evaluate the integral using trigonometric substitution.\n int\frac{x^{2}}{sqrt{81 - x^{2}}}dx \n(use…

Explanation:

Step1: Make trigonometric substitution

Let $x = 9\sin\theta$, then $dx=9\cos\theta d\theta$. And $\sqrt{81 - x^{2}}=\sqrt{81-81\sin^{2}\theta}=9\cos\theta$. Also, $x^{2}=81\sin^{2}\theta$.

Step2: Rewrite the integral

The integral $\int\frac{x^{2}}{\sqrt{81 - x^{2}}}dx$ becomes $\int\frac{81\sin^{2}\theta}{9\cos\theta}\cdot9\cos\theta d\theta=\int81\sin^{2}\theta d\theta$.

Step3: Use the double - angle formula

Since $\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}$, the integral is $81\int\frac{1 - \cos(2\theta)}{2}d\theta=\frac{81}{2}\int(1-\cos(2\theta))d\theta$.

Step4: Integrate term - by - term

$\frac{81}{2}\left(\int 1d\theta-\int\cos(2\theta)d\theta\right)$. The integral of $1$ with respect to $\theta$ is $\theta$, and the integral of $\cos(2\theta)$ with respect to $\theta$ is $\frac{1}{2}\sin(2\theta)$. So we have $\frac{81}{2}\left(\theta-\frac{1}{2}\sin(2\theta)\right)+C$.

Step5: Use the double - angle formula for $\sin(2\theta)$

$\frac{81}{2}\left(\theta-\sin\theta\cos\theta\right)+C$.

Step6: Back - substitute

Since $x = 9\sin\theta$, then $\sin\theta=\frac{x}{9}$ and $\theta=\arcsin\left(\frac{x}{9}\right)$. Also, $\cos\theta=\frac{\sqrt{81 - x^{2}}}{9}$. The integral is $\frac{81}{2}\arcsin\left(\frac{x}{9}\right)-\frac{x\sqrt{81 - x^{2}}}{2}+C$.

Answer:

$\frac{81}{2}\arcsin\left(\frac{x}{9}\right)-\frac{x\sqrt{81 - x^{2}}}{2}+C$