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Solution to a integration problem

B.Sudhakar
23/01/2017 0 0

∫ sin‾¹ ( x ) dx 

Let's solve this using integration by parts

Let u = sin‾¹ x ∴ du = 1/√( 1 - x² ) dx      let dv = dx   ∴ v = ∫ dx = x

Using the formula ∫ u dv = uv - ∫ v du

On substituting u , v , dv and du

∫ sin‾¹ x dx = sin‾¹ x • x - ∫ x • 1/√ ( 1 - x² ) dx  ---------------( 1 )

[ Let's integrate ∫ (x • 1)/√ ( 1 - x² ) dx of  ( 1 )  using integration by u substitution ]

[ Let u =( 1 - x² ) ∴ du = - 2 x dx   ∴ ∫ x • 1/√( 1 - x² ) dx =∫ x • 1• du / ( √u ) • ( - 2 ) x

= ( - 1/2 ) ∫ u–½ du = -1/2 • ( u ½ / (1/2) = (-1/2 • 2/1 • √u ) = - √u = - √ ( 1 - x² ) ]

∫ sin‾¹ x dx = x • sin‾¹ x - [ - √ ( 1 - x² ) ]

∴ ∫ sin‾¹ x dx = [ x sin‾¹ x + √ ( 1 - x² ) + C ]

 

 

 

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