# calculus antiderivativewhat is antiderivative of (25-25y^2)^(1/2)?

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Calculate integral of the function to find the antiderivative.

`int sqrt(25-25y^2)dy`

Factor 25 in the expression `25-25y^2` => `25-25y^2 = 25(1-y^2) =gt sqrt (25-25y^2) = sqrt(25(1-y^2)) = 5sqrt (1-y^2)`

`int sqrt(25-25y^2)dy = 5*int sqrt (1-y^2)dy`

Put `y = sin x =gtsqrt (1-y^2) = sqrt (1-sin^2 x) = +- cos x`

Find dx => dy = cos x dx

`5* int sqrt (1-sin^2 x)*cos x dx = 5* int cos x*cos x dx = 5*int cos^2 x dx`

Use the formula `cos ^2x = (1+cos 2x)/2`

`5*int cos^2 x dx =(5/2)*int(1+cos 2x) dx`

`5*int cos^2 x dx = (5/2)*int dx + (5/2)*int cos 2x dx`

`5*int cos^2 x dx = (5/2)*x + (5/2)*[(sin 2x)/2] + c`

y = sin x => x = arcsin y

`int sqrt(25-25y^2)dy = (5/2)*(arcsin y) + (5/4)*(sin 2arcsin y) + c`

**ANSWER: Antiderivative **

**`int sqrt(25-25y^2)dy = (5/2)*(arcsin y) + (5/4)*(sin 2arcsin y) + c` **