Question

Решение интегралов
С решением

Answer 1

1 вариант

1.

$\int\limits^{ \frac{\pi}{2} } _ {0}2 \sin {}^{5} (x) \cos(x)dx = 2 \int\limits^{ \frac{\pi}{2} } _ {1} \sin {}^{5} (x)d( \sin(x)) = \\ = \frac{2 \sin {}^{6} (x) }{6} | ^{ \frac{\pi}{2} } _ {0} = \frac{ \sin {}^{6} (x) }{3} | ^{ \frac{\pi}{2} } _ {0} = \\ = \frac{1}{3} (1 - 0) = \frac{1}{3}$

2.

$\int\limits^{ 0 } _ { - 1}3 {x}^{2}( {x}^{3} + 5) {}^{2} dx = \int\limits^{ 0 } _ { - 1}( {x}^{3} + 5) {}^{2} d( {x}^{3} + 5) = \\ = \frac{ {( {x}^{3} + 5)}^{3} }{3} | ^{ 0 } _ { - 1} = \frac{ {5}^{3} }{3} - \frac{ {4}^{3} }{3} = \frac{125 - 64}{3} = \\ = \frac{61}{3}$

3.

$\int\limits^{ \frac{\pi}{8} } _ {0}x \sin(2x) dx \\ \\ \\ u = x \: \: \: \: du = dx \\ dv = \sin(2x) dx \: \: \: v = - \frac{1}{2} \cos(2x ) \\ \\ - \frac{x}{2} \cos(2x) + \frac{1}{2} \int\limits \cos(2x) dx + c= \\ = - \frac{x}{2} \cos(2x) + \frac{1}{4} \sin(2x) + c \\ \\ ( - \frac{x}{2} \cos(2x) + \frac{1}{4} \sin(2x)) | ^{ \frac{\pi}{8} } _ {0} = \\ = - \frac{\pi}{16} \cos( \frac{\pi}{4} ) + \frac{1}{4} \sin( \frac{\pi}{4} ) + 0 - 0 = \\ = - \frac{\pi}{16} \times \frac{ \sqrt{2} }{2} + \frac{1}{4} \times \frac{ \sqrt{2} }{2} = - \frac{\pi \sqrt{2} }{32} + \frac{ \sqrt{2} }{8}$

3 вариант

1.

$\int\limits^{ 0} _ { - 1}(1 + 2x) {}^{4} dx = \frac{1}{2} \int\limits^{ 0 } _ { - 1} {(1 + 2x) {}^{4} }^{} d(1 + 2x) = \\ = \frac{ {(1 + 2x)}^{5} }{10} | ^{ 0} _ { - 1} = \frac{1}{10} - ( - \frac{1}{10} ) = \frac{1}{5} = 0.2$

2.

$\int\limits^{ 0 } _ { - 1}x( {x}^{2} + 3) {}^{5} dx = \frac{1}{2} \int\limits^{ 0} _ { - 1}2x( {x}^{2} + 3) {}^{5} dx = \\ = \frac{1}{2} \int\limits^{ 0 } _ {1}( {x}^{2} + 3) {}^{5} d( {x}^{2} + 3) = \frac{ {( {x}^{2} + 3)}^{6} }{12} | ^{ 0} _ { - 1} = \\ = \frac{ {3}^{6} }{12} - \frac{4 {}^{6} }{12} = \frac{729 - 4096}{12} = - \frac{3367}{12}$

3.

$\int\limits^{ 1 } _ {0}arctgxdx \\ \\ u = arctgx \: \: \: \: \: \: \: du = \frac{dx}{1 + {x}^{2} } \\ dv = dx \: \: \: \: \: \: \: \: \: v = x \\ \\ xarctgx - \int\limits \frac{xdx}{x {}^{2} + 1} = xarctgx - \frac{1}{2} \int\limits \frac{2xdx}{ {x}^{2} + 1} = \\ = xarctgx - \int\limits\frac{d( {x}^{2} + 1)}{ {x}^{2} + 1 } = xarctgx - ln( | {x}^{2} + 1| ) + c \\ \\ (xarctgx - ln( | {x}^{2} + 1 | ) )| ^{ 1} _ {0} = \\ = 1 \times \frac{\pi}{4} - ln(2) - 0 + 0 = \frac{ \pi}{4} - ln(2)$