Chapter 5 Techniques of Integrals(1/3)

1. $\int (1+x)^5dx$
2. $\int x^2(1+x^3)^5dx$
3. $\int 2x \sqrt{1+x^2} \; dx$
4. $\int \sqrt{1+2x} \; dx$

方法一:
Let $u = 1+x$, $\frac{du}{dx} = \frac{d}{dx}(1+x) \Rightarrow \frac{du}{dx} = 1 \Rightarrow du = dx$
$\int (1+x)^5dx = \int u^5du = \frac{1}{6}u^6+k = \frac{1}{6}(1+x)^6+k$
方法二 (直接寫就好):
$\int (1+x)^5 \; dx = \int (1+x)^5 \; d(1+x) = \frac{1}{6}(1+x)^6+k$

方法一:
Let $u = 1+x^3$, $\frac{du}{dx} = \frac{d}{dx}(1+x^3) \Rightarrow \frac{du}{dx} = 3x^2 \Rightarrow du = 3x^2dx$
$\int (1+x^3)^5x^2dx = \int \frac{1}{3} u^5du = \frac{1}{18}u^6+k = \frac{1}{18}(1+x)^6+k$
方法二 (直接寫就好):
$\int (1+x^3)^5x^2dx = \int \frac{1}{3}(1+x^3)^5 \; d(1+x^3) = \frac{1}{18}(1+x)^6+k$

$\int 2x \sqrt{1+x^2} \; dx = \int (1+x^2)^{\frac{1}{2}}d(1+x^2) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}+k$

$\int \sqrt{1+2x} \; dx = \int (1+2x)^{\frac{1}{2}}\frac{1}{2}d(1+2x) = \frac{1}{3}(1+2x)^{\frac{3}{2}}+k$

1. $\int 2^x \; dx$
2. $\int e^{sinx}cosx \; dx$

$2^x = e^{ln2^{x}} = e^{xln2}.$
Let $u = xln2 \Rightarrow du = ln2dx$
$\int 2^x \; dx = \int e^{xln2}dx = \frac{1}{ln2}\int e^u du = \frac{1}{ln2}e^u+k=\frac{1}{ln2}e^{xln2}+k = \frac{1}{ln2}2^x+k$

$\int e^{sinx}cosx \; dx = \int e^{sinx}d(sinx) = e^{sinx}+k$

$\int x^5\sqrt{1+x^2} dx$

Let $u=1+x^2$, $du=2xdx$ and $x^2=u-1$
$\int x^5\sqrt{1+x^2} dx = \int u^{\frac{1}{2}}x^4(\frac{1}{2}du)=\frac{1}{2}u^{\frac{1}{2}}(u-1)^2du = \frac{1}{2}\int (u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}})du = \frac{1}{2}(\frac{2}{7}u^{\frac{7}{2}}-\frac{4}{5}u^{\frac{5}{2}}+\frac{2}{3}u^{\frac{3}{2}})+k = \frac{1}{7}(1+x)^{\frac{7}{2}}-\frac{5}{2}(1+x^2)^{\frac{2}{5}}+\frac{1}{3}(1+x^2)^{\frac{3}{2}}+k$

1. $\int xsinxdx$
2. $\int x^2sinxdx$

Let $f(x) = x, \; g'(x) = sinx \\ \Rightarrow f'(x) = 1 \; g(x) = -cosx \\ \int xsinxdx = -xcosx+\int cosxdx = -xcosx + sinx + k$

Let $f(x) = x^2, \; g'(x) = sinx$
$\Rightarrow f'(x) = 2x \; g(x) = -cosx$
$\int x^2sinxdx = -x^2cosx+2\int xcosx dx = -x^2cosx+2(xsinx-\int sinx dx) = -x^2cosx +2xsinx+2cosx+k$

1. $\int lnx \, dx$
2. $\int x^2lnx \, dx$

Let $f(x) = lnx, \; g'(x)=1$
$\Rightarrow f'(x) = \frac{1}{x}, \; g(x) = x$
$\int lnx \, dx = xlnx - \int \frac{1}{x} \cdot x dx= xlnx - x +k$

Let $f(x) = lnx, \; g'(x)=x^2$
$\Rightarrow f'(x) = \frac{1}{x}, \; g(x) = \frac{1}{3}x^3$
$\int lnx \, dx = xlnx - \int \frac{1}{3}x^2 dx= xlnx - \frac{1}{9}x^3 +k$

1. $\int tan^{-1}xdx$
2. $\int sin^{-1}xdx$

Let $f(x) = tan^{-1}x, \; g'(x) = 1$
$\Rightarrow f'(x) = \frac{1}{1+x^2}, \; g(x) = x$
$\int tan^{-1}xdx = xtan^{-1}x - \int \frac{x}{1+x^2}dx = xtan^{-1}x - \frac{1}{2}\int \frac{1}{1+x^2}d(1+x^2) = xtan^{-1}x - \frac{1}{2}ln(1+x^2)+k$

Let $f(x) = sin^{-1}x dx \; g'(x) = 1$
$\Rightarrow f'(x) = \frac{1}{\sqrt{1-x^2}} \; g(x) = x$
$\int sin^{-1}x dx = xsin^{-1}x - \int \frac{x}{\sqrt{1-x^2}}dx = xsin^{-1}x + \frac{1}{2}\int \frac{1}{\sqrt{1-x^2}}d(1-x^2) = xsin^{-1}x + \sqrt{1-x^2}+k$

$\int sin^2x dx = \frac{1}{2} \int (1-cos(2x)) dx = \frac{1}{2} (\int 1 dx - \int cos(2x) dx ) = \frac{1}{2} ( x- \frac{1}{2} sin(2x))+k = \frac{1}{2}x - \frac{1}{4}sin(2x)+k$

$\int sin^3x dx = \int sin^2x \cdot sinxdx = -\int (1-cos^2x)d(cosx) = -cosx + \frac{1}{3}cos^3x + k$

$\int sin^4x dx = \int (\frac{1-cos(2x)}{2})^2dx = \frac{1}{4} \int (1-2cos(2x)+cos^2(2x))dx = \frac{1}{4}(x - sin(2x))+ \int \frac{1+cos(4x)}{8} dx = \frac{3}{8}x - \frac{1}{4}sin(2x) + \frac{1}{32}sin(4x)+k$

$\int tanx dx = \int \frac{sinx}{cosx} dx = - \int \frac{1}{cosx} d(cosx) = -ln|cosx| + k$

$\int tan^2x dx = \int (sec^2x - 1) dx = tanx -x + k$

$\int tan^3x dx = \int tan^2x \cdot tanx dx = \int sec^2xtanx \; dx - \int tanx \; dx = \int secx \; d(secx) - \int tanx dx = \frac{1}{2} sec^2x + ln|cosx| + k$

$\int tan^4x dx = \int tan^2x \cdot tan^2x dx = \int sec^2xtan^2x \; dx - \int tan^2x \; dx = \int tan^2x \; d(tanx) - \int tanx^2 dx = \frac{1}{3}tan^3x -(tanx -x + k) = \frac{1}{3}tan^3x -tanx +x + k$

$\int secx dx = \int secx \cdot \frac{secx + tanx}{secx + tanx}dx = \int \frac{sec^2x+tanxsecx}{secx+tanx}dx = \int \frac{1}{secx+tanx}d(tanx+secx) = ln|secx+tanx|+k$

$\int sec^2 dx = tanx+k$

$\int sec^3x \; dx = \int sec^2x \cdot secx \; dx$
Let $g'(x) = sec^2x, f(x) = secx$
$\Rightarrow g(x) = tanx, f'(x) = tanxsecx$

$\int sec^2x \cdot secx \; dx = secxtanx - \int tanx \cdot tanxsecx \; dx = secxtanx - \int tan^2xsecx \; dx = secxtanx - \int (sec^2x-1)^2secx \; dx = secxtanx - \int secx^3 \; dx + \int secx \; dx$
$\Rightarrow 2\int sec^3x \; dx = secxtanx + \int secx \; dx$
$\Rightarrow \int sec^3x \; dx = \frac{1}{2}secxtanx + \frac{1}{2}ln|secx+tanx| + k$

$\int sec^4x \; dx = \int sec^2x \cdot sec^2x \; dx = \int (tan^2x+1) \; d(tanx) = \frac{1}{3}tan^3x + tanx + k$

$\int sin^5xcos^2x \; dx = \int sin^4xcos^2x \; d(-cosx) = -\int (1-cos^2x)^2cos^2x \; d(cosx) = - \int cos^2x - 2cos^4x + cos^6 \; dx = -\frac{1}{3}cos^3x + \frac{2}{5}cos^5x - \frac{1}{7}cos^7x +k$

$\int tan^6xsec^4x \; dx = \int tan^6xsec^2x \; d(tanx) = \int tan^6x(tan^2x+1) \; d(tanx) = \int tan^8x + tan^6x \; d(tanx) = \frac{1}{9}tan^9x+\frac{1}{7}tan^7x +k$

$\int tan^5xsec^7x \; dx = \int tan^4x sec^6x \cdot tanxsecx \; dx = \int (sec^2x-1)^2sec^6x \; d(secx) = \int sec^{10}x-2sec^8x +sec^6x \; d(secx) = \frac{1}{11}sec^{11}x-\frac{2}{9}sec^9 + \frac{1}{7} sec^7x+k$

$\int \frac{cos^3x}{\sqrt{sinx}} dx = \int cos^3x(sinx)^{-\frac{1}{2}} dx = \int cos^2x(sinx)^{-\frac{1}{2}}d(sinx) = \int (1-sin^2x)^2(sinx)^{-\frac{1}{2}}dx= \int (sinx)^{-\frac{1}{2}}-(sinx)^{\frac{3}{2}}d(sinx) = 2(sinx)^{\frac{1}{2}} - \frac{2}{5}(sinx)^{\frac{5}{2}} +k$