Solve ∫ ((5-y^2)^2-1^2) wrt y

$$\int ( ( 5 - y ^ { 2 } ) ^ { 2 } - 1 ^ { 2 } ) d y$$

Evaluate

$$\frac{y^{5}}{5}-\frac{10y^{3}}{3}+24y+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-y^{2}\right)^{2}.

$$\int 25-10y^{2}+\left(y^{2}\right)^{2}-1^{2}\mathrm{d}y$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int 25-10y^{2}+y^{4}-1^{2}\mathrm{d}y$$

Calculate 1 to the power of 2 and get 1.

$$\int 25-10y^{2}+y^{4}-1\mathrm{d}y$$

Subtract 1 from 25 to get 24.

$$\int 24-10y^{2}+y^{4}\mathrm{d}y$$

Integrate the sum term by term.

$$\int 24\mathrm{d}y+\int -10y^{2}\mathrm{d}y+\int y^{4}\mathrm{d}y$$

Factor out the constant in each of the terms.

$$\int 24\mathrm{d}y-10\int y^{2}\mathrm{d}y+\int y^{4}\mathrm{d}y$$

Find the integral of 24 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$24y-10\int y^{2}\mathrm{d}y+\int y^{4}\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{2}\mathrm{d}y with \frac{y^{3}}{3}. Multiply -10 times \frac{y^{3}}{3}.

$$24y-\frac{10y^{3}}{3}+\int y^{4}\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{4}\mathrm{d}y with \frac{y^{5}}{5}.

$$24y-\frac{10y^{3}}{3}+\frac{y^{5}}{5}$$

If F\left(y\right) is an antiderivative of f\left(y\right), then the set of all antiderivatives of f\left(y\right) is given by F\left(y\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$24y-\frac{10y^{3}}{3}+\frac{y^{5}}{5}+С$$

Differentiate w.r.t. y

$$y^{4}-10y^{2}+24$$