What is the integral of (1-y^2)^2 ?

The integral of (1-y^2)^2 is y-(2y^3)/3+(y^5)/5+C

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integral of (1-y^2)^2

\int (1 - y^{2})^{2} d y

y - \frac{2 y ^{3}}{3} + \frac{y ^{5}}{5} + C

Solution steps

\int (1 - y^{2})^{2} d y

Expand

(1 - y^{2})^{2} : 1 - 2 y^{2} + y^{4}

= \int 1 - 2 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 1 d y - \int 2 y^{2} d y + \int y^{4} d y

\int 1 d y = y

\int 2 y^{2} d y = \frac{2 y ^{3}}{3}

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

= y - \frac{2 y ^{3}}{3} + \frac{y ^{5}}{5}

Add a constant to the solution

= y - \frac{2 y ^{3}}{3} + \frac{y ^{5}}{5} + C

y = \frac{1}{( 7 x + 12 ) ^{3}}

\frac{d}{d x} (\frac{1}{3} x^{3} + x^{2} - x + 5)

n \to \infty lim (sin (\frac{nπ}{2}))

\int \frac{e ^{2 x}}{1 - e ^{2 x}} d x

What is the integral of (1-y^2)^2 ?
The integral of (1-y^2)^2 is y-(2y^3)/3+(y^5)/5+C