Integration of some elementary integrals
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\title{Integration of Some Elementary Integrals}
\author{Adrian D'Costa}
\begin{document}
\maketitle
\subsection{1}
\begin{align*}&\int x\cos(ax)\,\,dx\\&\text{Let }u = x \therefore du = dx \text{ and } dv = \cos(ax)dx \therefore v = \frac{1}{a}\sin(ax)\\&=\frac{x\sin(ax)}{a}-\int \frac{1}{a}\sin{(ax)}\,\,dx\\&=\frac{x\sin(ax)}{a}+\frac{1}{a^2}\cos{(ax)}+C\end{align*}
\subsection{2}
\begin{align*}&\int\sin^{2}(x)\cos^{2}(x)\,\,dx\\&=\int (1 + \cos{(2x)} )( 1 - \cos{(2x)})\,\,dx\\&=\int ( 1 - \cos^{2}(2x))\,\,dx\\&=\frac{1}{2}\int (2-2\cos^{2}{(2x)}\,\,dx\\&=\frac{1}{2} \int (\cos{(4x)} +4)\,\,dx\\&=\frac{1}{32} (\sin{(4x)} +4x)\,\,dx + C\end{align*}
\subsection{3}
\begin{align*}&\int \frac{7x+1}{x^2 -x - 2}\,\,dx\\ &\frac{7x+1}{x^2-x-2} = \frac{7x+1} {(x-2) (x+1)}\\&7x+1 = \frac{A}{x-2} + \frac{B}{x+1}\\7x+1 &= Ax+A + Bx - 2B \\7x+1 &= (A+ B)x +(A - 2B)\\&\therefore A+ B = 7.....\text{(i)}\\&A- 2B = 1....\text{(ii)}\\&\text{subtracting equation (ii) from (i) gives}\Rightarrow 3B = 6\therefore B = 2\text{ and } A = 5\end{align*}
\begin{align*}\int \frac{7x+1}{x^2 -x - 2}\,\,dx &= \int \frac{5}{x-2}\,\,dx + \int \frac{2}{x+1}\,\,dx\\&=5\ln|x-2| + 2\ln|x+1|+C \end{align*}
\subsection{4}
\begin{align*}&\int \frac{1}{(x+1)^2}\,\,dx\\&\text{Let }u = x + 1 \therefore du = dx\\&=\int \frac{1}{u^2}\,\,du\\&= -u^{-1} + C \\&= -\frac{1}{(x + 1)} + C \end{align*}
\subsection{5}
\begin{align*}\int \sin^{6}(x)\cos^{3}(x)\,\,dx &= \int \sin^{6}(x)\cos^{2}(x)\cos(x)\,\,dx\\&= \int \sin^{6}(x) (1 - \sin^{2}(x))\cos(x)\,\,dx\\&= \int \sin^{6}(x)\cos(x)\,\,dx - \int \sin^{8}\cos(x)\,\,dx\\\text{Let } u = \sin(x) \therefore du = \cos(x)\\& \int u^6\,\,du - \int u^8\,\,du\\&= \frac{u^7}{7} -\frac{u^9}{9}+C\\&=\frac{\sin^{7}(x)}{7} - \frac{\sin^{9}(x)}{9} + C\end{align*}
\subsection{6}
\begin{align*}&\int \sin(\theta)\ln(\cos(\theta))\,\,d\theta\\&\text{Let } u = \ln(\cos(\theta))\therefore du = -\tan(\theta)\,\,d\theta\\&dv = \sin(\theta)\,\,d\theta \therefore v = -\cos(\theta)\end{align*}
\begin{align*}\int \sin(\theta)\ln(\cos(\theta)\,\,d\theta=& -\cos(\theta)\ln(\cos(\theta)) - \int \sin(\theta)d\theta\\&= -\cos(\theta)\ln(\cos(\theta)) + \cos(\theta)+ C\end{align*}
\subsection{7}
\begin{align*}&\int \sec^{4}(x)\tan^{3}(x)\,\,dx\\&= \int \sec^{3}(x) \tan^{2}(x) \sec(x) \tan(x)\,\,dx \\ &= \int \sec^{3}(x)(\sec^{2}(x) - 1)\sec(x)\tan(x)\,\,dx\\ &= \int \sec^{5}(x) \sec(x)\tan(x)\,\,dx - \int \sec^{3}(x) \sec(x)\tan(x)\,\,dx\\ &\text{Let } u = \sec(x) \therefore du = \sec(x)\tan(x)\,\,dx\\ &=\int u^{5}\,\,du - \int u^{3}\,\,du \\ &= \frac{u^{6}}{6} - \frac{u^{4}}{4} + C\\ &= \frac{\sec^{6}}{6} - \frac{\sec^{4}}{4} + C \end{align*}
\subsection{8}
\begin{align*}&\int xe^{5x}\,\,dx\\&\text{Let } u = x \therefore du = dx\\& \text{ and } dv = e^{5x} \therefore v = \frac{e^{5x}}{5}\\&\frac{xe^{5x}}{5} - \int \frac{e^{5x}}{5}\,\,dx\\=&\frac{xe^{5x}}{5} - \frac{e^{5x}}{25}+ C\end{align*}
\subsection{9}
\begin{align*}&\int\sec^{3}(\theta)\,\,d\theta\\&=\int \sec{(x)} \sec^{2}{(x)}\,\,dx\\&\text{Let } u = \sec{(x)} \therefore du = \sec{(x)}\tan{(x)}\,\,dx\text{ and } dv = \sec^{2}{(x)}\,\,dx \therefore v = \tan{(x)}\\&= \sec{(x)}\tan{(x)} - \int \sec{(x)}\tan^{2}{(x)}\,\,dx\\&=\sec{(x)}\tan{(x)} - \int \sec{(x)}(\sec^{2}{(x)} - 1)\,\,dx\\&= \sec{(x)}\tan{(x)} - \int \sec^{3}{(x)}\,\,dx + \int \sec{(x)}\,\,dx\\&=\frac{\sec{(x)}\tan{(x)}}{2} + \frac{\ln|\sec{(x)} + \tan{(x)}|}{2} + C \end{align*}
\subsection{10}
\begin{align*}&\int \tan^{3}(x)\sec^{4}(x)\,\,dx\\&=\int \sec{(x)}\tan{(x)}\sec^{3}{(x)}\tan^{2}(x)\,\,dx\\&= \int \sec^{5}{(x)} \sec{(x)}\tan{(x)}\,\,dx - \int \sec^{3}{(x)} \sec{(x)}\tan{(x)}\,\,dx \\&\text{Let } u = \sec{(x)}\,\, du = \sec{(x)}\tan{(x)}\,\,dx\\&=\int u^{5}\,\,du - \int u^{3}\,\,du\\&= \frac{u^{6}}{6} - \frac{u^{4}}{4} + C\\&= \frac{\sec^{6}{(x)}}{6} -\frac{\sec^{4}{(x)}}{4} + C\end{align*}
\subsection{11}
\begin{align*}&\text{Evaluate where } E \text{ is the region under the plane that lies in the first octant.}\\&\int_0^3\int_0^{\frac{-2x}{3}+2}\int_0^{6-2x-3y}2x\,\,dz\,\,dy\,\,dx\\=&\int_0^3\int_0^{\frac{-2x}{3}+2}2xz\Bigg]_0^{6-2x-3y}dy\,\,dx\\&=\int_0^3 \int_0^{\frac{-2x}{3}+2}2x(6-2x-3y)\,\,dy\,\,dx\\&=\int_0^3 \int_0^{\frac{-2x}{3}+2} (12x-4x^2-6xy)\,\,dy\,\,dx\\&=\int_0^3 (12xy - 4x^2y - 3xy^2)\Big]_0^{\frac{-2x}{3}+2}\,\,dx\\&=\int_0^3 \left(12x\frac{6-2x}{3}-4x^2\frac{6-2x}{3} -3x\frac{36-24x + 4x^2}{9} \right)\,\,dx\\&=\int_0^3 \left(\frac{36x(6-2x) - 12x^2(6-2x)-108x+72x^2 - 12x^3}{9}\right)\,\,dx\\&=\\&=\int_0^3\left(\frac{216x - 72x^{2} - 72x^2 + 24x^3 -108 x+72x^2 -12x^3}{9}\right)\,\,dx\\&=\int_0^3 \frac{-72x^{2} + 108x +12x^3}{9}\,\,dx\\&= \int_0^3 -8x^{2} + 12x + \frac{4x^3}{3}\\&=\Bigg[\frac{-8x^{3}}{3}+ 6x^2 + \frac{x^{4}}{3}\Bigg]_0^3\\&=9\end{align*}
\subsection{12}
$\displaystyle \text{Prove that } \frac{d(\sin^{-1}(x))}{dx}= \frac{1}{\sqrt{1-x^2}}$
Proof:
\begin{align*}&\text{Let } f(x) = \sin{(x)} \text{ and } g(x) = \sin^{-1}(x)\end{align*}
Then $g'(x) = \frac{1}{f'(g(x))}= \frac{1}{\cos{(\sin^{-1}(x))}}$
Let's $y = \sin^{-1}(x) \therefore x = \sin(y)$
Using this part of the definition:
$\cos(\sin^{-1}(x))=\cos(y)$
But we know that:
$\sin^{2}(y) + \cos^{2}(y) = 1 \therefore \cos{(y)} = \sqrt{1-\sin^{2}(y)} = \sqrt{1 - x^2}$
$\therefore \frac{d(\sin^{-1}(x))}{dx} = \frac{1}{\sqrt{1-x^2}}$
\subsection{13}
$\text{Prove that } \frac{d(\cos^{-1}(x))}{dx}= \frac{-1}{\sqrt{1-x^2}}$
Proof:
\begin{align*}&\text{Let } f(x) = \cos{(x)} \text{ and } g(x) = \cos^{-1}(x)\end{align*}
Then $g'(x) = \frac{1}{f'(g(x))}= \frac{-1}{\sin{(\cos^{-1}(x))}}$
Let's $y = \cos^{-1}(x) \therefore x = \cos(y)$
Using this part of the definition:
$\sin{(\cos^{-1}(x))}=\sin{(y)}$
But we know that:
$\sin^{2}(y) + \cos^{2}(y) = 1 \therefore \sin{(y)} = \sqrt{1-\cos^{2}(y)} = \sqrt{1 - x^2}$
$\therefore \frac{d(\cos^{-1}(x))}{dx} = \frac{-1}{\sqrt{1-x^2}}$
\end{document}