EXERCISE 7.4

 

EXERCISE 7.4

$1-26$ Evaluate the integral.

(1) $$ \int \sqrt{4-x^{2}} d x$$

$$
=\int \sqrt{2^{2}-x^{2}} —(1)d x
$$

Let

$$
x=2 \sin \theta—(2)
$$

$$
d x=2 \cos \theta d \theta
$$

then (1) becomes

$$
\begin{aligned}
& \int \sqrt{4-4 \sin ^{2} \theta} 2 \cos \theta d \theta \\ \\
= & 2 \int \sqrt{4\left(1-\sin ^{2} \theta\right)} \cos \theta d \theta \\ \\
= & 2 \int \sqrt{4 \cos ^{2} \theta} \cos \theta d \theta \\ \\
= & 2 \int 2 \cos ^{2} \cos \theta d \theta \\ \\
= & 4 \int \cos ^{2} \theta d \theta
\end{aligned}
$$

Since $\cos ^{2} \theta=\frac{1+\cos \theta}{2}$

Then the  last integral takes the form

$$
\begin{aligned}
& =4 \int \frac{1+\cos \theta}{2} d \theta \\ \\
& =2 \int(1+\cos \theta) d \theta
\end{aligned}
$$

$$
\begin{aligned}
& =2 \int d \theta+2 \int \cos 2 \theta d \theta \\ \\
& =2 \theta+2 \frac{\sin 2 \theta}{2}+c \\ \\
& =2 \theta+\sin 2 \theta+c \\ \\
& =2 \theta+2 \sin \theta \cos \theta+c—(3)
\end{aligned}
$$

From (2) we have

$$
\begin{aligned}
x & =2 \sin \theta \\ \\
\sin \theta & =x / 2 —(4)\\ \\
\theta & =\sin ^{-1}(x / 2)—(5)
\end{aligned}
$$

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

$$
\begin{aligned}
& \cos \theta=\sqrt{1-\frac{x^{2}}{4}} \\ \\
& \cos \theta=\sqrt{\frac{4-x^{2}}{4}} \\ \\
& \cos \theta=\frac{\sqrt{4-x^{2}}}{2}—(6)
\end{aligned}
$$

Substituting (4)-(6) into (3)

$$
\begin{aligned}
\int \sqrt{4-x^{2}} d x & =2 \sin ^{-1} x / 2+2 \cdot \frac{x}{2} \cdot \frac{\sqrt{4-x^{2}}}{2}+c \\
& =2 \sin ^{-1}(x / 2)+\frac{1}{2} x \sqrt{4-x^{2}}+c
\end{aligned}
$$

(2)

$$
\begin{aligned}
\int \sqrt{1-4 x^{2}} d x & \\ \\
= & \int \sqrt{1-(2 x)^{2}} d x—  (1)  \\ \\
\text { Let } 2 x & =\sin \theta— (2)  \\ \\
2 d x & =\cos \theta d \theta \\ \\
d x & =\frac{1}{2} \cos \theta d \theta
\end{aligned}
$$

Then (1) becomes

$$
\begin{aligned}
& \int \sqrt{1-\sin ^{2} \theta} \cdot \frac{1}{2} \cos \theta d \theta \\ \\
& =\frac{1}{2} \int \sqrt{\cos ^{2} \theta} \cos \theta d \theta \\ \\
& =\frac{1}{2} \int \cos \theta \cos \theta d \theta \\ \\
& =\frac{1}{2} \int \cos ^{2} \theta d \theta
\end{aligned}
$$

Since $\cos ^{2} \theta=\frac{1+\cos 2 \theta}{2}$

Then the last integral becomes

$$
\begin{aligned}
& =\frac{1}{2} \int \frac{1+\cos 2 \theta}{2} d \theta \\ \\
& =\frac{1}{4} \int d \theta+\frac{1}{4} \int \cos 2 \theta d \theta \\ \\
& =\frac{1}{4} \theta+\frac{1}{4} \cdot \frac{\sin 2 \theta}{2}+c \\ \\
& =\frac{1}{4} \theta+\frac{1}{8} \sin 2 \theta+c
\end{aligned}
$$

 

$$
\begin{aligned}
& =\frac{1}{4} \theta+\frac{1}{8} \cdot 2 \sin \theta \cos \theta+c \\ \\
& =\frac{1}{4} \theta+\frac{1}{4} \sin \theta \cos \theta+c—(3)
\end{aligned}
$$

From (2) we have

$$
\begin{aligned}
& 2 x=\sin \theta \\
& \theta=\sin ^{-1}(2 x)—(4)
\end{aligned}
$$

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

$$
\begin{aligned}
& \cos \theta=\sqrt{1-(2 x)^{2}} \\ \\
& \cos \theta=\sqrt{1-4 x^{2}}—(5)
\end{aligned}
$$

Substituting (2), (4) and (6) into (3)

$$
\begin{aligned}
\int \sqrt{1-4 x^{2}} d x & =\frac{1}{4} \sin ^{-1}(2 x)+\frac{1}{4}(2 x) \sqrt{1-4 x^{2}}+c \\ \\
& =\frac{1}{4} \sin ^{-1}(2 x)+\frac{1}{2} x \sqrt{1-4 x^{2}}+c
\end{aligned}
$$

(3)  $$\int \frac{x^{2}}{\sqrt{16-x^{2}}} d x —(1)$$

Let

$$
\begin{aligned}
x & =4 \sin \theta—(2) \\
d x & =4 \cos \theta d \theta
\end{aligned}
$$

then (1) becomes

$$
\int \frac{(4 \sin \theta)^{2}}{\sqrt{16-16 \sin ^{2} \theta}} \cdot 4 \cos \theta d \theta
$$

$$
\begin{aligned}
& =\int \frac{16 \sin ^{2} \theta}{\sqrt{16\left(1-\sin ^{2} \theta\right)}} 4 \cos \theta d \theta \\ \\
& =\int \frac{16 \sin ^{2} \theta}{\sqrt{16 \cos ^{2} \theta}} 4 \cos \theta d \theta=\int \frac{16 \sin ^{2} \theta}{4 \cos \theta} 4 \cos \theta d \theta \\ \\
& =16 \int \sin ^{2} \theta d \theta
\end{aligned}
$$

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

then the last integral becomes

$$
\begin{aligned}
& =16 \int \frac{1-\cos 2 \theta}{2} d \theta=8 \int d \theta-8 \int \cos 2 \theta d \theta \\ \\
& =8 \theta-8 \frac{\sin 2 \theta}{2}+c=8 \theta-4(2 \sin \theta \cos \theta)+c \\ \\
& =8 \theta-8 \sin \theta \cos \theta+c
\end{aligned}
$$

From (2) we have

$$
\begin{aligned}
& x=4 \sin \theta \\
& \sin \theta=x / 4 — (4)  \\ \\
& \theta=\sin ^{-1}(x / 4)—(5) \\ \\
& \cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-x^{2} / 16} \\ \\
& \cos \theta=\sqrt{\frac{16-x^{2}}{16}}=\frac{\sqrt{16-x^{2}}}{4}—(6)
\end{aligned}
$$

Substituting (4) – (6) into (3)

$$
\begin{aligned}
\int \frac{x^{2}}{\sqrt{16-x^{2}}} d x & =8 \sin ^{-1}(x / 4)-8\left(\frac{x}{4}\right)\left(\frac{\sqrt{16-x^{2}}}{4}\right)+c \\ \\
& =8 \sin ^{-1}(x / 4)-\frac{1}{2} x \sqrt{16-x^{2}}+c
\end{aligned}

$$