Worksheet__1.7 - d dx sin-1 ( x ) = 1 cos (sin-1 ( x )) = 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 dened on the closed interval [-1 , 1], its antiderivatives x sin-1 ( x ) + 1 2 1-x 2 + C are only dierentiable on the open interval (-1 , 1). 1...
View Full Document

Ask a homework question - tutors are online