Solve ∫ (1-x^2)^3 wrt x

$$\int ( 1 - x ^ { 2 } ) ^ { 3 } d x$$

Evaluate

$$-\frac{x^{7}}{7}+\frac{3x^{5}}{5}-x^{3}+x+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(1-x^{2}\right)^{3}.

$$\int 1-3x^{2}+3\left(x^{2}\right)^{2}-\left(x^{2}\right)^{3}\mathrm{d}x$$

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

$$\int 1-3x^{2}+3x^{4}-\left(x^{2}\right)^{3}\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

$$\int 1-3x^{2}+3x^{4}-x^{6}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 1\mathrm{d}x+\int -3x^{2}\mathrm{d}x+\int 3x^{4}\mathrm{d}x+\int -x^{6}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}x-3\int x^{2}\mathrm{d}x+3\int x^{4}\mathrm{d}x-\int x^{6}\mathrm{d}x$$

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

$$x-3\int x^{2}\mathrm{d}x+3\int x^{4}\mathrm{d}x-\int x^{6}\mathrm{d}x$$

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 -3 times \frac{x^{3}}{3}.

$$x-x^{3}+3\int x^{4}\mathrm{d}x-\int x^{6}\mathrm{d}x$$

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}. Multiply 3 times \frac{x^{5}}{5}.

$$x-x^{3}+\frac{3x^{5}}{5}-\int x^{6}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply -1 times \frac{x^{7}}{7}.

$$x-x^{3}+\frac{3x^{5}}{5}-\frac{x^{7}}{7}$$

Simplify.

$$-\frac{x^{7}}{7}+\frac{3x^{5}}{5}-x^{3}+x$$

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

$$-\frac{x^{7}}{7}+\frac{3x^{5}}{5}-x^{3}+x+С$$

Differentiate w.r.t. x

$$-\left(x^{2}-1\right)^{3}$$