Integration by parts, reduction formulae and series

Find (a) $\int x{e}^{2x}dx\backslash int\; x\; e^\{2x\}\backslash ,dx$ (b) $\int \ln xdx\backslash int\; \backslash ln\; x\backslash ,dx$ (c) $\int x^2\cos xdx\backslash int\; x^2\backslash cos\; x\backslash ,dx$.

Solution

1. $u=xu\; =\; x$, $dv=e^{2x}dxdv\; =\; e^\{2x\}dx$; so $du=dxdu\; =\; dx$, $v=\frac{1}{2}{e}^{2x}v\; =\; \backslash frac\{1\}\{2\}e^\{2x\}$. $$\int x{e}^{2x}dx=\frac{x}{2}{e}^{2x}-\int \frac{1}{2}{e}^{2x}dx=\frac{x}{2}{e}^{2x}-\frac{1}{4}{e}^{2x}+C\mathrm{.}\backslash int\; x\; e^\{2x\}\backslash ,dx\; =\; \backslash frac\{x\}\{2\}e^\{2x\}\; -\; \backslash int\; \backslash frac\{1\}\{2\}e^\{2x\}\backslash ,dx\; =\; \backslash frac\{x\}\{2\}e^\{2x\}\; -\; \backslash frac\{1\}\{4\}e^\{2x\}\; +\; C\; .$$
2. $u=\ln xu\; =\; \backslash ln\; x$, $dv=dxdv\; =\; dx$; so $du=\frac{dx}{x}du\; =\; \backslash frac\{dx\}\{x\}$, $v=xv\; =\; x$. $$\int \ln xdx=x\ln x-\int x\cdot \frac{1}{x}dx=x\ln x-x+C\mathrm{.}\backslash int\; \backslash ln\; x\backslash ,dx\; =\; x\backslash ln\; x\; -\; \backslash int\; x\backslash cdot\backslash frac\{1\}\{x\}\backslash ,dx\; =\; x\backslash ln\; x\; -\; x\; +\; C\; .$$
3. $u=x^{2}u\; =\; x^2$, $dv=\cos xdxdv\; =\; \backslash cos\; x\backslash ,dx$: $\int x^2\cos xdx=x^2\sin x-\int 2x\sin xdx\backslash int\; x^2\backslash cos\; x\backslash ,dx\; =\; x^2\backslash sin\; x\; -\; \backslash int\; 2x\backslash sin\; x\backslash ,dx$. Apply parts again with $u=2xu\; =\; 2x$, $dv=\sin xdxdv\; =\; \backslash sin\; x\backslash ,dx$: $\int 2x\sin xdx=-2x\cos x+2\sin x\backslash int\; 2x\backslash sin\; x\backslash ,dx\; =\; -2x\backslash cos\; x\; +\; 2\backslash sin\; x$. Hence $$\int x^2\cos xdx=x^2\sin x+2x\cos x-2\sin x+C\mathrm{.}\backslash int\; x^2\backslash cos\; x\backslash ,dx\; =\; x^2\backslash sin\; x\; +\; 2x\backslash cos\; x\; -\; 2\backslash sin\; x\; +\; C\; .$$

Find (a) $I=\int e^x\sin xdxI\; =\; \backslash int\; e^x\backslash sin\; x\backslash ,dx$ (b) $\int {\tan }^{-1}xdx\backslash int\; \backslash tan^\{-1\}x\backslash ,dx$.

Solution

1. With $u=\sin xu\; =\; \backslash sin\; x$, $dv=e^{x}dxdv\; =\; e^x\; dx$: $I=e^x\sin x-\int e^x\cos xdxI\; =\; e^x\backslash sin\; x\; -\; \backslash int\; e^x\backslash cos\; x\backslash ,dx$. Again with $u=\cos xu\; =\; \backslash cos\; x$, $dv=e^{x}dxdv\; =\; e^x\; dx$: $\int e^x\cos xdx=e^x\cos x+\int e^x\sin xdx=e^x\cos x+I\backslash int\; e^x\backslash cos\; x\backslash ,dx\; =\; e^x\backslash cos\; x\; +\; \backslash int\; e^x\backslash sin\; x\backslash ,dx\; =\; e^x\backslash cos\; x\; +\; I$. So $$I=e^x\sin x-e^x\cos x-I⟹I=\frac{e^x(\sin x-\cos x)}{2}+C\mathrm{.}I\; =\; e^x\backslash sin\; x\; -\; e^x\backslash cos\; x\; -\; I\; \backslash quad\backslash Longrightarrow\backslash quad\; I\; =\; \backslash frac\{e^x(\backslash sin\; x\; -\; \backslash cos\; x)\}\{2\}\; +\; C\; .$$ The integral returned with the opposite sign, so we solved for it as an unknown. Keep the same type of $uu$ (here the trigonometric factor) in both steps, or the work undoes itself.
2. $u={\tan }^{-1}xu\; =\; \backslash tan^\{-1\}x$, $dv=dxdv\; =\; dx$: $du=\frac{dx}{1+x^2}du\; =\; \backslash frac\{dx\}\{1\; +\; x^2\}$, $v=xv\; =\; x$. $$\int {\tan }^{-1}xdx=x{\tan }^{-1}x-\int \frac{x}{1+x^2}dx=x{\tan }^{-1}x-\frac{1}{2}\ln (1+x^2)+C\mathrm{.}\backslash int\; \backslash tan^\{-1\}x\backslash ,dx\; =\; x\backslash tan^\{-1\}x\; -\; \backslash int\; \backslash frac\{x\}\{1\; +\; x^2\}\backslash ,dx\; =\; x\backslash tan^\{-1\}x\; -\; \backslash frac\{1\}\{2\}\backslash ln(1\; +\; x^2)\; +\; C\; .$$

(a) For $n\ge 2n\; \backslash ge\; 2$, show that $I_n=\int {\sin }^{n}xdxI\_n\; =\; \backslash int\; \backslash sin^n\; x\backslash ,dx$ satisfies $I_n=-\frac{1}{n}{\sin }^{n-1}x\cos x+\frac{n-1}{n}{I}_{n-2}I\_n\; =\; -\backslash frac\{1\}\{n\}\backslash sin^\{n-1\}x\backslash cos\; x\; +\; \backslash frac\{n\; -\; 1\}\{n\}I\_\{n-2\}$. (b) Hence find $\int {\sin }^{4}xdx\backslash int\; \backslash sin^4\; x\backslash ,dx$. (c) Use $I_n=x^n{e}^x-n{I}_{n-1}I\_n\; =\; x^n\; e^x\; -\; nI\_\{n-1\}$ to find $\int x^3{e}^{x}dx\backslash int\; x^3\; e^x\backslash ,dx$.

Solution

1. Write ${\sin }^{n}x={\sin }^{n-1}x\cdot \sin x\backslash sin^n\; x\; =\; \backslash sin^\{n-1\}x\; \backslash cdot\; \backslash sin\; x$. Take $u={\sin }^{n-1}xu\; =\; \backslash sin^\{n-1\}x$, $dv=\sin xdxdv\; =\; \backslash sin\; x\backslash ,dx$; then $du=(n-1){\sin }^{n-2}x\cos xdxdu\; =\; (n\; -\; 1)\backslash sin^\{n-2\}x\backslash cos\; x\backslash ,dx$ and $v=-\cos xv\; =\; -\backslash cos\; x$: $$I_n=-{\sin }^{n-1}x\cos x+(n-1)\int {\sin }^{n-2}x{\cos }^{2}xdx\mathrm{.}I\_n\; =\; -\backslash sin^\{n-1\}x\backslash cos\; x\; +\; (n\; -\; 1)\backslash int\; \backslash sin^\{n-2\}x\backslash cos^2\; x\backslash ,dx\; .$$ Replace ${\cos }^{2}x\backslash cos^2\; x$ by $1-{\sin }^{2}x1\; -\; \backslash sin^2\; x$: the last integral is $I_{n-2}-I_{n}I\_\{n-2\}\; -\; I\_n$. So $I_n=-{\sin }^{n-1}x\cos x+(n-1)I_{n-2}-(n-1)I_{n}I\_n\; =\; -\backslash sin^\{n-1\}x\backslash cos\; x\; +\; (n\; -\; 1)I\_\{n-2\}\; -\; (n\; -\; 1)I\_n$, and collecting $I_{n}I\_n$: $$n{I}_n=-{\sin }^{n-1}x\cos x+(n-1)I_{n-2}\mathrm{.}nI\_n\; =\; -\backslash sin^\{n-1\}x\backslash cos\; x\; +\; (n\; -\; 1)I\_\{n-2\}\; .$$
2. $I_0=xI\_0\; =\; x$. Then $I_2=-\frac{1}{2}\sin x\cos x+\frac{1}{2}xI\_2\; =\; -\backslash frac\{1\}\{2\}\backslash sin\; x\backslash cos\; x\; +\; \backslash frac\{1\}\{2\}x$, and $$I_4=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{4}{I}_2=-\frac{1}{4}{\sin }^{3}x\cos x-\frac{3}{8}\sin x\cos x+\frac{3x}{8}+C\mathrm{.}I\_4\; =\; -\backslash frac\{1\}\{4\}\backslash sin^3\; x\backslash cos\; x\; +\; \backslash frac\{3\}\{4\}I\_2\; =\; -\backslash frac\{1\}\{4\}\backslash sin^3\; x\backslash cos\; x\; -\; \backslash frac\{3\}\{8\}\backslash sin\; x\backslash cos\; x\; +\; \backslash frac\{3x\}\{8\}\; +\; C\; .$$
3. $I_0=e^{x}I\_0\; =\; e^x$, $I_1=x{e}^x-e^{x}I\_1\; =\; xe^x\; -\; e^x$, $I_2=x^2{e}^x-2I_{1}I\_2\; =\; x^2e^x\; -\; 2I\_1$, $I_3=x^3{e}^x-3I_{2}I\_3\; =\; x^3e^x\; -\; 3I\_2$. Altogether $$\int x^3{e}^{x}dx=e^x(x^3-3x^2+6x-6)+C\mathrm{.}\backslash int\; x^3\; e^x\backslash ,dx\; =\; e^x\backslash left(x^3\; -\; 3x^2\; +\; 6x\; -\; 6\backslash right)\; +\; C\; .$$

(a) Find $\int e^{-x^2}dx\backslash int\; e^\{-x^2\}\backslash ,dx$ as a series. (b) Let $FF$ be the antiderivative with $F(0)=0F(0)\; =\; 0$. Estimate $F(1)F(1)$ with five terms, and bound the error.

Solution

1. Put $-x^2-x^2$ for $xx$ in $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots e^x\; =\; 1\; +\; x\; +\; \backslash frac\{x^2\}\{2!\}\; +\; \backslash frac\{x^3\}\{3!\}\; +\; \backslash cdots$, valid for all $xx$: $$e^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\frac{x^8}{4!}-\cdots e^\{-x^2\}\; =\; 1\; -\; x^2\; +\; \backslash frac\{x^4\}\{2!\}\; -\; \backslash frac\{x^6\}\{3!\}\; +\; \backslash frac\{x^8\}\{4!\}\; -\; \backslash cdots$$ $$\int e^{-x^2}dx=C+x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+\frac{x^9}{216}-\cdots =C+\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{x}^{2k+1}}{k!(2k+1)}\mathrm{.}\backslash int\; e^\{-x^2\}\backslash ,dx\; =\; C\; +\; x\; -\; \backslash frac\{x^3\}\{3\}\; +\; \backslash frac\{x^5\}\{10\}\; -\; \backslash frac\{x^7\}\{42\}\; +\; \backslash frac\{x^9\}\{216\}\; -\; \backslash cdots\; =\; C\; +\; \backslash sum\_\{k=0\}^\{\backslash infty\}\backslash frac\{(-1)^k\; x^\{2k+1\}\}\{k!\backslash ,(2k\; +\; 1)\}\; .$$
2. $F(0)=0F(0)\; =\; 0$ gives $C=0C\; =\; 0$, and $$F(1)\approx 1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\frac{1}{216}\approx \mathrm{0.7475.}F(1)\; \backslash approx\; 1\; -\; \backslash frac\{1\}\{3\}\; +\; \backslash frac\{1\}\{10\}\; -\; \backslash frac\{1\}\{42\}\; +\; \backslash frac\{1\}\{216\}\; \backslash approx\; 0.7475\; .$$ The series alternates with terms decreasing to 0, so the error is less than the first term left out, $\frac{1}{5!\times 11}=\frac{1}{1320}\approx 0.0008\backslash frac\{1\}\{5!\; \backslash times\; 11\}\; =\; \backslash frac\{1\}\{1320\}\; \backslash approx\; 0.0008$. (The true value is $0.746820.746\backslash ,82$.) In lesson 4 you will see that $F(1)F(1)$ is the area under $y=e^{-x^2}y\; =\; e^\{-x^2\}$ from 0 to 1.