Exercise 10.4

Exercise 10.4

1. Use the partial fractions method to evaluate the following integrals.

(a) $\int \frac{1}{2x^2+5x+3}dx$

Solution:

$$ \frac{1}{2x^2+5x+3} = \frac{1}{(2x+3)(x+1)} = \frac{A}{2x+3} + \frac{B}{x+1} $$

$$ 1 = A(x+1) + B(2x+3) = (A+2B)x + (A+3B) $$

By equating the coefficients of corresponding powers of x,

$$ A+2B=0 \tag{1} $$ $$ A+3B=1 \tag{2} $$

If eq(2)-eq(1), $B=1$.

If $B=1$ in eq(1), $A=-2$.

$$ \begin{aligned} \text{Therefore, } \int \frac{1}{(2x+3)(x+1)}dx &= \int (\frac{-2}{2x+3} + \frac{1}{x+1})dx \\ &= \ln|x+1| - \ln|2x+3| + C \\ &= \ln|\frac{x+1}{2x+3}| + C \end{aligned} $$

(b) $\int \frac{2x-1}{(x-3)^2}dx$

Solution:

$$ \frac{2x-1}{(x-3)^2} = \frac{A}{x-3} + \frac{B}{(x-3)^2} $$

$$ 2x-1 = A(x-3) + B = Ax - (3A-B) $$

By equating the coefficients of corresponding powers of x, $A=2$

$$ 3A-B=1 \tag{1} $$

If $A=2$ in eq (1), $B=5$.

$$ \begin{aligned} \text{Therefore, } \int \frac{2x-1}{(x-3)^2}dx &= \int (\frac{2}{x-3} + \frac{5}{(x-3)^2})dx \\ &= 2 \int \frac{1}{x-3}dx + 5 \int(x-3)^{-2}dx \\ &= 2\ln|x-3| - \frac{5}{x-3} + C \end{aligned} $$

(c) $\int \frac{x+1}{(2x+5)(x+4)}dx$

Solution:

$$ \frac{x+1}{(2x+5)(x+4)} = \frac{A}{2x+5} + \frac{B}{x+4} $$

$$ x+1 = A(x+4) + B(2x+5) = (A+2B)x + (4A+5B) $$

By equating coefficients:

$$ A+2B=1 \tag{1} $$ $$ 4A+5B=1 \tag{2} $$

If eq(2)-eq(1)x4, $B=-1$. If $B=-1$ in eq(1), $A=1$.

$$ \begin{aligned} \text{Therefore, } \int \frac{x+1}{(2x+5)(x+4)}dx &= \int (\frac{1}{2x+5} - \frac{1}{x+4})dx \\ &= \frac{1}{2}\ln|2x+5| - \ln|x+4| + C \\ &= \ln \frac{\sqrt{|2x+5|}}{|x+4|} + C \end{aligned} $$

(d) $\int \frac{2x^2-1}{x^2-1}dx$

Solution:

First, we write $$ \int \frac{2x^2-1}{x^2-1}dx = \int (2 + \frac{1}{x^2-1})dx $$

First, we write $$ \frac{1}{x^2-1} = \frac{1}{(x+1)(x-1)} = \frac{A}{x+1} + \frac{B}{x-1} $$

and so $$ 1 = A(x-1) + B(x+1) $$

When $x=1$, $1=B(2)$, so $B=1/2$.

When $x=-1$, $1=A(-2)$, so $A=-1/2$.

$$ \begin{aligned} \text{Therefore, } \int \frac{2x^2-1}{x^2-1}dx &= \int 2 dx - \int \frac{\frac{1}{2}}{x+1}dx + \int \frac{\frac{1}{2}}{x-1}dx \\ &= 2x - \frac{1}{2}\ln|x+1| + \frac{1}{2}\ln|x-1| + C \\ &= 2x + \frac{1}{2}\ln|\frac{x-1}{x+1}| + C \end{aligned} $$

2. Find the function $f(x)$ that satisfying the equation $f'(x) = \sin 4x \cos 2x$ with $f(\frac{\pi}{2}) = 0$.

Solution:

By integrating, $$ f(x) = \int \sin 4x \cos 2x dx = \frac{1}{2} \int (\sin 6x + \sin 2x) dx $$

$$ = \frac{1}{2} (-\frac{1}{6}\cos 6x - \frac{1}{2}\cos 2x) + C $$

Since $f(\frac{\pi}{2}) = 0$:

$$ \frac{1}{2} (-\frac{1}{6}\cos 6(\frac{\pi}{2}) - \frac{1}{2}\cos 2(\frac{\pi}{2})) + C = 0 $$

$$ -\frac{1}{12}\cos 3\pi - \frac{1}{4}\cos\pi + C = 0 $$

$$ C = \frac{1}{12}\cos 3\pi + \frac{1}{4}\cos\pi = \frac{1}{12}(-1) + \frac{1}{4}(-1) = -\frac{1}{3} $$

Therefore, $$ f(x) = -\frac{1}{12}\cos 6x - \frac{1}{4}\cos 2x - \frac{1}{3} $$

3. Find the function $g(x)$ that satisfying the equation $g'(x) = x^2 e^{x^3}$ with $g(0) = -\frac{2}{3}$.

Solution:

By integrating, $$ g(x) = \int x^2 e^{x^3} dx $$

Let $u = x^3$, so $\frac{1}{3}du = x^2 dx$.

$$ g(x) = \int \frac{1}{3}e^u du = \frac{1}{3} e^u + C = \frac{1}{3} e^{x^3} + C $$

Since $g(0) = -\frac{2}{3}$:

$$ \frac{1}{3} e^{0} + C = -\frac{2}{3} \text{, so } \frac{1}{3} + C = -\frac{2}{3} \text{, so } C = -1 $$

Therefore, $$ g(x) = \frac{1}{3}e^{x^3} - 1 $$

4. Find the function $h(x)$, that satisfying the equation $h'(x) = \frac{x}{x^2-1}$ with $h(2) = \frac{1}{2}$.

Solution:

By integrating, $$ h(x) = \int \frac{x}{x^2-1} dx $$

Let $u = x^2-1$, so $\frac{1}{2}du = x dx$.

$$ h(x) = \int \frac{1}{2u} du = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln|x^2-1| + C $$

Since $h(2) = \frac{1}{2}$:

$$ \frac{1}{2}\ln|2^2-1| + C = \frac{1}{2} \text{, so } \frac{1}{2}\ln|3| + C = \frac{1}{2} $$

$$ \ln|3| + 2C = 1 \text{, so } C = \frac{1 - \ln 3}{2} $$

Therefore, $$ h(x) = \frac{1}{2}\ln|x^2-1| + \frac{1 - \ln 3}{2} $$

5. Find the function $f(x)$ satisfying the equation $f''(x) = 2x - 1$ with $f(0) = -1$ and $f'(1) = 2$.

Solution:

By integrating, $$ f'(x) = \int (2x-1)dx = x^2 - x + C $$

Since $f'(1) = 2$: $1^2 - 1 + C = 2$, so $C = 2$.

Therefore $f'(x) = x^2 - x + 2$.

By integrating, we get $$ f(x) = \int (x^2 - x + 2)dx = \frac{x^3}{3} - \frac{x^2}{2} + 2x + C $$

Since $f(0) = -1$: $\frac{0}{3} - \frac{0}{2} + 2(0) + C = -1$, so $C = -1$.

Therefore, $$ f(x) = \frac{x^3}{3} - \frac{x^2}{2} + 2x - 1 $$

6. Find the function $g(x)$ that satisfying the equation $g''(x) = x \sin x$ with $g(\frac{\pi}{2}) = 0$ and $g'(0) = 0$.

Solution:

By integrating, $$ g'(x) = \int x \sin x dx $$

Let $u=x$, $dv = \sin x dx$, then $du=dx$, $v=-\cos x$.

$$ g'(x) = -x \cos x - \int (-\cos x) dx = -x \cos x + \sin x + C $$

Since $g'(0) = 0$: $-(0) \cos 0 + \sin 0 + C = 0$, so $C = 0$.

Therefore, $g'(x) = -x \cos x + \sin x$.

By integrating, $$ g(x) = \int (-x \cos x + \sin x)dx = - \int x \cos x dx + \int \sin x dx $$

Let $u=x$, $dv = \cos x dx$, then $du=dx$, $v=\sin x$.

$$ \therefore g(x) = -(x \sin x - \int \sin x dx) + \int \sin x dx = -x \sin x + 2 \int \sin x dx $$

$$ = -x \sin x - 2\cos x + C $$

Since $g(\frac{\pi}{2}) = 0$:

$$ -\frac{\pi}{2}\sin(\frac{\pi}{2}) - 2\cos(\frac{\pi}{2}) + C = 0 \text{, so } -\frac{\pi}{2}(1) - 2(0) + C = 0 \text{, so } C = \frac{\pi}{2} $$

Therefore, $$ g(x) = -x \sin x - 2 \cos x + \frac{\pi}{2} $$

Exercise 10.3

Exercise 10.3

Use the integration by parts to evaluate the following integrals.

(a) $\int s e^{-2s} ds$

Solution:

Let $u = s, \quad dv = e^{-2s} ds$

$du = ds, \quad v = -\frac{1}{2} e^{-2s}$

$$ \begin{aligned} \int s e^{-2s} ds &= s(-\frac{1}{2} e^{-2s}) - \int (-\frac{1}{2} e^{-2s}) ds \\ &= -\frac{1}{2} s e^{-2s} + \frac{1}{2} \int e^{-2s} ds \\ &= -\frac{1}{2} s e^{-2s} + \frac{1}{2} (-\frac{1}{2}) e^{-2s} + C \\ &= -\frac{1}{2} s e^{-2s} - \frac{1}{4} e^{-2s} + C \end{aligned} $$

(b) $\int \ln(x+1) dx$

Solution:

Let $u = \ln(x+1), \quad dv = dx$

$du = \frac{1}{x+1} dx, \quad v = x$

$$ \begin{aligned} \int \ln(x+1) dx &= \ln(x+1) \cdot x - \int x \cdot \frac{1}{x+1} dx \\ &= x \ln(x+1) - \int \frac{x+1-1}{x+1} dx \\ &= x \ln(x+1) - \int (1 - \frac{1}{x+1}) dx \\ &= x \ln(x+1) - x + \ln|x+1| + C \end{aligned} $$

(c) $\int t \sin 2t dt$

Solution:

Let $u = t, \quad dv = \sin 2t dt$

$du = dt, \quad v = -\frac{1}{2} \cos 2t$

$$ \begin{aligned} \int t \sin 2t dt &= t(-\frac{1}{2} \cos 2t) - \int (-\frac{1}{2} \cos 2t) dt \\ &= -\frac{1}{2} t \cos 2t + \frac{1}{2} \int \cos 2t dt \\ &= -\frac{1}{2} t \cos 2t + \frac{1}{2} \cdot \frac{1}{2} \sin 2t + C \\ &= -\frac{1}{2} t \cos 2t + \frac{1}{4} \sin 2t + C \end{aligned} $$

(d) $\int x 2^x dx$

Solution:

Let $u = x, \quad dv = 2^x dx$

$du = dx, \quad v = \frac{2^x}{\ln 2}$

$$ \begin{aligned} \int x 2^x dx &= x \cdot \frac{2^x}{\ln 2} - \int \frac{2^x}{\ln 2} dx \\ &= \frac{x \cdot 2^x}{\ln 2} - \frac{1}{\ln 2} \int 2^x dx \\ &= \frac{x \cdot 2^x}{\ln 2} - \frac{1}{\ln 2} \cdot \frac{2^x}{\ln 2} + C \\ &= \frac{x \cdot 2^x}{\ln 2} - \frac{2^x}{(\ln 2)^2} + C \end{aligned} $$

(e) $\int x \cos 5x dx$

Solution:

Let $u = x, \quad dv = \cos 5x dx$

$du = dx, \quad v = \frac{1}{5} \sin 5x$

$$ \begin{aligned} \int x \cos 5x dx &= x \cdot \frac{1}{5} \sin 5x - \int \frac{1}{5} \sin 5x dx \\ &= \frac{1}{5} x \sin 5x - \frac{1}{5} \int \sin 5x dx \\ &= \frac{1}{5} x \sin 5x - \frac{1}{5} (-\frac{1}{5} \cos 5x) + C \\ &= \frac{1}{5} x \sin 5x + \frac{1}{25} \cos 5x + C \end{aligned} $$

(f) $\int e^x \cos x dx$

Solution:

Let $u = \cos x, \quad dv = e^x dx$

$du = -\sin x dx, \quad v = e^x$

$$ \int e^x \cos x dx = \cos x \cdot e^x - \int e^x (-\sin x dx) $$ $$ \int e^x \cos x dx = \cos x \cdot e^x + \int e^x \sin x dx \quad (1) $$

For $\int e^x \sin x dx$,

Let $u = \sin x, \quad dv = e^x dx$

$du = \cos x dx, \quad v = e^x$

$$ \int e^x \sin x dx = \sin x \cdot e^x - \int e^x (\cos x dx) $$

By substituting $\int e^x \sin x dx = \sin x \cdot e^x - \int e^x \cos x dx$ in (1), we get

$$ \int e^x \cos x dx = \cos x \cdot e^x + \sin x \cdot e^x - \int e^x \cos x dx $$ $$ 2 \int e^x \cos x dx = \cos x \cdot e^x + \sin x \cdot e^x + C_1 $$ $$ \therefore \int e^x \cos x dx = \frac{1}{2} (\cos x \cdot e^x + \sin x \cdot e^x) + C, \text{ where } C = \frac{1}{2}C_1 $$

Exercise 10.2

Exercise 10.2: Integration by Substitution

1. Integrate the following functions using the given substitutions.

(a) \(\int 4x^3 \sqrt{x^4-1} \,dx; \quad u = x^4 - 1\)

Solution

\( u = x^4 - 1 \)

Then \( du = 4x^3 \,dx \)

\(\int 4x^3 \sqrt{x^4-1} \,dx = \int \sqrt{u} \,du = \int u^{\frac{1}{2}} \,du \)

\(= \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} + C \)

\(= \frac{2}{3} u^{\frac{3}{2}} + C \)

\(= \frac{2}{3} (x^4 - 1)^{\frac{3}{2}} + C \)

(b) \(\int \cos^3 x \sin x \,dx; \quad u = \cos x\)

Solution

\( u = \cos x \)

Then \( du = -\sin x \,dx \implies -du = \sin x \,dx \)

\(\int \cos^3 x \sin x \,dx = \int u^3 (-du) \)

\(= -\int u^3 \,du \)

\(= -\frac{u^{3+1}}{3+1} + C \)

\(= -\frac{1}{4} u^4 + C \)

\(= -\frac{1}{4} \cos^4 x + C \)

(c) \(\int \frac{1}{x \ln|x|} \,dx; \quad u = \ln|x|\)

Solution

\( u = \ln|x| \)

Then \( du = \frac{1}{x} \,dx \)

\(\int \frac{1}{x \ln|x|} \,dx = \int \frac{1}{u} \,du \)

\(= \ln|u| + C \)

\(= \ln|\ln|x|| + C \)

(d) \(\int \sin^5 x \cos x \,dx; \quad u = \sin x\)

Solution

\( u = \sin x \)

Then \( du = \cos x \,dx \)

\(\int \sin^5 x \cos x \,dx = \int u^5 \,du \)

\(= \frac{u^{5+1}}{5+1} + C \)

\(= \frac{u^6}{6} + C \)

\(= \frac{\sin^6 x}{6} + C \)

(e) \(\int \frac{\ln x}{x} \,dx, x > 0; \quad u = \ln x\)

Solution

\( u = \ln x \)

Then \( du = \frac{1}{x} \,dx \)

\(\int \frac{\ln x}{x} \,dx = \int u \,du \)

\(= \frac{u^{1+1}}{1+1} + C \)

\(= \frac{1}{2} u^2 + C \)

\(= \frac{1}{2} (\ln x)^2 + C \)

(f) \(\int x^3 e^{x^4} \,dx; \quad u = x^4\)

Solution

\( u = x^4 \)

Then \( du = 4x^3 \,dx \implies \frac{1}{4} du = x^3 \,dx \)

\(\int x^3 e^{x^4} \,dx = \int e^u (\frac{1}{4} du) \)

\(= \frac{1}{4} \int e^u \,du \)

\(= \frac{1}{4} e^u + C \)

\(= \frac{1}{4} e^{x^4} + C \)

2. Use the substitution method to evaluate the following integrals.

(a) \(\int x \sqrt{1-x} \,dx\)

Solution

Let \( u = 1 - x \implies x = 1 - u \)

Then \( dx = -du \)

\(\int x \sqrt{1-x} \,dx = \int (1-u)\sqrt{u}(-du) \)

\(= \int (u-1)u^{\frac{1}{2}} \,du = \int (u^{\frac{3}{2}} - u^{\frac{1}{2}}) \,du \)

\(= \frac{2}{5}u^{\frac{5}{2}} - \frac{2}{3}u^{\frac{3}{2}} + C \)

\(= \frac{2}{5}(1-x)^{\frac{5}{2}} - \frac{2}{3}(1-x)^{\frac{3}{2}} + C \)

(b) \(\int (2x+1)(x^2+x)^7 \,dx\)

Solution

Let \( u = x^2+x \)

Then \( du = (2x+1) \,dx \)

\(\int (2x+1)(x^2+x)^7 \,dx = \int u^7 \,du \)

\(= \frac{1}{8} u^8 + C \)

\(= \frac{1}{8} (x^2+x)^8 + C \)

(c) \(\int \sin^3 x \,dx\)

Solution

\(\int \sin^3 x \,dx = \int \sin^2 x \cdot \sin x \,dx \)

\(= \int (1 - \cos^2 x) \sin x \,dx \)

Let \( u = \cos x \implies du = -\sin x \,dx \)

\(= \int (1-u^2) (-du) = \int (u^2-1) \,du \)

\(= \frac{1}{3}u^3 - u + C \)

\(= \frac{1}{3}\cos^3 x - \cos x + C \)

(d) \(\int x^2 \sqrt{x^3 - 2} \,dx\)

Solution

Let \( u = x^3 - 2 \)

Then \( du = 3x^2 \,dx \implies \frac{1}{3}du = x^2 \,dx \)

\(\int x^2 \sqrt{x^3 - 2} \,dx = \int \sqrt{u} (\frac{1}{3} du) \)

\(= \frac{1}{3} \int u^{\frac{1}{2}} \,du = \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}} + C \)

\(= \frac{2}{9} u^{\frac{3}{2}} + C \)

\(= \frac{2}{9} (x^3-2)^{\frac{3}{2}} + C \)

(e) \(\int \frac{\sec^2 x}{\tan x} \,dx\)

Solution

Let \( u = \tan x \)

Then \( du = \sec^2 x \,dx \)

\(\int \frac{\sec^2 x}{\tan x} \,dx = \int \frac{1}{u} \,du \)

\(= \ln|u| + C \)

\(= \ln|\tan x| + C \)

(f) \(\int \frac{x}{\sqrt{x+1}} \,dx\)

Solution

Let \( u = x+1 \implies x = u-1 \)

Then \( dx = du \)

\(\int \frac{x}{\sqrt{x+1}} \,dx = \int \frac{u-1}{\sqrt{u}} \,du \)

\(= \int (u^{\frac{1}{2}} - u^{-\frac{1}{2}}) \,du \)

\(= \frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}} + C \)

\(= \frac{2}{3}(x+1)^{\frac{3}{2}} - 2(x+1)^{\frac{1}{2}} + C \)

3. Evaluate the integral \(\int \frac{x}{(x^2+1)\ln(x^2+1)} \,dx\).

Solution

Let \( u = \ln(x^2+1) \)

Then \( du = \frac{1}{x^2+1} \cdot 2x \,dx \implies \frac{1}{2}du = \frac{x}{x^2+1} \,dx \)

\(\int \frac{x}{(x^2+1)\ln(x^2+1)} \,dx = \int \frac{1}{u} (\frac{1}{2} du) \)

\(= \frac{1}{2} \int \frac{1}{u} \,du \)

\(= \frac{1}{2} \ln|u| + C \)

\(= \frac{1}{2} \ln|\ln(x^2+1)| + C \)

Exercise 10.1

Evaluate the following integrals.

(a) $\int 4x^8 dx$

$$ = 4\int x^8 dx = 4\frac{x^{8+1}}{8+1} + C = \frac{4}{9}x^9 + C $$

(b) $\int \frac{3}{2}x^{\frac{3}{2}}\sqrt[3]{x} dx$

$$ = \frac{3}{2}\int x^{\frac{7}{3}} dx = \frac{3}{2}\frac{x^{\frac{7}{3}+1}}{\frac{7}{3}+1} + C = \frac{9}{20}x^{\frac{10}{3}} + C $$

(c) $\int (5^x + 2) dx$

$$ = \int 5^x dx + \int 2 dx = \frac{5^x}{\ln 5} + 2x + C $$

(d) $\int \sin^2 x dx$

$$ = \int\frac{1-\cos 2x}{2} dx = \frac{1}{2}\int(1-\cos 2x) dx = \frac{1}{2}(x - \frac{\sin 2x}{2}) + C $$

(e) $\int \frac{x+3}{\sqrt{x}} dx$

$$ = \int (x^{\frac{1}{2}} + 3x^{-\frac{1}{2}}) dx = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 3\frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = \frac{2}{3}x^{\frac{3}{2}} + 6x^{\frac{1}{2}} + C $$

(f) $\int (\frac{1}{x} + 5) dx$

$$ = \int \frac{1}{x} dx + 5\int dx = \ln|x| + 5x + C $$

(g) $\int (e^x + \frac{2}{x}) dx$

$$ = \int e^x dx + 2\int\frac{1}{x} dx = e^x + 2\ln|x| + C $$

(h) $\int (\frac{1}{x^5} + 4e^x) dx$

$$ = \int x^{-5} dx + 4\int e^x dx = \frac{x^{-5+1}}{-5+1} + 4e^x + C = -\frac{1}{4x^4} + 4e^x + C $$

(i) $\int (\frac{3}{x} + e^x + 10) dx$

$$ = 3\int\frac{1}{x} dx + \int e^x dx + 10\int dx = 3\ln|x| + e^x + 10x + C $$

(j) $\int \sin^2 3x dx$

$$ = \int\frac{1-\cos(6x)}{2}dx = \frac{1}{2}\int(1-\cos 6x)dx = \frac{1}{2}(x - \frac{1}{6}\sin 6x) + C = \frac{1}{2}x - \frac{1}{12}\sin 6x + C $$

(k) $\int \sin 5x \sin 2x dx$

$$ = -\frac{1}{2}\int(\cos 7x - \cos 3x)dx = -\frac{1}{2}(\frac{\sin 7x}{7} - \frac{\sin 3x}{3}) + C = \frac{\sin 3x}{6} - \frac{\sin 7x}{14} + C $$

(l) $\int \cos 7x \cos 4x dx$

$$ = \frac{1}{2}\int(\cos 11x + \cos 3x)dx = \frac{1}{2}(\frac{\sin 11x}{11} + \frac{\sin 3x}{3}) + C = \frac{\sin 11x}{22} + \frac{\sin 3x}{6} + C $$

Problem 2

Evaluate the following integrals.

(a) $\int (1-2x)^3 dx$

$$ = -\frac{1}{2}\frac{(1-2x)^{3+1}}{3+1} + C = -\frac{1}{8}(1-2x)^4 + C $$

(b) $\int \sin(2\pi x + 7) dx$

$$ = \frac{1}{2\pi}(-\cos(2\pi x + 7)) + C = -\frac{1}{2\pi}\cos(2\pi x + 7) + C $$

(c) $\int \cos(3x-7) dx$

$$ = \frac{1}{3}\sin(3x-7) + C $$

(d) $\int 3^{5x-2} dx$

$$ = \frac{3^{5x-2}}{5\ln 3} + C $$

(e) $\int \frac{1}{7x-6} dx$

$$ = \frac{1}{7}\ln|7x-6| + C $$

(f) $\int \frac{\sin 2x}{\sin x} dx$

$$ = \int \frac{2\sin x \cos x}{\sin x} dx = 2\int\cos x dx = 2\sin x + C $$

(g) $\int \sec^2(2x+3) dx$

$$ = \frac{1}{2}\tan(2x+3) + C $$

(h) $\int e^{7x-3} dx$

$$ = \frac{1}{7}e^{7x-3} + C $$

(i) $\int (1 + \tan^2 2\pi x) dx$

$$ = \int\sec^2 2\pi x dx = \frac{1}{2\pi}\tan 2\pi x + C $$

Exercise 1.1

1. Solve the following equations.

(a) Solution

$$ x^2 - 6x + 10 = 0 $$ $$ x^2 - 6x + 9 = -1 $$ $$ (x - 3)^2 = i^2 $$ $$ x - 3 = \pm i $$ $$ x = 3 \pm i $$

So there are two solutions: \( x = 3 + i \) and \( x = 3 - i \).

(b) Solution

$$ -2x^2 + 4x - 3 = 0 $$ $$ x^2 - 2x + \frac{3}{2} = 0 $$ $$ x^2 - 2x + 1 = -\frac{1}{2} $$ $$ (x - 1)^2 = \frac{1}{2}i^2 $$ $$ x - 1 = \pm \sqrt{\frac{1}{2}}i $$ $$ x = 1 \pm \frac{\sqrt{2}}{2}i $$

So there are two solutions: \( x = 1 + \frac{\sqrt{2}}{2}i \) and \( x = 1 - \frac{\sqrt{2}}{2}i \).

(c) Solution

$$ 5x^2 - 2x + 1 = 0 $$ $$ x^2 - \frac{2}{5}x + \frac{1}{5} = 0 $$ $$ x^2 - \frac{2}{5}x + \frac{1}{25} = -\frac{4}{25} $$ $$ (x - \frac{1}{5})^2 = \frac{4}{25}i^2 $$ $$ x - \frac{1}{5} = \pm \frac{2}{5}i $$ $$ x = \frac{1}{5} \pm \frac{2}{5}i $$

So there are two solutions: \( x = \frac{1}{5} + \frac{2}{5}i \) and \( x = \frac{1}{5} - \frac{2}{5}i \).

(d) Solution

$$ 3x^2 + 7x + 5 = 0 $$ $$ x^2 + \frac{7}{3}x + \frac{5}{3} = 0 $$ $$ x^2 + \frac{7}{3}x + \frac{49}{36} = -\frac{11}{36} $$ $$ (x + \frac{7}{6})^2 = \frac{11}{36}i^2 $$ $$ x + \frac{7}{6} = \pm \frac{\sqrt{11}}{6}i $$ $$ x = -\frac{7}{6} \pm \frac{\sqrt{11}}{6}i $$

So there are two solutions: \( x = -\frac{7}{6} + \frac{\sqrt{11}}{6}i \) and \( x = -\frac{7}{6} - \frac{\sqrt{11}}{6}i \).

2. Solve the following equations and check your answers.

(a) Solution

$$x^2 - 2x + 4 = 0$$ $$x^2 - 2x + 1 = -3$$ $$(x - 1)^2 = 3i^2$$ $$x - 1 = \pm \sqrt{3}i$$ $$x = 1 \pm \sqrt{3}i$$

So the two solutions are \( x = 1 + \sqrt{3}i \) and \( x = 1 - \sqrt{3}i \).

For \(x = 1 + \sqrt{3}i\),

\((1 + \sqrt{3}i)^2 - 2(1 + \sqrt{3}i) + 4\)

\(= (1 + 2\sqrt{3}i - 3) - 2 - 2\sqrt{3}i + 4\)

\(= -2 - 2 + 4 = 0\)

For \(x = 1 - \sqrt{3}i\),

\((1 - \sqrt{3}i)^2 - 2(1 - \sqrt{3}i) + 4\)

\(= (1 - 2\sqrt{3}i - 3) - 2 + 2\sqrt{3}i + 4\)

\(= -2 - 2 + 4 = 0\)

(b) Solution

$$x^2 - 4x + 5 = 0$$ $$x^2 - 4x + 4 = -1$$ $$(x - 2)^2 = i^2$$ $$x - 2 = \pm i$$ $$x = 2 \pm i$$

So the two solutions are \( x = 2 + i \) and \( x = 2 - i \).

For \(x = 2 + i\),

\((2+i)^2 - 4(2+i) + 5\)

\(= (4 + 4i - 1) - 8 - 4i + 5\)

\(= 3 - 8 + 5 = 0\)

For \(x = 2 - i\),

\((2-i)^2 - 4(2-i) + 5\)

\(= (4 - 4i - 1) - 8 + 4i + 5\)

\(= 3 - 8 + 5 = 0\)

3. Find the value of \(i^n\) for every positive integer \(n\).

Solution (Method 1)

The powers of \(i\) follow a repeating cycle. We can find this pattern by calculating the first few powers step-by-step, using the definition \(i^2 = -1\).

• Step 1: Calculate \(i^2\).
  \(i^2 = -1\) (by definition)
• Step 2: Calculate \(i^3\).
  \(i^3 = i^2 \cdot i = (-1) \cdot i = -i\)
• Step 3: Calculate \(i^4\).
  \(i^4 = i^2 \cdot i^2 = (-1)(-1) = 1\)
• Step 4: Calculate \(i^5\).
  \(i^5 = i^4 \cdot i = (1) \cdot i = i\)

The pattern \(i, -1, -i, 1\) begins to repeat. This means the value of \(i^n\) depends on which category the exponent \(n\) falls into. We can express this using an integer \(k\).

• If \(n\) is of the form \(4k+1\) (e.g., 1, 5, 9,...), then \(i^n = i\). (for \(k \ge 0\))
• If \(n\) is of the form \(4k+2\) (e.g., 2, 6, 10,...), then \(i^n = -1\). (for \(k \ge 0\))
• If \(n\) is of the form \(4k+3\) (e.g., 3, 7, 11,...), then \(i^n = -i\). (for \(k \ge 0\))
• If \(n\) is of the form \(4k\) (e.g., 4, 8, 12,...), then \(i^n = 1\). (for \(k \ge 1\))

This can be written as a piecewise-defined function where \(k\) is an integer satisfying the conditions above:

$$ i^n = \begin{cases} i & \text{if } n = 4k+1 \\ -1 & \text{if } n = 4k+2 \\ -i & \text{if } n = 4k+3 \\ 1 & \text{if } n = 4k \end{cases} $$

Solution (Method 2: )

An alternative method is to consider whether the exponent \(n\) is even or odd.

If \(n\) is an even integer, we can write \(n=2k\). Then \(i^n = i^{2k} = (i^2)^k = (-1)^k\). Since \(k=\frac{n}{2}\), this gives \(i^n = (-1)^{\frac{n}{2}}\).

If \(n\) is an odd integer, we can write \(n=2k+1\). Then \(i^n = i^{2k+1} = i^{2k} \cdot i = (-1)^k \cdot i\). Since \(k=\frac{n-1}{2}\), this gives \(i^n = (-1)^{\frac{n-1}{2}}i\).

This can be summarized in the following piecewise-defined function:

$$ i^n = \begin{cases} (-1)^{\frac{n}{2}} & \text{if } n \text{ is even} \\ (-1)^{\frac{n-1}{2}}i & \text{if } n \text{ is odd}\end{cases}$$

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Axis of Symmetry နှင့် ဖြတ်မှတ်နှစ်မှတ် ပေးထားသော Parabola ပုစ္ဆာများ

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၁။ Axis of symmetry ကို ကြည့်၍ ညီမျှခြင်းကို $y^2$ သို့မဟုတ် $x^2$ ဟု ခွဲခြားနိုင်ပါသည်။

$y = k$ ဖြစ်လျှင်

$(y - k)² = \pm4p (x - h)$ ဖြစ်မည်။

$x= h$ ဖြစ်လျှင်

$(x - h)² = \pm4p (y - k)$ ဖြစ်မည်။

၂။ ညီမျှခြင်း မှ $\pm$ ကို ဆုံးဖြတ်ရန်အတွက် ပေးရင်း အမှတ် နှစ်မှတ်ကို အသုံးပြုပါမည်။

ဝင်ရိုး (Axis of Symmetry) မှ ဝေးရာသို့ ရွေ့လျားလေလေ၊ Parabola သည် ၎င်း၏ ဖွင့်သည့်ဘက်သို့ ပို၍ ရွေ့လျားသွားလေလေ ဖြစ်သည်။

$(y - k)² = \pm4p (x - h)$ အတွက်

ဝင်ရိုး Axis of symmetry မှ အဝေးဆုံးအမှတ် တွင် $x$ တန်ဖိုး ပိုငယ်နေလျှင် Parabola opens to the left ,

Equation: $(y - k)² = -4p (x - h)$.

ဝင်ရိုး Axis of symmetry မှ အဝေးဆုံးအမှတ် တွင် $x$ တန်ဖိုး ပိုကြီးနေလျှင် Parabola opens to the right ,

Equation: $(y - k)² = 4p (x - h)$.

$(x - h)² = \pm 4p (y - k)$ အတွက်

ဝင်ရိုး Axis of symmetry မှ အဝေးဆုံးအမှတ် တွင် $y$ တန်ဖိုး ပိုငယ်နေလျှင် Parabola opens down ,

Equation: $(x - h)² = -4p (y - k)$.

ဝင်ရိုး Axis of symmetry မှ အဝေးဆုံးအမှတ် တွင် $y$ တန်ဖိုး ပိုကြီးနေလျှင် Parabola opens up ,

Equation: $(x - h)² = 4p (y - k)$.

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Examples

Find an equation of the parabola satisfying the given conditions.

(1) Axis : $y = 0$ passes through $(3, 2)$ and $(2, -3)$

Thinking

-3 သည် Axis of Symmetry: $y = 0$ မှ အဝေးဆုံးအမှတ်ဖြစ်ပြီး $x$ တန်ဖိုးမှာ ငယ်သည်။ ထို့ကြောင့် Parabola opens to the left ဖြစ်သည်။

Equation: $(y - k)² = -4p (x - h)$ ကို အသုံးပြုရမည်။

(2) Axis : $x = 0$ passes through $(2, -1)$ and $(-4, 5)$

Thinking

-4 သည် Axis of Symmetry: $x = 0$ မှ အဝေးဆုံးအမှတ်ဖြစ်ပြီး $y$ တန်ဖိုးမှာ ကြီးသည်။ ထို့ကြောင့် Parabola opens up ဖြစ်သည်။

Equation: $(x - h)² = 4p (y - k)$. ကို အသုံးပြုရမည်။

(3) Axis : $y = 1$ passes through $(-4, -2)$ and $(-3, -4)$

Thinking

-4 သည် Axis of Symmetry: $y = 1$ မှ အဝေးဆုံးအမှတ်ဖြစ်ပြီး $x$ တန်ဖိုးမှာ ကြီးသည်။ ထို့ကြောင့် Parabola opens to the right ဖြစ်သည်။

Equation: $(y - k)² = 4p (x - h)$ ကို အသုံးပြုရမည်။