How do you integrate $\int \frac{x^{3}}{\sqrt{144 - x^{2}}} d x$ $int x^3/sqrt(144-x^2)dx$ using trigonometric substitution?

Question

How do you integrate $\int \frac{x^{3}}{\sqrt{144 - x^{2}}} d x$ $int x^3/sqrt(144-x^2)dx$ using trigonometric substitution?

1 Answer

Impact of this question

4191 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-07T21:36:09

$- \frac{1}{3} \sqrt{144 - x^{2}} (x^{2} + 288) + C$ $-1/3sqrt(144-x^2)(x^2+288)+C$

Explanation:

Use the substitution $x = 12 sin θ$ $x=12sintheta$ . From here we see that $d x = 12 cos θ d θ$ $dx=12costhetad theta$ . Substituting:

$\int \frac{x^{3}}{\sqrt{144 - x^{2}}} d x = \int \frac{1728 {sin}^{3} θ (12 cos θ)}{\sqrt{144 - 144 {sin}^{2} θ}} d θ$ $intx^3/sqrt(144-x^2)dx=int(1728sin^3theta(12costheta))/sqrt(144-144sin^2theta)d theta$

Factoring $\sqrt{\frac{1}{144}} = \frac{1}{12}$ $sqrt(1/144)=1/12$ from the square root:

$= \int \frac{1728 {sin}^{3} θ (12 cos θ)}{12 \sqrt{1 - {sin}^{2} θ}} d θ$ $=int(1728sin^3theta(12costheta))/(12sqrt(1-sin^2theta))d theta$

Canceling the $12$ $12$ s and recalling that since ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$ , we know that $cos θ = \sqrt{1 - {sin}^{2} θ}$ $costheta=sqrt(1-sin^2theta)$ :

$= 1728 \int \frac{{sin}^{3} θ cos θ}{cos θ} d θ = 1728 \int {sin}^{3} θ d θ$ $=1728int(sin^3thetacostheta)/costhetad theta=1728intsin^3thetad theta$

We will use here ${sin}^{2} θ = 1 - {cos}^{2} θ$ $sin^2theta=1-cos^2theta$ :

$= 1728 \int {sin}^{2} θ sin θ d θ = 1728 \int (1 - {cos}^{2} θ) sin θ d θ$ $=1728intsin^2thetasinthetad theta=1728int(1-cos^2theta)sinthetad theta$

Now, using the substitution $u = cos θ$ $u=costheta$ , so that $d u = - sin θ d θ$ $du=-sinthetad theta$ :

$= - 1728 \int (1 - u^{2}) d u = 1728 \int u^{2} d u - 1728 \int d u$ $=-1728int(1-u^2)du=1728intu^2du-1728intdu$

Integrating using the power rule for integration:

$= 1728 (\frac{u^{3}}{3}) - 1728 u = 576 u^{3} - 1728 u$ $=1728(u^3/3)-1728u=576u^3-1728u$

Reverse substituting with $u = cos θ$ $u=costheta$ :

$= 576 {cos}^{3} θ - 1728 cos θ$ $=576cos^3theta-1728costheta$

Because our substitution is $sin θ = \frac{x}{12}$ $sintheta=x/12$ , we should write this in terms of sine functions. That is, $cos θ = \sqrt{1 - {sin}^{2} θ}$ $costheta=sqrt(1-sin^2theta)$ and ${cos}^{3} θ = {(1 - {sin}^{2} θ)}^{\frac{3}{2}}$ $cos^3theta=(1-sin^2theta)^(3/2)$ :

$= 576 {(1 - {sin}^{2} θ)}^{\frac{3}{2}} - 1728 \sqrt{1 - {sin}^{2} θ}$ $=576(1-sin^2theta)^(3/2)-1728sqrt(1-sin^2theta)$

Factoring:

$= \sqrt{1 - {sin}^{2} θ} (576 (1 - {sin}^{2} θ) - 1728)$ $=sqrt(1-sin^2theta)(576(1-sin^2theta)-1728)$

Now using $sin θ = \frac{x}{12}$ $sintheta=x/12$ :

$= \sqrt{1 - \frac{x^{2}}{144}} (576 (1 - \frac{x^{2}}{144}) - 1728)$ $=sqrt(1-x^2/144)(576(1-x^2/144)-1728)$

$= \sqrt{\frac{144 - x^{2}}{144}} (576 - 4 x^{2} - 1728)$ $=sqrt((144-x^2)/144)(576-4x^2-1728)$

Factoring $\sqrt{\frac{1}{144}} = \frac{1}{12}$ $sqrt(1/144)=1/12$ from the square root and $- 4$ $-4$ from the parentheses:

$= - \frac{1}{3} \sqrt{144 - x^{2}} (x^{2} + 288) + C$ $=-1/3sqrt(144-x^2)(x^2+288)+C$

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{x^{3}}{\sqrt{144 - x^{2}}} d x$ $int x^3/sqrt(144-x^2)dx$ using trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫x3√144−x2dxint x^3/sqrt(144-x^2)dx using trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{x^{3}}{\sqrt{144 - x^{2}}} d x$ $int x^3/sqrt(144-x^2)dx$ using trigonometric substitution?