# How do you integrate int sqrt(9-x^2) by trigonometric substitution?

Aug 3, 2016

int sqrt(9-x^2)d x=?

$\text{Substitute } x = 3 \cdot \sin \theta$

$d x = 3 \cdot \cos \theta \cdot d \theta \text{ ; } \theta = a r c \sin \left(\frac{x}{3}\right)$

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

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

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

$\text{So ;} 1 - {\sin}^{2} \theta = {\cos}^{2} \theta$

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

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

$\text{Now use the reduction formula...}$
$\int {\cos}^{n} \theta d \theta = \frac{n - 1}{n} \int {\cos}^{n - 2} \theta \cdot d \theta + \frac{{\cos}^{n - 1} \theta \cdot \sin \theta}{n}$

$\text{use } n = 2$

$\int {\cos}^{2} \theta d \theta = \frac{2 - 1}{2} \int {\cos}^{2 - 2} \theta \cdot d \theta + \frac{{\cos}^{2 - 1} \theta \cdot \sin \theta}{2}$

$\int {\cos}^{2} \theta \cdot d \theta = \frac{1}{2} \int d \theta + \frac{1}{2} \left(\cos \theta \cdot \sin \theta\right)$

$\int d \theta = \theta \text{ ; } \theta = a r c \sin \left(\frac{x}{3}\right)$

$\text{So ;"x=3*sin theta;" ; } \sin \theta = \frac{x}{3}$

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

$\cos \theta = \sqrt{1 - {x}^{2} / 9}$

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

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

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

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

$\text{The problem has solved...}$

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