Let ∫ (x¹/²)/(√1-x³) dx = ⅔ gof(x) + c and f(x) = x³/², then the value of g(0) is) + c and f(x) = x³/², then the value of g(0) is
Let ∫ (x¹/²)/(√1-x³) dx = ⅔ gof(x) + c and f(x) = x³/², then the value of g(0) is) + c and f(x) = x³/², then the value of g(0) is
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1 Answer
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Put x³/² = t
√xdx = (2/3)dt
⇒ (2/3) ∫ dt/√ (1-t²) = (2/3)sin? ¹t + c
= (2/3)sin? ¹x³/² + c
⇒ g (x) = sin? ¹x
g (0) = 0
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