Worksheet__1.7

# Worksheet__1.7 - d dx sin-1 x = 1 cos(sin-1 x = 1 q 1-x 2...

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Calculus II Spring 2009 Worksheet 1.7 In terms of elementary functions evaluate the indeﬁnite integral Z sin - 1 ( x ) dx where it is understood that sin - 1 ( x ) is deﬁned for - 1 < x < 1. Solution. Using the integration by parts formula Z udv = uv - Z vdu and using the fact that d dx sin - 1 ( x ) = 1 1 - x 2 , - 1 < x < 1 one writes Z sin - 1 ( x ) dx = Z sin - 1 ( x ) · 1 dx = Z sin - 1 ( x ) d dx xdx = x sin - 1 ( x ) - Z x 1 1 - x 2 dx = x sin - 1 ( x ) + 1 2 Z x 1 1 - x 2 ( - 2 x ) dx = x sin - 1 ( x ) + 1 2 (1 - x 2 ) 1 2 2 + C = x sin - 1 ( x ) + 1 2 p 1 - x 2 + C. sin(sin - 1 ( x )) = x differentiate cos (sin - 1 ( x )) d dx sin - 1 ( x ) = 1
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Unformatted text preview: d dx sin-1 ( x ) = 1 cos (sin-1 ( x )) = 1 q 1-x 2 . Thus R sin-1 ( x ) dx = x sin-1 ( x ) + 1 2 √ 1-x 2 + C. It is noted that even though the function sin-1 ( x ) is deﬁned on the closed interval [-1 , 1], its antiderivatives x sin-1 ( x ) + 1 2 √ 1-x 2 + C are only diﬀerentiable on the open interval (-1 , 1). 1...
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