How do you integrate $\int \sqrt{9 - x^{2}}$ $int sqrt(9-x^2)$ by trigonometric substitution?

Question

How do you integrate $\int \sqrt{9 - x^{2}}$ $int sqrt(9-x^2)$ by trigonometric substitution?

1 Answer

Impact of this question

1754 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-03T11:04:36

$\int \sqrt{9 - x^{2}} d x = ?$ $int sqrt(9-x^2)d x=?$

$Substitute x = 3 \cdot sin θ$ $"Substitute "x=3*sin theta$

$d x = 3 \cdot cos θ \cdot d θ; θ = a r c sin (\frac{x}{3})$ $d x=3*cos theta *d theta" ; "theta =arc sin(x/3)$

$x^{2} = 9 \cdot {sin}^{2} θ$ $x^2=9*sin^2 theta$

$\int \sqrt{9 - x^{2}} d x = \int \sqrt{9 - 9 \cdot {sin}^{2} θ} \cdot 3 \cdot cos θ \cdot d θ$ $int sqrt(9-x^2)d x=int sqrt(9-9*sin^2 theta)* 3*cos theta * d theta$

$\int \sqrt{9 - x^{2}} d x = \int \sqrt{9 (1 - {sin}^{2} θ)} \cdot 3 \cdot cos θ \cdot d θ$ $int sqrt(9-x^2)d x=int sqrt(9(1-sin^2 theta))*3*cos theta*d theta$

$So ; 1 - {sin}^{2} θ = {cos}^{2} θ$ $"So ;" 1-sin^2 theta =cos ^2theta$

$\int \sqrt{9 - x^{2}} d x = \int 3 \sqrt{{cos}^{2} θ} \cdot 3 \cdot cos θ \cdot d θ$ $int sqrt(9-x^2)d x=int 3 sqrt(cos^2 theta)*3*cos theta*d theta$

$\int \sqrt{9 - x^{2}} d x = 9 \int {cos}^{2} θ \cdot d θ$ $int sqrt(9-x^2)d x=9 int cos^2 theta *d theta$

$Now use the reduction formula...$ $"Now use the reduction formula..."$
$\int {cos}^{n} θ d θ = \frac{n - 1}{n} \int {cos}^{n - 2} θ \cdot d θ + \frac{{cos}^{n - 1} θ \cdot sin θ}{n}$ $int cos^n theta d theta=(n-1)/n int cos^(n-2) theta*d theta+(cos^(n-1)theta*sin theta)/n$

$use n = 2$ $"use " n=2$

$\int {cos}^{2} θ d θ = \frac{2 - 1}{2} \int {cos}^{2 - 2} θ \cdot d θ + \frac{{cos}^{2 - 1} θ \cdot sin θ}{2}$ $int cos^2 theta d theta=(2-1)/2 int cos^(2-2) theta*d theta +(cos^(2-1) theta*sin theta)/2$

$\int {cos}^{2} θ \cdot d θ = \frac{1}{2} \int d θ + \frac{1}{2} (cos θ \cdot sin θ)$ $int cos^2 theta* d theta=1/2 int d theta+1/2(cos theta*sin theta)$

$\int d θ = θ; θ = a r c sin (\frac{x}{3})$ $int d theta =theta" ; "theta=arc sin (x/3)$

$So ; x = 3 \cdot sin θ;; sin θ = \frac{x}{3}$ $"So ;"x=3*sin theta;" ; "sin theta=x/3$

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

$cos θ = \sqrt{1 - \frac{x^{2}}{9}}$ $cos theta=sqrt(1-x^2/9)$

$\int \sqrt{9 - x^{2}} d x = 9 [\frac{1}{2} \cdot a r c sin (\frac{x}{3}) + \frac{1}{2} (\sqrt{1 - \frac{x^{2}}{9}} \cdot \frac{x}{3})]$ $int sqrt(9-x^2)d x=9[1/2*arc sin (x/3)+1/2(sqrt(1-x^2/9 )*x/3)]$

$\int \sqrt{9 - x^{2}} d x = \frac{9 \cdot a r c sin (\frac{x}{3})}{2} + \frac{1}{2} (\sqrt{1 - \frac{x^{2}}{9}} \cdot 3 x)$ $int sqrt(9-x^2)d x=(9 *arc sin (x/3))/2+1/2(sqrt(1-x^2/9)*3x)$

$\int \sqrt{9 - x^{2}} d x = \frac{9 a r c sin (\frac{x}{3})}{2} + \frac{1}{2} (\sqrt{\frac{9 - x^{2}}{9}} \cdot 3 x)$ $int sqrt(9-x^2)d x=(9 arc sin (x/3))/2+1/2(sqrt((9-x^2)/9)*3x)$

$\int \sqrt{9 - x^{2}} d x = \frac{9 a r c sin (\frac{x}{3})}{2} + \frac{1}{2} (\sqrt{9 - x^{2}} \cdot x)$ $int sqrt(9-x^2)d x=(9 arc sin (x/3))/2+1/2(sqrt(9-x^2)*x)$

$The problem has solved...$ $"The problem has solved..."$

$\int \sqrt{9 - x^{2}} d x = \frac{9 a r c sin (\frac{x}{3}) + x \cdot \sqrt{9 - x^{2}}}{2} + C$ $int sqrt(9-x^2)d x=(9 arc sin (x/3)+x*sqrt(9-x^2))/2+C$

Science

Math

Humanities

... and beyond

How do you integrate $\int \sqrt{9 - x^{2}}$ $int sqrt(9-x^2)$ by trigonometric substitution?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫√9−x2int sqrt(9-x^2) by trigonometric substitution?

1 Answer

Related questions

Impact of this question

How do you integrate $\int \sqrt{9 - x^{2}}$ $int sqrt(9-x^2)$ by trigonometric substitution?