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High School
Calculus and Analysis
Calculus 2
Applications of integrals

kihanja20

Answered question

2021-12-11

Evaluate the following integrals.

\int \frac{x^{3}}{\sqrt{4 - x^{2}}} d x

Answer & Explanation

Neunassauk8

Beginner2021-12-12Added 30 answers

Step 1
Let the given integral is

\int \frac{x^{3}}{\sqrt{4 - x^{2}}} d x

Step 2
Explanation:
Let the integral is

\int \frac{x^{3}}{\sqrt{4 - x^{2}}} d x

Put

u = \sqrt{4 - x^{2}}

\Rightarrow d u = - \frac{x}{\sqrt{4 - x^{2}}} d x

\Rightarrow d x = - \frac{d u \sqrt{4 - x^{2}}}{x}

\int \frac{x^{3}}{\sqrt{4 - x^{2}}} d x = \int \frac{x^{3}}{\sqrt{4 - x^{2}}} \cdot \frac{- \sqrt{4 - x^{2}}}{x} d u

= \int - x^{2} d u

\int \frac{x^{3}}{\sqrt{4 - x^{2}}} d x = \int (u^{2} - 4) d u

\Rightarrow \int (u^{2} - 4) d u = \int u^{2} d u - \int 4 d u

By using the power rule,

\Rightarrow \int u^{2} - 4 d u = \frac{u^{3}}{3} - 4 u

resubstitute

u = \sqrt{4 - x^{2}}

= \frac{{(\sqrt{4 - x^{2}})}^{3}}{3} - 4 \sqrt{4 - x^{2}}

Adding the integral constant

\int \frac{x^{3}}{\sqrt{4 - x^{2}}} d x = \frac{{(\sqrt{4 - x^{2}})}^{3}}{3} - 4 \sqrt{4 - x^{2}} + C

poleglit3

Beginner2021-12-13Added 32 answers

$\int \frac{x^{3}}{\sqrt{4 - x^{2}}} d x$
Transform
$\int \frac{t - 4}{2 \sqrt{t}} d t$
$\frac{1}{2} * \int \frac{t - 4}{\sqrt{t}} d t$
$\frac{1}{2} * \int \frac{t - 4}{t^{\frac{1}{2}}} d t$
$\frac{1}{2} * \int \frac{t}{t^{\frac{1}{2}}} - \frac{4}{t^{\frac{1}{2}}} d t$
$\frac{1}{2} * \int t^{\frac{1}{2}} - \frac{4}{t^{\frac{1}{2}}} d t$
$\frac{1}{2} \cdot (\int t^{\frac{1}{2}} d t - \int \frac{4}{t^{\frac{1}{2}}} d t)$
Evaluate
$\frac{1}{2} \cdot (\frac{2 t \sqrt{t}}{3} - 8 \sqrt{t})$
Tranform back
$\frac{1}{2} \cdot (\frac{2 (4 - x^{2}) \sqrt{4 - x^{2}}}{3} - 8 \sqrt{4 - x^{2}})$
$\frac{(4 - x^{2}) \sqrt{4 - x^{2}}}{3} - 4 \sqrt{4 - x^{2}}$
Answer:
$\frac{(4 - x^{2}) \sqrt{4 - x^{2}}}{3} - 4 \sqrt{4 - x^{2}} + C$

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