Solution to a trigonometric problem.

∫ sin‾¹ x dx 

Let's use integration by parts

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

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

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

[ now let us use " integration by   u   substitution"  to solve ( ∫ x/√1 - x²)dx ]

let u = 1 - x² ∴ du = - 2 x dx or dx = - du/2x. ( by substituting u & du ) 

                      = x sin‾¹ x - ∫ (-x/√u • 2 • x du     ( x cancels out )

                      = x sin‾¹ x + 1/2 ∫ u¯½ du.          [ by power rule, x^n = (x^n+1)/(n+1)]

                      = x sin‾¹ x + 1/2 [ u½/½ ] +C

                      = x sin‾¹ x + 1/2 • 2/1 [ √u ] + C

                      = x sin‾¹ x + 1 • √u + C.               ( Substituting back u = 1 - x² )

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