Question

Вычислите интеграл, преобразуя подынтегральные функции

Answer 1

1) $6*2sin(\frac{ \pi }{8}-x)*cos(\frac{ \pi }{8}-x)=6sin(2*(\frac{ \pi }{8}-x))=6sin(\frac{ \pi }{4}-2x)$
$\int\limits^ \frac{3 \pi }{8}_\frac{\pi }{8} {6sin(\frac{ \pi }{4}-2x)} \, dx=-6*0.5*\int\limits^ \frac{3 \pi }{8}_\frac{\pi }{8} {sin(\frac{ \pi }{4}-2x)} \, d(\frac{ \pi }{4}-2x)=3*cos(\frac{ \pi }{4}-2x)$$=3*cos(\frac{ \pi }{4}-\frac{6\pi }{4})-3*cos(\frac{ \pi }{4}-\frac{\pi }{4})=3*cos(-\frac{5\pi }{4})-3=$$=3*(cos(\frac{5\pi }{4})-1)=3*(cos( \pi +\frac{\pi }{4})-1)=3*(-cos(\frac{\pi }{4})-1)=$$=-3*(\frac{\sqrt{2}}{2}-\frac{2}{2})=-\frac{3*(\sqrt{2}-2)}{2}=\frac{3*(2-\sqrt{2})}{2}$

2) $cos^{2}(x+\frac{ \pi }{3})-sin^{2}(x+\frac{ \pi }{3})=cos(2*(x+\frac{ \pi }{3}))=cos(2x+\frac{2\pi }{3})$

$\int\limits^\frac{\pi }{3}_\frac{\pi }{6} {cos(2x+\frac{2\pi }{3})} \, dx=0.5*\int\limits^\frac{\pi }{3}_\frac{\pi }{6} {cos(2x+\frac{2\pi }{3})} \, d(2x+\frac{2\pi }{3})=0.5*sin(2x+\frac{2\pi }{3})=0.5sin(\frac{2\pi }{3}+\frac{2\pi }{3})-0.5*sin(\frac{2\pi }{6}+\frac{2\pi }{3})=0.5*sin(\frac{4\pi }{3})-0.5*sin(\frac{\pi }{3}+\frac{2\pi }{3})=0.5*sin( \pi+\frac{\pi }{3})-0.5*sin( \pi)=-0.5*sin(\frac{\pi }{3})=-\frac{\sqrt{3}}{4}$