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?dx / Root 2ax-x^2 = a^n sin^-1 [x/a-1] ,Then find the value of n Jee Question plz help

?dx / Root 2ax-x^2 = a^n sin^-1 [x/a-1] ,Then find the value of n 

Jee Question plz help



Grade:12

1 Answers

Mohammed Patel
31 Points
10 years ago

how mahn...shw the step ^

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