--- ## How to integrate "by parts" - **Integration by parts** or **partial integration** is a process that makes it easier to find the ==**antiderivative of equations that can be represented as *the product of two or more individual functions.*** == - Deciding which term should be assigned to a and b *(shown below)* is very important. Since integration by parts requires you to take the integral of **the derivative of a multiplied by the integral of b**, "a" should be chosen such that its derivative simplifies this product into a single term *(Ideally, its derivative is a constant).* $\newline$ $\large\text{Integration by parts}$ $\newline$ $F(x)=\int f(x)\cdot g(x)\longrightarrow\int a\cdot c\space\space\space\space\space\small\color{grey}\therefore a\text{ represents }f(x)\text{ and }c\text{ represents }g(x)$ $\large\text{Let }b=\int c\;\;\;\;\;\;\;\;\;\;\;\;\large\color{LightCyan} F(x)=a\cdot b\space -\int a'\cdot b$ $\newline$ >[!quote] >##### *Integration by parts examples* >![[Antiderivativeintegrationbypartsvisual.png|850]] - Since **integration by parts** requires you to solve another integral which **itself is the product of two functions, $a'$ and $c$** you may have to use this technique *inside of itself* multiple times. As a result its important to keep an eye out for situations where the integral of $a'$ and c **will never simplify into a single term** no matter *how many times* you recursively utilize integration by parts. $\newline$ $\text{Nested integration by parts: }\int a\cdot b\longrightarrow \;\;\;\;a\cdot c-\int\,\biggl(a'\cdot d-\int\,\biggl(a''\cdot n-\int...$ $\small\color{grey}\text{where }\biggl(c = \int b\biggl)\,\text{ and }\biggl(d=\int c\biggl)\,\text{ and }\biggl(n=\int d\biggl)...$ $\newline$ - U substitution is **often used in tandem with integration by parts since it allows us to simplify complex equations**. It should be used prior to integration by parts if it allows you to introduce a simplified term, whose derivative is equal to a constant *(usually this is u or u to some power)*. Likewise u substitution can be used to solve the secondary integral created created through integration by parts if it simplifies $a'$ and $b$ into a single term *(or a term that is more ideal when performing a **nested integration by parts**)*. - If $a$ and $b$ are ==both trigonometric functions their product will **never be equal to a single term**.== Therefore, you must either perform a **u substitution on the initial integrand *(getting rid of a trig term)* or plan ahead so that $a'$ and $c$ can be simplified into a single term using some trigonometric identity.** --- ## Example problems >[!example] >##### *U sub and Integration by parts example* >--- >$\small\text{First use u substitution:}$ >$\int x^3 \cos(x^2)\longrightarrow \Big(u=x^2\Big)\;\text{\small\color{grey} therefore }\;\Big(du=2x\,dx\Big)$ >$\int x^2\cancel{x^3}\cos(u)\,\frac{du}{2\cancel{x}}\;\longrightarrow\;\int u\cos(u)\,du$ >$\newline$ >$\small\text{Now using integration by parts:}$ >$a=u\;\text{\small\color{grey} and }\; b=\int\cos(u)=-\sin(u)\;\longrightarrow \biggl(-u\sin(u)-\int -1\sin(u)\biggl)$ >$\therefore\;-u\sin(u)-\cos(u)\longrightarrow\color{Bisque}\large \biggl(-x^2\cdot\sin(x^2)-\cos(x^2)\biggl)$ >[!example] >##### *Nested integration by parts example* >--- >$\small\text{Initial integration by parts:}$ >$\int x^2\cdot e^{4x}\,dx\;\;\;\;\;\;\color{grey}a=x^2\;\;\;\; b=\int e^{4x}=\frac{1}{4}e^{4x}$ >$\therefore \frac{1}{4}x^2\cdot e^{4x}-\int \frac{1}{2}x\cdot e^{4x}\;\;\;\;\;\;\color{grey} a= \frac{1}{2}x\;\;\;\;b=\int e^{4x}=\frac{1}{4}e^{4x}$ >$\newline$ >$\small\text{Nested integration by parts:}$ >$\therefore\frac{1}{8}x\cdot e^{4x}-\int\frac{1}{8}e^{4x}\longrightarrow\color{Bisque}\large\left(\frac{1}{4}x^2\cdot e^{4x}-\frac{1}{8}x\cdot e^{4x}-\frac{1}{32}e^{4x}\right)$ --- ## Tips/tricks ##### *Integration by parts trick #1* - There are some very specific situations ==**where integration by parts introduces a new integral into the equation that is identical to the *original* integral**.== When this happens you can **get rid of it by adding a new instance of the integral to both sides of the equation**, *(the other side original integral you are trying to solve)*. You can then use algebra to isolate the original integral on to one side of the equation, leaving you with its solution on the other. *This works because all sub integrals created through integration by parts are negative.* - This really only happens when the integrand contains trigonometric terms, exponential terms or terms involving the constant $e$. >[!example] >##### *Identical sub integral example problem* >--- >$\int\sin(x)e^x\,dx\;\;\;\;\;\;\color{grey}a=\sin(x)\;\;\;b=\int e^x=e^x$ >$\therefore e^x \cdot sin(x) -\int e^x\cdot cos(x)\;\;\;\;\;\;\color{grey} a_1=cos(x)\;\;\;\;b_2=\int e^x = e^x$ >$\newline$ >$\small\text{By using integration by parts inside itself:}$ >$\therefore e^x\cdot\sin(x)-\Big(e^x\cdot\cos(x)-\int e^x\cdot\sin(x)\Big)\;\;\;\;\color{grey}+\int e^x\cdot\sin(x)$ >$\therefore e^x\cdot\sin(x)-\Big(e^x\cdot\cos(x)-\cancel{\int e^x\cdot\sin(x)}\Big)=\int\sin(x)e^x\;\;\;\color{grey}+\int e^x\cdot\sin(x)$ >$\newline$ >$\small\text{By adding the original integral to both sides: }$ >$\therefore 2\int e^x\cdot\sin(x)\,dx = e^x\cdot\sin(x)-e^x\cdot\cos(x)=\color{Bisque}\large\frac{e^x\cdot\sin(x)-e^x\cdot\cos(x)}{2}$ #### *Integration by parts trick #2* - It is possible *(and sometimes useful)* to use *integration by parts* in situations where the **integrand contains only one term**. In these situations, the term $b$ is **set equal to one** which allows you to use integration by parts *without changing the integrand.* Since the integral of 1 is x there are some very niche situations where this can be helpful. - This technique is almoast exclusively used when natural log is in the integrand. There are some very rare situations where you may want to use it with trig functions as well. $\newline$ $\small\text{Evaluating the integral of the natural log of x}$$\int\ln(x)\,dx\longrightarrow\,{\color{grey}a =\ln(x)\;\;\;b=\int 1=x}\,\longrightarrow x\cdot\ln(x)-\int x\cdot\frac{1}{x}\;=\color{LightCyan}\; x\cdot\ln(x)-x$ ---