--- ## How to perform a u substitution - U substitution, also known as the **u rule** is a technique that can help you find the antiderivative of equations that contain **nested functions *(functions inside other functions)***. This is done by first, setting a term equal to $u$ . Then, by setting the derivative of this term: $dx$ equal to $du$ *(and bringing that reciprocal of that derivative back into the equation)* you can replace the instance of this term in the integrand with $u$. In some cases, this will allow you to simplify the integrand making it easier to integrate. At the end of this process, u is replaced with it's original term and the bounds are plugged into the resulting antiderivative $F(x)$ to solve the integral. $\newline$ $\color{LightCyan}\Large\text{U substitution:}$ $\newline$ $\large\int f\Big[g(x)\Big]\cdot g'(x)\space dx =\color{LightCyan}\int f(u)\space du = F(u) + C$ $\large\text{where: }u = g(x)\space\space\space\space\text{and: }du = g'(x)$ $\newline$ - You can substitute more than one term *(in the integrand)* with $u$ if those two terms are the same *(or through algebraic manipulation you can **make** those two terms the same)*. - Substitution is **extremely useful for solving integrals that contain fractions**. This is because fractions are often just nested functions that are visually split apart. - This is because the **denominator of any fraction is technically just the product of the numerator and itself to the negative first power** $\text{Recall: }\space\frac{a}{b}=a(b)^{-1}\space\space\space\space\space\text{ thus, use substitution or integration by parts}$ $\small\color{grey}\text{substitution works well if }a=b'$ - An important aspect of substitution is choosing which term to substitute with u. Ideally the term **whose differential also occurs in the integrand, which will allow you to cancel it out.** - Its important **to integrate *only* after you have fully substituted $g(x)$ for $u$ and $dx$ with $\frac{du}{g'(x)}$.** After that, treat the integral as you would any other and solve it normally, treating $u$ as a variable. - You may need to use **integration by parts** or **substitution again** to solve the integral you made as a result of using technique. As a result, its important to keep your work clear and organized so you always know how to collapse everything back into the final equation. --- ## Common situations >[!quote] >##### *Common u substitution situations* >--- >$\text{Term cancelation}$ >$\int 2x\cdot\ln(x^2)\space dx\longrightarrow u = x^2\longrightarrow du=2x\space dx$ >$\large\color{LightCyan}\int ln(u)\space\cancel{2x\space dx}\longrightarrow\int\ln(u)du$ >--- >$\text{Advanced term cancelation}$ >$\int\frac{x^2}{x^3+3}\space dx\longrightarrow u = x^3\longrightarrow du = 3x^2\space dx\longrightarrow\frac{du}{3}=x^2\space dx$ >$\large\color{LightCyan}\int\frac{\cancel{x^2}}{u+3}\space\frac{du}{3}\longrightarrow\frac{1}{3}\int\frac{1}{u+3}\space du$ >--- >$\text{Multiple substitution}$ >$\int\frac{x+7}{x-2}\longrightarrow u = x-2\longrightarrow du = dx$ >$\large\color{LightCyan}\int\frac{u+9}{u}\space\cancel{dx}\longrightarrow\int 1 + 9u^{-1}\space du$ --- ## U substitution example problems >[!example] >##### *Substitution examples* >--- >$\textit{Example 1 - evaluate: }\int\sqrt{2x+1}dx$ >$\textit{\color{grey}Let }\space u=2x+1\space\space\textit{\color{grey} thus, }\space du=2dx\longrightarrow dx=\frac{1}{2}du$ >$\newline$ >$\therefore\int\sqrt{u}\cdot\frac{1}{2}du=\frac{1}{2}\int (u)^{\frac{1}{2}}du\longrightarrow\left(\frac{1}{2}\cdot\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C\right)=\left(\frac{1}{3}u^{\frac{3}{2}}+C\right)\space\textit{\color{grey}\small Now substitute!}$ >$\large\therefore\int\sqrt{2x+1}dx=\color{Bisque}\frac{1}{3}(2x+1)^{\frac{3}{2}}+C$ >--- >$\textit{Example 2 - evaluate: }\int\frac{x}{\sqrt{1-4x^2}}dx$ >$\textit{\color{grey}Let }\space u=1-4x^2\space\text{\color{grey} thus, }\space du=-8x\space dx\longrightarrow x\cdot dx=-\frac{1}{8}du$ >$\newline$ >$\therefore\int\frac{x}{\sqrt{1-4x^2}}dx=\left(-\frac{1}{8}\int u^{-\frac{1}{2}}du\right)\longrightarrow\left(-\frac{1}{8}(2\sqrt{u}+C\right)\space\textit{\color{grey}\small Now substitute!}$ >$\large\therefore\int\frac{x}{\sqrt{1-4x^2}}dx=\color{Bisque}-\frac{1}{4}\sqrt{1-4x^2}+C$ >--- >![[Antiderivativesubstitution example.png|850]] >![[Antiderivativesubstitution example 2.png|850]] --- ## U substitution tips/tricks ##### *U sub trick #1* - In some situations *(especially those involving a square root)* you can substitute **both the original term and its derivative with u**. After doing this you can use *integration by parts* to finish evaluating the integral. An example of this is shown below. $\int\sin(\sqrt{x})\;\;\;\;\;\;\color{grey}u=\sqrt{x}\;\;\;\;\;\;du=\frac{1}{2\sqrt{x}}dx\rightarrow du\cdot 2\sqrt{x}=dx$ $\therefore\;2\int\sqrt{x}\cdot \sin(u)\;du\longrightarrow 2\int u\sin(u)\;du$ $\newline$ $\color{grey}\small\text{Now you can evaluate the integral using integration by parts...}$ $\newline$ ##### *U sub trick #2* - In some situations it may be to **algebraically re-arrange the u substitution equation *(the one that looks like $u=x$)* if it allows you to simplify the integrand more**. - *NOTE: In the example below the simplification from $\sin^{-1}(\sin(u))$ to $u$ only works if the bounds of the integral or x is between negative and positive $\frac{\pi}{2}$)* $\int\sin^{-1}(x)\;dx\;\;\;\;\color{grey}u=\sin^{-1}(x)\longrightarrow\frac{u}{\sin^{-1}}=\frac{\cancel{\sin^{-1}}(x)}{\cancel{\sin^{-1}}}$ ${\color{grey}\therefore x=\sin(u)\;\;\;dx=cos(u)}\longrightarrow\int\sin^{-1}\Big(\sin(u)\Big)\cos(u)\,du\longrightarrow\color{Bisque}\int u\cos(u)\,du$ $\newline$ --- #mainpage