\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Precalculus - Technique of Substitution




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The technique of substitution is something you need to understand while you learn calculus. It is used a lot, so it is important that you have a good grasp on how to use it.
There are actually two contexts when talking about substitution. First, using substitution to change an equation so that we can factor it or otherwise manipulate it more easily, which is discussed on this page. Another context is to use substitution to solve systems of equations. You can find that discussion on the solving systems of equations page.

The idea is to replace part of an equation with a temporary variable so that you can perform some operation more easily. One example is factoring. Let's say you have an expression that looks like this

\( e^{2n} + 2e^n + 1 \)

and you are asked to factor this. I don't know about you but nothing really jumps out at me and I don't even know if it is factorable. However, let's use substitution and see if that helps. I know how to factor polynomials, so if I replace \( e^n \) with \( t \), i.e. let \( t = e^n \), the result is \( t^2 + 2t + 1 \) which factors into \( (t + 1)(t + 1) = (t+1)^2 \). Substituting back in we end up with

\( e^{2n} + 2e^n + 1 = (e^n+1)^2 \)

Now, let's discuss the details of substitution, using the last example. First, we need to use a variable different from what is given in the problem. In this case, the variable was n, so we used t. We could have used x or y or almost any other letter. We could have even used a greek letter like \( \theta \). The point is to use something other than n. Note: We never use the letter o or O since they are easily confused with the number zero.

Secondly, what to choose to substitute for is a matter of experience. However, as you are learning, try different things until something works. Then sit back and notice patterns in your work.

Third, once you have chosen the substitution, you need to substitute appropriately everywhere in the expression. So, in the previous example, it wouldn't help to substitute at one place and not the other, i.e. \( e^{2n} + 2t + 1 \) doesn't really help us and I can't really do anything with it.

Fourth, never substitute something like \( n = t \). This substitution doesn't help at all.

Finally, once you have finished all adjustments or algebra or calculus on the expression, remember to undo the substitution by substituting back, i.e. if you initially substitute \( t \) for \( e^n \), you need to go back and replace all places where you have \( t \) with \( e^n \). Your final answer should contain the same variables as the original expression.

Sometimes this last step is easy, other times it involves more work. In the example above, we just factored, so we had \( t \) in the expression and we could just do a straight substitution back. But sometimes, you may need to do some algebra on the expression to know how to substitute back.

For example, if you initially substituted \( t = e^n \) but ended up with something like \( n + 1 \), you need to do the following.

\( \begin{array}{rcl} t & = & e^n \\ \ln(t) & = & \ln(e^n) \\ \ln(t) & = & n \end{array} \)

Then substitute \( \ln(t) \) for \( n \) to get \( n + 1 = \ln(t) + 1 \). This kind of substitution is something you may see in derivatives and integrals.

Another thing to remember when doing the initial substitution is that you do not need to do anything with constants in the problem, UNLESS the constants are dependent on the variable.

Really UNDERSTAND Precalculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)


\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)


\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)


\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)


\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)


\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)


\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)


\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)


\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)


\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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