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 ]
