Question

1-неопределенный интеграл
2-определенный интеграл​

Answer 1

Ответ:

1

$\int\limits \frac{ {ctg}^{3}(6x) }{ \sin {}^{2} (6x) } dx = \frac{1}{6}\int\limits \frac{ {ctg}^{3}(6x) }{ \sin {}^{2} (6x) } d(6x) = \\ = - \frac{1}{6} \int\limits {ctg}^{3} (6x)d(ctg(6x)) = \\ = - \frac{1}{6} \times \frac{ {ctg}^{4}(6x) }{4} +C = - \frac{ {ctg}^{4}(6x) }{24} + C$

3

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