Solve ∫ (from 0 to 2 pi) of (x^2-1)^2 wrt x

$$\int _ { 0 } ^ { 2 \pi } ( x ^ { 2 } - 1 ) ^ { 2 } d x$$

Evaluate

$$\frac{32\pi ^{5}}{5}-\frac{16\pi ^{3}}{3}+2\pi \approx 1799.442358971$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-1\right)^{2}.

$$\int _{0}^{2\pi }\left(x^{2}\right)^{2}-2x^{2}+1\mathrm{d}x$$

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

$$\int _{0}^{2\pi }x^{4}-2x^{2}+1\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{4}-2x^{2}+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int -2x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x-2\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{x^{5}}{5}-2\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{x^{5}}{5}-\frac{2x^{3}}{3}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}-\frac{2x^{3}}{3}+x$$

Simplify.

$$x-\frac{2x^{3}}{3}+\frac{x^{5}}{5}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$2\pi -\frac{2}{3}\times \left(2\pi \right)^{3}+\frac{1}{5}\times \left(2\pi \right)^{5}-\left(0-\frac{2}{3}\times 0^{3}+\frac{0^{5}}{5}\right)$$

Simplify.

$$2\pi -\frac{16\pi ^{3}}{3}+\frac{32\pi ^{5}}{5}$$