Question

Вычислить определенные интегралы:

Answer 1

$\int\limits^{ 16} _ {0} \sqrt{256 - {x}^{2} } dx \\ \\ x = 16 \sin(t) \\ t = arcsin( \frac{x}{16} ) \\ dx = 16 \cos(t) dt \\ t1 = arcsin( \frac{16}{16} ) = \frac{\pi}{2} \\ t2 = arcsin(0 )= 0 \\ \\ \int\limits^{ \frac{\pi}{2} } _ {0} \sqrt{256 - 256 \sin {}^{2} (t) } \times 16 \cos(t) dt = \\ = 16\int\limits^{ \frac{\pi}{2} } _ {0}16 \cos(t) \times \cos( t ) dt = 256\int\limits^{ \frac{\pi}{2} } _ {0} \cos {}^{2} (t) dt = \\ 256\int\limits^{ \frac{\pi}{2} } _ {0} \frac{1 + \cos(2t) }{2} dt = 128\int\limits^{ \frac{\pi}{2} } _ {0}dt + 64\int\limits^{ \frac{\pi}{2} } _ {0} \cos(2t) d(2t) = \\ =( 128t + 64 \cos(2t) | ^{ \frac{\pi}{2} } _ {0} = 64\pi + 64 \cos(\pi) - 0 - 64 = \\ = 64\pi - 64 - 64 = 64\pi - 128$