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)

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