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Integration by trig substitution

B.Sudhakar
06/01/2017 0 0

 Let's use trig substitution to solve a problem.

∫ x² / (√ 1 - x²) dx 

Let x = sin θ.       dx = cos θ dθ.   Substituting

∫ sin² θ cos θ / √ 1 - sin² θ dθ

= ∫ sin² θ cos θ / √ cos² θ dθ

,= ∫ sin² θ cos θ / cos θ dθ.                    ( cos θ cancels out )

∫ sin² θ dθ.                                            { sin² θ =1/2 ( 1 - cos 2θ ) }

= 1/2 ∫ (1 - cos 2θ ) dθ 

= 1/2 ∫ dθ - ∫ cos 2θ dθ

= 1/2 ( θ - 1/2 sin 2θ ) + C

= 1/2 [  arc sin x - 1/2 ( 2 sin θ cos θ ) ] +C

= 1/2 arc sin x - 1/2 x ( √1 - x² ) + C

 

 

 

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