Question

Нужна в решении)) ( см закрепленное фото)

Answer 1

а

$\int\limits^{ - 9} _ { - 10} \frac{ \sqrt{x + 10} }{ \sqrt{x + 10} + 1} dx \\ \\ \sqrt{x + 10} = t \\ x + 10 = {t}^{2} \\ dx = 2tdt \\ t_1 = \sqrt{10 - 9} = 1 \\ t_2 = \sqrt{10 - 10} = 0 \\ \\ \int\limits^{ 1 } _ {0} \frac{t \times 2tdt}{t + 1} = 2\int\limits^{ 1} _ {0} \frac{ {t}^{2} dt}{t + 1} = \\ = 2\int\limits^{ 1 } _ {0}(t - 1 + \frac{1}{t + 1}) dt = \\ = 2( \frac{ {t}^{2} }{2} - t + ln |1 + t| ) | ^{ 1} _ {0} = \\ = 2( \frac{1}{2} - 1 + ln(2) - 0 - 0 - 0) = \\ = 1 - 2 + 2 ln(2) = 2 ln(2) - 1$

б

$\int\limits^{ 2} _ { \sqrt{2} } \frac{dx}{ \sqrt{ {( {x}^{2} - 1) }^{3} } } \\ \\ x {}^{2} = \frac{1}{ \cos {}^{2} (t) } \\ t = arccos( \frac{1}{ {x}^{} } ) \\ {x }^{2} - 1 = \frac{1}{ \cos {}^{2} (t) } - 1 = {tg}^{2} t \\ dx = - ( \cos(t) ) {}^{ - 2} \times ( - \sin(t)) dt \\ dx = \frac{tgt}{ \cos(t) } dt \\ t_1 = arccos( \frac{1}{2}) = \frac{\pi}{3} \\ t_2 = arccos( \frac{1}{ \sqrt{2} } ) = \frac{\pi}{4} \\ \\ \int\limits^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } \frac{1}{ {tg}^{3}t } \times \frac{tgt}{ \cos(t) } = \int\limits^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } \frac{1}{ {tg}^{2} t} \times \frac{dt}{ \cos(t) } = \\ = \int\limits^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } \frac{1}{ \frac{1}{ \cos {}^{2} (t) } - 1 } \times \frac{dt}{ \cos(t) } = \int\limits^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } \frac{ \cos {}^{2} (t)}{ \sin {}^{2} (t) } \times \frac{dt}{ \cos(t) } = \\ = \int\limits^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } \frac{ \cos(t) dt}{ \sin {}^{2} (t) } = \int\limits^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } \frac{d( \sin {}^{} (t) )}{ \sin{}^{2} (t) } = \\ = - \frac{ (\sin(t)) {}^{ - 1} }{ - 1} |^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } = - \frac{1}{ \sin(t) } |^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } = \\ = - \frac{1}{ \sin( \frac{\pi}{3} ) } + \frac{1}{ \sin( \frac{\pi}{4} ) } = - \frac{2}{ \sqrt{3} } + \frac{2}{ \sqrt{2} } = \\ = \frac{2}{ \sqrt{3} } - \sqrt{2}$