a \( \displaystyle \int{{\tan x\,dx}}\)

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The first question about this problem is probably why is it here? Substitution rule problems generally require more than a single function. The key to this problem is to realize that there really are two functions here. All we need to do is remember the definition of tangent and we can write the integral as,

\[\int{{\tan x\,dx}} = \int{{\frac{{\sin x}}{{\cos x}}\,dx}}\]

Written in this way we can see that the following substitution will work for us,

\[u = \cos x\hspace{0.25in}du = - \sin x\,dx \hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\sin x\,dx = - du\]

The integral is then,

\[\begin{align*}\int{{\tan x\,dx}} & = - \int{{\frac{1}{u}\,du}}\\ & = - \ln \left| u \right| + c\\ & = - \ln \left| {\cos x} \right| + c\end{align*}\]

Now, while this is a perfectly serviceable answer that minus sign in front is liable to cause problems if we aren’t careful. So, let’s rewrite things a little. Recalling a property of logarithms we can move the minus sign in front to an exponent on the cosine and then do a little simplification.

\[\begin{align*}\int{{\tan x\,dx}} & = - \ln \left| {\cos x} \right| + c\\ & = \ln {\left| {\cos x} \right|^{ - 1}} + c\\ & = \ln \frac{1}{{\left| {\cos x} \right|}} + c\\ & = \ln \left| {\sec x} \right| + c\end{align*}\]

This is the formula that is typically given for the integral of tangent.

Note that we could integrate cotangent in a similar manner.

b \( \displaystyle \int{{\sec y\,dy}}\)

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This problem also at first appears to not belong in the substitution rule problems. This is even more of a problem upon noticing that we can’t just use the definition of the secant function to write this in a form that will allow the use of the substitution rule.

This problem is going to require a technique that isn’t used terribly often at this level but is a useful technique to be aware of. Sometimes we can make an integral doable by multiplying the top and bottom by a common term. This will not always work and even when it does it is not always clear what we should multiply by but when it works it is very useful.

Here is how we’ll use this idea for this problem.

\[\int{{\sec y\,dy}} = \int{{\frac{{\sec y}}{1}\frac{{\left( {\sec y + \tan y} \right)}}{{\left( {\sec y + \tan y} \right)}}\,dy}}\]

First, we will think of the secant as a fraction and then multiply the top and bottom of the fraction by the same term. It is probably not clear why one would want to do this here but doing this will actually allow us to use the substitution rule. To see how this will work let’s simplify the integrand somewhat.

\[\int{{\sec y\,dy}} = \int{{\frac{{{{\sec }^2}y + \tan y\sec y}}{{\sec y + \tan y}}\,dy}}\]

We can now use the following substitution.

\[u = \sec y + \tan y\hspace{0.25in}du = \left( {\sec y\tan y + {{\sec }^2}y} \right)dy\]

The integral is then,

\[\begin{align*}\int{{\sec y\,dy}} & = \int{{\frac{1}{u}\,du}}\\ & = \ln \left| u \right| + c\\ & = \ln \left| {\sec y + \tan y} \right| + c\end{align*}\]

Sometimes multiplying the top and bottom of a fraction by a carefully chosen term will allow us to work a problem. It does however take some thought sometimes to determine just what the term should be.

We can use a similar process for integrating cosecant.

c \( \displaystyle \int{{\frac{{\cos \left( {\sqrt x } \right)}}{{\sqrt x }}\,dx}}\)

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This next problem has a subtlety to it that can get us in trouble if we aren’t paying attention. Because of the root in the cosine it makes some sense to use the following substitution.

\[u = {x^{\frac{1}{2}}} \hspace{0.5in} du = \frac{1}{2}{x^{ - \,\,\frac{1}{2}}}dx\]

This is where we need to be careful. Upon rewriting the differential we get,

\[2du = \frac{1}{{\sqrt x }}dx\]

The root that is in the denominator will not become a \(u\) as we might have been tempted to do. Instead it will get taken care of in the differential.

The integral is,

\[\begin{align*}\int{{\frac{{\cos \left( {\sqrt x } \right)}}{{\sqrt x }}\,dx}} & = 2\int{{\cos \left( u \right)\,du}}\\ & = 2\sin \left( u \right) + c\\ & = 2\sin \left( {\sqrt x } \right) + c\end{align*}\]

d \( \displaystyle \int{{{{\bf{e}}^{t + {{\bf{e}}^t}}}\,dt}}\)

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With this problem we need to very carefully pick our substitution. As the problem is written we might be tempted to use the following substitution,

\[u = t + {{\bf{e}}^t}\hspace{0.25in}du = \left( {1 + {{\bf{e}}^t}} \right)dt\]

However, this won’t work as you can probably see. The differential doesn’t show up anywhere in the integrand and we just wouldn’t be able to eliminate all the \(t\)’s with this substitution.

In order to work this problem we will need to rewrite the integrand as follows,

\[\int{{{{\bf{e}}^{t + {{\bf{e}}^t}}}\,dt}} = \int{{{{\bf{e}}^t}{{\bf{e}}^{{{\bf{e}}^t}}}\,dt}}\]

We will now use the substitution,

\[u = {{\bf{e}}^t} \hspace{0.5in} du = {{\bf{e}}^t}dt\]

The integral is,

\[\begin{align*}\int{{{{\bf{e}}^{t + {{\bf{e}}^t}}}\,dt}} & = \int{{{{\bf{e}}^u}\,du}}\\ & = {{\bf{e}}^u} + c\\ & = {{\bf{e}}^{{{\bf{e}}^t}}} + c\end{align*}\]

Some substitutions can be really tricky to see and it’s not unusual that you’ll need to do some simplification and/or rewriting to get a substitution to work.

e \( \displaystyle \int{{2{x^3}\sqrt {{x^2} + 1} \,dx}}\)

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This last problem in this set is different from all the other substitution problems that we’ve worked to this point. Given the fact that we’ve got more than an \(x\) under the root it makes sense that the substitution pretty much has to be,

\[u = {x^2} + 1 \hspace{0.5in} du = 2x\,dx\]

At first glance it looks like this might not work for the substitution because we have an \({x^3}\) in front of the root. However, if we first rewrite \(2{x^3} = {x^2}\left( {2x} \right)\) we could then move the \(2x\) to the end of the integral so at least the \(du\) will show up explicitly in the integral. Doing this gives the following,

\[\begin{align*}\int{{2{x^3}\sqrt {{x^2} + 1} \,dx}} & = \int{{{x^2}\sqrt {{x^2} + 1} \,\left( {2x} \right)dx}}\\ & = \int{{{x^2}{u^{\frac{1}{2}}}\,du}}\end{align*}\]

This is a real problem. Our integrals can’t have two variables in them. Normally this would mean that we chose our substitution incorrectly. However, in this case we can rewrite the substitution as follows,

\[{x^2} = u - 1\]

and now, we can eliminate the remaining \(x\)’s from our integral. Doing this gives,

\[\begin{align*}\int{{2{x^3}\sqrt {{x^2} + 1} \,dx}} & = \int{{\left( {u - 1} \right){u^{\frac{1}{2}}}\,du}}\\ & = \int{{{u^{\frac{3}{2}}} - {u^{\frac{1}{2}}}\,du}}\\ & = \frac{2}{5}{u^{\frac{5}{2}}} - \frac{2}{3}{u^{\frac{3}{2}}} + c\\ & = \frac{2}{5}{\left( {{x^2} + 1} \right)^{\frac{5}{2}}} - \frac{2}{3}{\left( {{x^2} + 1} \right)^{\frac{3}{2}}} + c\end{align*}\]

Sometimes, we will need to use a substitution more than once.

This kind of problem doesn’t arise all that often and when it does there will sometimes be alternate methods of doing the integral. However, it will often work out that the easiest method of doing the integral is to do what we just did here.