# please help

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### 1 Answer

Denote the function under integral as a(t) and its antiderivative as A(t):

`a(t) = 3sqrt(t)` , `A'(t) = a(t)`

Then `F(x) = A(sin(x)) - A(0)`

and therefore `F'(x) = A'(sin(x))*[sin(x)]' = a(sin(x))*cos(x).`

(so no need for A(x)).

The answer is `F'(x) = 3sqrt(sin(x))*cos(x).`

This formula is valid only for those x from 0 until sin(x) >= 0, i.e. for x from 0 to `pi`

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