Question

Вычислите ...................................

Answer 1

Ответ:

1.

$\int\limits^{ \frac{\pi}{2} } _ {0} \cos {}^{2} ( \frac{x}{2} )dx = \int\limits^{ \frac{\pi}{2} } _ {0} \frac{1 + \cos(x) }{2}dx = \\ = \frac{1}{2}( \int\limits^{ \frac{\pi}{2} } _ {0} \: dx + \int\limits^{ \frac{\pi}{2} } _ {0} \cos(x) dx) = \\ = ( \frac{x}{2} + \frac{1}{2} \sin(x))| ^{ \frac{\pi}{2} } _ {0} = \frac{\pi}{4} + \frac{1}{2} - 0 - 0 = \frac{\pi}{4} + \frac{1}{2}$

2.

$\int\limits^{ 0} _ { - \frac{\pi}{2} } \sin {}^{2} ( \frac{x}{2} )dx = \int\limits^{ 0 } _ { - \frac{\pi}{2} } \frac{1 - \cos(x) }{2}dx = \\ = \frac{1}{2} ( \int\limits^{ 0 } _ { - \frac{\pi}{2} }dx - \int\limits^{ 0} _ {1} \cos(x) dx) = \\ = ( \frac{x}{2} - \frac{1}{2} \sin(x)) | ^{ 0 } _ { - \frac{\pi}{2} } = \\ = 0 - 0 - ( - \frac{\pi}{4} + \frac{1}{2} ) = \frac{\pi}{4} - \frac{1}{2}$

3.

$\int\limits^{ \pi } _ {0}3 \cos {}^{2} (2x) dx = 3\int\limits^{ \pi } _ {0} \frac{1 + \cos(4x) }{2}dx = \\ = \frac{3}{2} \int\limits^{ \pi } _ {0}dx + \int\limits^{ \pi} _ {0} \cos(4x) dx) = \\ = ( \frac{3x}{2} + \frac{3}{8} \sin(4x)) | ^{ \pi } _ {0} = \\ = \frac{3\pi}{2} + 0 - 0 - 0 = \frac{3\pi}{2}$

4.

$\int\limits^{ \frac{\pi}{8} } _ {0} \frac{1}{4} \sin {}^{2} (4x) = \frac{1}{4} \int\limits^{ \frac{\pi}{8} } _ {0} \frac{1 - \cos(8x) }{2}dx = \\ = \frac{1}{4} ( \int\limits^{ \frac{\pi}{8} } _ {0}dx - \int\limits^{ \frac{\pi}{8} } _ {0} \cos(8x) dx) = \\ = ( \frac{x}{4} - \frac{1}{32} \sin(8x)) | ^{ \frac{\pi}{8} } _ {0} = \\ = \frac{\pi}{32} - 0 - 0 + 0 = \frac{\pi}{32}$