# How to solve \intx^2\sqrt(9-x^2)dx with integration by parts?

## Does it involve trig substitution?

May 2, 2018

the answer $\int {\sin}^{2} \theta - {\sin}^{4} \theta \cdot d \left(\theta\right) = - \frac{\sin \left(4 {\sin}^{-} 1 \left(x\right)\right) - 4 \cdot {\sin}^{-} 1 \left(x\right)}{32}$

#### Explanation:

show the steps

$\int {x}^{2} \setminus \sqrt{9 - {x}^{2}} \mathrm{dx}$

suppose
${a}^{2} = 9 \mathmr{and} a = 3$

${b}^{2} = 1 \mathmr{and} b = 1$

$x = \frac{a}{b} \cdot \sin \theta = \sin \theta$

$\mathrm{dx} = \cos \theta \cdot d \left(\theta\right)$

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

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

$\int {\sin}^{2} \theta \cdot \left(1 - {\sin}^{2} \theta\right) \cdot d \left(\theta\right)$

$\int {\sin}^{2} \theta - {\sin}^{4} \theta \cdot d \left(\theta\right)$

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

$\int {\sin}^{4} \theta \cdot d \left(\theta\right) = \frac{\sin \left(4 \cdot \theta\right) - 8 \cdot \sin \left(2 \cdot \theta\right) + 12 \cdot \theta}{32}$

$\int {\sin}^{2} \theta - {\sin}^{4} \theta \cdot d \left(\theta\right) = - \frac{\sin \left(4 \theta\right) - 4 \cdot \theta}{32}$

$\theta = {\sin}^{-} 1 x$

$\int {\sin}^{2} \theta - {\sin}^{4} \theta \cdot d \left(\theta\right) = - \frac{\sin \left(4 {\sin}^{-} 1 \left(x\right)\right) - 4 \cdot {\sin}^{-} 1 \left(x\right)}{32}$