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1)

$\int{ arcsin^2{2x} } \, dx = x \cdot arcsin^2{2x} - \int{x} \, d(arcsin^2{2x}) =$

$= x \cdot arcsin^2{2x} - \int{ 2 x \cdot arcsin{2x} } \, d(arcsin{2x}) =$

$= x \cdot arcsin^2{2x} - \int{ arcsin{2x} \cdot \sin{ ( arcsin{2x} ) } } \, d(arcsin{2x}) =$

$= x \cdot arcsin^2{2x} + \int{ arcsin{2x} } \, d \cos{ ( arcsin{2x} ) } =$

$= x \cdot arcsin^2{2x} + arcsin{2x} \cdot \cos{ ( arcsin{2x} ) } - \int{ \cos{ ( arcsin{2x} ) } } \, d(arcsin{2x}) =$

$= x \cdot arcsin^2{2x} + arcsin{2x} \cdot \sqrt{ 1 - 4x^2 } - \sin{ ( arcsin{2x} ) } + C =$

$= arcsin{2x} ( x \cdot arcsin{2x} + \sqrt{ 1 - 4x^2 } ) - 2x + C \ ;$

поскольку вещественная $D ( arcsin^2{2x} ) \equiv [ -\frac{1}{2} ; \frac{1}{2} ] \ ,$ то значение интеграла на вещественных значениях функции:

$\int\limits_0^1{ arcsin^2{2x} } \, dx \ \Rightarrow \ \int\limits_0^{1/2}{ arcsin^2{2x} } \, dx = \\\\ = arcsin{2x} ( x \cdot arcsin{2x} + \sqrt{ 1 - 4x^2 } ) |_0^{1/2} + 2x |_{1/2}^0 = \\\\ = ( arcsin{ ( 2 \cdot \frac{1}{2} ) } \cdot ( \frac{1}{2} \cdot arcsin{ ( 2 \cdot \frac{1}{2} ) } + \sqrt{ 1 - 4 ( \frac{1}{2} )^2 } ) - \\ - arcsin{ ( 2 \cdot 0 ) } \cdot ( 0 \cdot arcsin{ ( 2 \cdot 0 ) } + \sqrt{ 1 - 4 \cdot 0^2 } ) ) + 2 \cdot ( 0 - \frac{1}{2} ) = \\\\ = ( arcsin{1} \cdot ( \frac{1}{2} \cdot arcsin{1} + \sqrt{ 1 - 4 \cdot \frac{1}{4} } ) - arcsin{0} \cdot ( 0 \cdot arcsin{0} + \sqrt{ 1 - 4 \cdot 0 } ) ) - 2 \cdot \frac{1}{2} = \\\\ = ( \frac{ \pi }{2} ( \frac{1}{2} \cdot \frac{ \pi }{2} + \sqrt{ 1 - 1 } ) - 0 \cdot ( 0 \cdot 0 + \sqrt{ 1 - 0 } ) ) - 1 = \\\\ = ( \frac{ \pi }{2} ( \frac{ \pi }{4} + \sqrt{ 0 } ) - 0 \cdot ( 0 + \sqrt{1} ) ) - 1 = ( \frac{ \pi }{2} ( \frac{ \pi }{4} + 0 ) - 0 \cdot ( 0 + 1 ) ) - 1 = \\\\ = ( \frac{ \pi }{2} \cdot \frac{ \pi }{4} - 0 \cdot 1 ) - 1 = ( \frac{ \pi }{2} \cdot \frac{ \pi }{4} - 0 ) - 1 = \frac{ \pi }{2} \cdot \frac{ \pi }{4} - 1 = \frac{ \pi^2 }{8} - 1 \ ;$

О т в е т : $\int\limits_0^1{ arcsin^2{2x} } \, dx \ \Rightarrow \ \int\limits_0^{1/2}{ arcsin^2{2x} } \, dx = \frac{ \pi^2 }{8} - 1 \ ;$

2)

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