# Evaluate the integral: {eq}\displaystyle \int x^2 \sqrt{1 - x^2}\ dx {/eq}

## Question:

Evaluate the integral:

{eq}\displaystyle \int x^2 \sqrt{1 - x^2}\ dx {/eq}

## Indefinite Integral:

Integration is the most useful method of mathematics. It can be classified into two parts.

Definite Integral

Indefinite Integral

The indefinite integral is the set of all antiderivatives of a model and definite integral gives the exact value of a function.

Given

• The integral is {eq}\int {{x^2}\sqrt {1 - {x^2}} dx} {/eq}.

Suppose {eq}x = \sin u {/eq}, then {eq}dx = \cos udu {/eq}.

Thus, the integral can be re-written in form as,

{eq}\begin{align*} \int {{x^2}\sqrt {1 - {x^2}} dx} &= \int {{{\left( {\sin u} \right)}^2}\sqrt {1 - {{\sin }^2}u} \left( {\cos udu} \right)} \\ \int {{x^2}\sqrt {1 - {x^2}} dx} &= \int {{{\left( {\sin u} \right)}^2}\sqrt {{{\cos }^2}u} \left( {\cos udu} \right)} \\ \int {{x^2}\sqrt {1 - {x^2}} dx} &= \int {{{\sin }^2}u{{\cos }^2}udu} \\ \int {{x^2}\sqrt {1 - {x^2}} dx} &= \int {\frac{{\left( {1 - \cos \left( {4u} \right)} \right)}}{8}du} \\ \int {{x^2}\sqrt {1 - {x^2}} dx} &= \int {\frac{1}{8}du} - \int {\frac{{\cos \left( {4u} \right)}}{8}du} \\ \int {{x^2}\sqrt {1 - {x^2}} dx} &= \frac{1}{8}\int {du} - \frac{1}{8}\int {\cos \left( {4u} \right)du} \end{align*} {/eq}

Integrate the above integral.

{eq}\int {{x^2}\sqrt {1 - {x^2}} dx} = \frac{1}{8}u - \frac{1}{{32}}\sin \left( {4u} \right) + C {/eq}

Substitute {eq}{\sin ^{ - 1}}x = u {/eq} in the above solution of the integral.

{eq}\int {{x^2}\sqrt {1 - {x^2}} dx} = \frac{1}{8}\left( {{{\sin }^{ - 1}}x} \right) - \frac{1}{{32}}\sin \left( {4\left( {{{\sin }^{ - 1}}x} \right)} \right) + C {/eq}

Therefore, the solution of the given integral is {eq}\frac{1}{8}\left( {{{\sin }^{ - 1}}x} \right) - \frac{1}{{32}}\sin \left( {4\left( {{{\sin }^{ - 1}}x} \right)} \right) + C {/eq}.