Example. Evaluate the definite integral

$$
\int_{0}^{3 / 2} \frac{x^{2}}{\sqrt{9-x^{2}}} d x
$$
Solution.

1. To get rid of the radical, we use the substitution
$x=3 \sin t$. Then

$$
d x=3 \cos t d t, \quad t(0)=\arcsin 0=0, \quad t\left(\frac{3}{2}\right)=\arcsin \frac{1}{2}=\frac{\pi}{6}
$$

and $\sqrt{9-x^{2}}=|3 \cos t|=3 \cos t$, since $\cos t>0$ for $t \in[0, \pi / 6]$.

2. We make a change of variable in the definite integral:

$$
\int_{0}^{3 / 2} \frac{x^{2}}{\sqrt{9-x^{2}}} d x=\int_{0}^{\pi / 6} \frac{9(\sin t)^{2} \cdot 3 \cos t}{\sqrt{9-9(\sin t)^{2}}} d t=9 \int_{0}^{\pi / 6} \sin ^{2} t d t
$$

3. Applying the formulas for reducing the degree, we get

$9 \int_{0}^{\pi / 6} \sin ^{2} t d t=\frac{9}{2} \int_{0}^{\pi / 6}(1-\cos 2 t) d t=\frac{3}{4} \pi-\frac{9}{2} \int_{0}^{\pi / 6} \cos 2 t d t=$

$$
=\frac{3}{4} \pi-\left.\frac{9}{4} \sin 2 t\right|_{0} ^{\pi / 6}=\frac{3}{4} \pi-\frac{9 \sqrt{3}}{8}
$$

Answer. $\quad \int_{0}^{3 / 2} \frac{x^{2}}{\sqrt{9-x^{2}}} d x=\frac{3}{4} \pi-\frac{9 \sqrt{3}}{8}$.

Conditions of the problems. Calculate the definite integrals.

1. $\int_{0}^{1} \sqrt{1-x^{2}} d x$
2. $\int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x^{4}} d x$
3. $\quad \int^{1} \frac{\sqrt{1-x^{2}}}{x^{2}} d x$

$1 / 2$

7. $\int_{1}^{3} \frac{\sqrt{\left(9-x^{2}\right)^{3}}}{x^{6}} d x$.
8. $\int_{1}^{\sqrt{3}} \frac{\sqrt{1+x^{2}}}{x^{2}} d x$.
9. $\int_{0}^{3} x^{2} \sqrt{9-x^{2}} d x$
10. $\int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x} d x$
11. $\int_{0}^{1} \frac{1}{\sqrt{\left(1+x^{2}\right)^{3}}} d x$.
12. $\int_{3}^{5} \frac{1}{x^{2} \sqrt{x^{2}-9}} d x$.
13. $\int_{0}^{1} \frac{1}{x+\sqrt{1-x^{2}}}$.

Answers. 1. $\frac{\pi}{4} \cdot$ 2. $\frac{81 \pi}{16} . \quad$ 3. $\frac{\sqrt{3}}{8} \cdot$ 4. $\sqrt{3}-\frac{\pi}{3} \cdot$ 5. $\sqrt{3}-\frac{\pi}{3} \cdot$ 6. $\frac{\sqrt{2}}{2}$.
7. $\frac{128 \sqrt{2}}{45}$.
8. $\frac{4}{45}$.
9. $\frac{3 \sqrt{2}-2 \sqrt{3}}{3}+\ln \frac{2+\sqrt{3}}{1+\sqrt{2}}$.
10. $\frac{\pi}{4}$

## 8.7. Calculation of Areas

## in Cartesian Coordinates

Problem Statement. Calculate the area of the region bounded by the graphs of the functions $y=f_{1}(x)$ and $y=f_{2}(x)\left(f_{1}(x) \geq f_{2}(x)\right.$ or $f_{1}(x) \leq f_{2}(x)$ for all points in the region) and, possibly, the lines $x=a$ and $x=b$.

Plan of Solution. If the region $D$ is defined by the system of inequalities

$$
\left\{\begin{aligned}
a & \leq x \leq b \\
u(x) & \leq y \leq v(x)
\end{aligned}\right.
$$

then the area of the region $D$ is calculated by the formula

$$
S_{D}=\int_{a}^{b}(v(x)-u(x)) d x
$$

If the inequalities defining the region $D$ are unknown, i.e., $a$ and $b$ are unknown and it is unknown which of the functions $f_{1}(x)$ and $f_{2}(x)$ is greater on $(a, b)$, then we perform the following operations.

1. Find $a$ and $b$ as the abscissas of the points of intersection of the graphs of the functions $f_{1}(x)$ and $f_{2}(x)$, i.e., solve the equation

$$
f_{1}(x)=f_{2}(x)
$$

2. Investigate the sign of the difference $f_{1}(x)-f_{2}(x)$ on $[a, b]$. For this, it is sufficient to compute the value of $f_{1}(x)-f_{2}(x)$ at some point in $(a, b)$. If it is positive, then $f_{1}(x) \geq f_{2}(x)$ and, consequently, $u(x)=f_{2}(x)$ and $v(x)=f_{1}(x)$. If it is negative, then $f_{1}(x) \leq f_{2}(x)$ and, consequently, $u(x)=f_{1}(x)$ and $v(x)=f_{2}(x)$.
3. Apply formula (1) and find $S_{D}$.

Write the answer, not forgetting the units of measurement.