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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.

Answer and Explanation: 1


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}.


Learn more about this topic:

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Indefinite Integrals as Anti Derivatives

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Chapter 12 / Lesson 11
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Indefinite integrals, an integral of the integrand which does not have upper or lower limits, can be used to identify individual points at specific times. Learn more about the fundamental theorem, use of antiderivatives, and indefinite integrals through examples in this lesson.


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