First slide
Methods of integration
Question

 dxa2b2x23/2 equals

Easy
Solution

Put x=absinθdx=abcosθdθ
dxa2b2x23/2=abcosθdθa3cos3θ=1a2btanθ+C                               =1a2bsinθ1sin2θ+C                               =1a2bbxa1b2a2x2+C=1a2xa2b2x2+C

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