}}_{\frac{3\pi}{8}}-\sqrt{2}\cdot \frac{1}{4}\cdot sin4x|^{\frac{7\pi}{8}}_{\frac{5\pi}{8}}+\\ +\sqrt{2}\cdot \frac{1}{4}\cdot sin4x|^{\pi}_{\frac{7\pi}{8}}=\\ =\frac{\sqrt2}{4}(sin\frac{\pi}{2}-sin0 -sin\frac{3\pi}{2}+sin\frac{\pi}{2}+\\ +sin\frac{5\pi}{2}-sin\frac{3\pi}{2}-sin\frac{7\pi}{2}+sin\frac{5\pi}{2}+\\ +sin4\pi-sin\frac{7\pi}{2})=\\ =\frac{\sqrt2}{4}(1-0+1+1+1+1+1+1+0+1)=\\ =2\sqrt2$