What is the integral of (9-y^2)^2 ?

The integral of (9-y^2)^2 is 81y-6y^3+(y^5)/5+C

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integral of (9-y^2)^2

\int (9 - y^{2})^{2} d y

81 y - 6 y^{3} + \frac{y ^{5}}{5} + C

Solution steps

\int (9 - y^{2})^{2} d y

Expand

(9 - y^{2})^{2} : 81 - 18 y^{2} + y^{4}

= \int 81 - 18 y^{2} + y^{4} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 81 d y - \int 18 y^{2} d y + \int y^{4} d y

\int 81 d y = 81 y

\int 18 y^{2} d y = 6 y^{3}

\int y^{4} d y = \frac{y ^{5}}{5}

= 81 y - 6 y^{3} + \frac{y ^{5}}{5}

Add a constant to the solution

= 81 y - 6 y^{3} + \frac{y ^{5}}{5} + C

What is the integral of (9-y^2)^2 ?
The integral of (9-y^2)^2 is 81y-6y^3+(y^5)/5+C