Question

Вычислить неопределенные интегралы.​

Answer 1

4.

a)

$\int\limits \frac{(6 - x)dx}{ \sqrt{ 4 - {x}^{2} + 9x } } \\ \\ (4 - {x}^{2} + 9x)' = - 2x + 9$

делаем в числителе

$\frac{1}{2} \int\limits \frac{( - 2x + 12)dx}{ \sqrt{4 - {x}^{2} + 9x } } + \frac{1}{2} \int\limits \frac{( - 2x + 9 + 3)}{ \sqrt{4 - {x}^{2} + 9x} } dx = \\ = \frac{1}{2} \int\limits \frac{( - 2x + 9)dx}{ \sqrt{4 - {x}^{2} + 9x} } + \frac{3}{2} \int\limits \frac{dx}{ \sqrt{4 - {x}^{2} + 9x } } = \\ = \frac{1}{2} \int\limits \frac{d(4 - {x}^{2} + 9x) }{ \sqrt{4 - {x}^{2} + 9x} } + \frac{3}{2} \int\limits \frac{dx}{ \sqrt{4 - {x}^{2} + 9x } } \\\\ \\ 4 - {x}^{2} + 9x = - ( {x}^{2} - 9x - 4) = \\ = - ( {x}^{2} - 2 \times x \times \frac{9}{2} + \frac{81}{4} - \frac{97}{4} ) = \\ = - ( {(x - \frac{9}{2} )}^{2} - {( \frac{ \sqrt{97} }{2}) }^{2} ) = {( \frac{ \sqrt{97} } {2 }) }^{2} - {(x - \frac{9}{2}) }^{2} \\ \\\\ \frac{1}{2} \times \frac{ {(4 - {x}^{2} + 9x)}^{ \frac{1}{2} } }{ \frac{1}{2} } + \frac{3}{2} \int\limits \frac{d(x - \frac{9}{2}) }{ {( \frac{ \sqrt{97} }{2}) }^{2} - {(x - \frac{9}{2} )}^{2} } = \\ = \sqrt{4 - {x}^{2} + 9x } + \frac{3}{2} \times \frac{1}{2 \times \frac{ \sqrt{97} }{2} } ln( \frac{ \frac{ \sqrt{97} }{2} - x + \frac{9}{2} }{ \frac{ \sqrt{97} }{2} + x - \frac{9}{2} } ) + C= \\ = \sqrt{4 - {x}^{2} + 9x } + \frac{3}{2 \sqrt{97} } ln( \frac{ \sqrt{97} + 9 - 2x }{ \sqrt{97} - 9 + 2x} ) + C$

б)

$\int\limits \frac{x + 1}{ {x}^{2} + 17x + 9 } dx \\ \\ ( {x}^{2} + 17x + 9)' = 2x + 17 \\ \\ \frac{1}{2} \int\limits \frac{2x + 2}{ {x}^{2} + 17x + 9} dx = \frac{1}{2} \int\limits \frac{2x + 17 - 15}{ {x}^{2} + 17x + 9 } dx = \\ = \frac{1}{2} \int\limits \frac{2x + 17}{{x}^{2} + 17x + 9} dx - \frac{15}{2} \int\limits \frac{dx}{ {x}^{2} + 17x + 9 } = \\ = \frac{1}{2} \int\limits \frac{d( {x}^{2} + 17x + 9)}{ {x}^{2} + 17x + 9 } - \frac{15}{2} \int\limits \frac{dx}{ {x}^{2} + 17x + 9 } \\ \\ \\ {x}^{2} + 17x + 9 = {x}^{2} + 2 \times x \times \frac{17}{2} + \frac{289}{4} - \frac{253}{4} = \\ {(x + \frac{17}{2} )}^{2} - {( \frac{ \sqrt{253} }{2}) }^{2} \\ \\ \\ \frac{1}{2} ln( {x}^{2} + 17x + 9 ) - \frac{15}{2} \int\limits \frac{d(x + \frac{17}{2}) }{ {(x + \frac{17}{2}) }^{2} - {( \frac{ \sqrt{253} }{2}) }^{2} } = \\ = \frac{1}{2} ln( {x}^{2} + 17x + 9) - \frac{15}{2} \times \frac{1}{2 \times \frac{ \sqrt{253} }{2} } ln( \frac{x + \frac{17}{2} - \frac{ \sqrt{253} }{2} }{x + \frac{17}{2} + \frac{ \sqrt{235} }{2} } ) + C = \\ = \frac{1}{2} ln( {x}^{2} + 17x + 9) - \frac{15}{2 \sqrt{253} } ln( \frac{2x + 17 - \sqrt{253} }{2x + 17 + \sqrt{253} } ) + C$

5.

а)

$\int\limits \cos {}^{2} (24x) \sin {}^{2} (24x) dx = \\ = \frac{1}{4} \int\limits4\cos {}^{2} (24x) \sin {}^{2} (24x) dx = \\ = \frac{1}{4} \int\limits \sin {}^{2} (48x) dx = \\ = \frac{1}{4} \int\limits \frac{1 - \cos(72x) }{2} dx = \\ = \frac{1}{8} (\int\limits \: dx - \frac{1}{72} \cos(72x) dx) = \\ = \frac{1}{8} x - \frac{1}{576} \sin(72x) + C$

б)

$\int\limits \cos {}^{5} (6 - 7x) dx = - \frac{1}{7} \int\limits \cos {}^{5} (6 - 7x) d(6 - 7x)\\ = - \frac{1}{7} \int\limits \cos {}^{4} (6 - 7x) \times \cos(6 - 7x) dx = \\ = - \frac{1}{7} \int\limits\cos {}^{4} (6 - 7x) d( \sin(6 - 7x)) = \\ = - \frac{1}{7} \int\limits {(1 - \sin {}^{2} (6 - 7x) )}^{2} d( \sin(6 - 7x)) = \\ = - \frac{1}{7} \int\limits(1 - 2 \sin {}^{2} (6 - 7x) + \sin {}^{4} (6 - 7x) )d (\sin(6 - 7x) ) = \\ = - \frac{1}{7}( \int\limits \: d( \sin(6 - 7x)) - 2\int\limits \sin {}^{2} (6 - 7x) d\sin((6 - 7x)) + \\ + \int\limits \sin {}^{4} (6 - 7x)) d(sin(6 - 7x)) = \\ = - \frac{1}{7} \sin(6 - 7x) + \frac{2 \sin {}^{3} (6 - 7x) }{27} - \frac{ \sin {}^{5} (6 - 7x) }{28} + C$