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Section 7.5 Integrating Powers of Trigonometric Functions

Let’s remind ourselves of two example problems that we’ve done in the past.
In Example 5.6.6, we performed a \(u\)-substitution, but needed to work to rewrite our whole integrand in terms of \(u\text{.}\) Specifically, we found that in the numerator, there was an \(x^3\text{,}\) but \(du=2x\;dx\text{.}\) We were substituting out a linear function of \(x\) in the numerator, but the actual function was cubic. This wasn’t a problem: we rewrote \(x^3=x^2\cdot x\text{,}\) and noticed that the extra \(x^2\) was able to be substituted, since we could rewrite our substitution rule: we noted that \(u=x^2+1\) is equivalent to \(x^2=u-1\text{.}\) This meant that even though we had an extra factor of \(x^2\) “in” the part that we were using for substituting in the differential \(du\text{,}\) we were still able to translate the whole function to be written in terms of \(u\text{.}\)
Then, more recently, in Example 7.4.9, we noted that we could use a mix of methods to integrate this:
\begin{equation*} \int \sin(x)\cos(x)\;dx\text{.} \end{equation*}
On one hand, we can look at the structure of the integrand and notice that we have a product of two functions! Integration by parts was a fine strategy to employ, and that’s what we did in the example. On the other hand, we noticed that since we have this function-derivative pairing, a \(u\)-substitution also was appropriate.
In this section, we’ll explore more combinations of trigonometric functions and build a strategy for antidifferentiating them that includes some ideas from both of these previous examples.

Subsection Building a Strategy for Powers of Sines and Cosines

Activity 7.5.1 Compare and Contrast.

Let’s do a quick comparison of two integrals, keeping the above examples in mind. Consider these two integrals:
\begin{equation*} \int \sin^4(x)\cos(x)\;dx \end{equation*}
\begin{equation*} \int \sin^4(x)\cos^3(x)\;dx \end{equation*}

(a)

Consider the first integral,\(\displaystyle\int\sin^4(x)\cos(x)\;dx\text{.}\) Think about and set up a good technique for antidifferentiating. Without actually solving the integral, explain why this technique will work.
Hint.
It might be helpful to notice that \(\sin^4(x)\) can be rewritten as \(\left(\sin(x)\right)^4\text{.}\) Does this help reveal something important about the structure of this integrand?

(b)

Now consider the second integral,\(\displaystyle\int\sin^4(x)\cos^3(x)\;dx\text{.}\) Does the same integration strategy work here? What happens when you apply the same thing?
Hint.
Let \(u=\sin(x)\) again, and \(du=\cos(x)\;dx\text{.}\) What happens with the cosine functions? How many are “left” after applying our substitution?

(c)

We know that \(\sin(x)\) and \(\cos(x)\) are related to each other through derivatives (each is the derivative of the other, up to a negative). Is there some other connection that we have between these functions? We might especially notice that we have a \(\cos^2(x)\) left over in our integral. Can we write this in terms of \(\sin(x)\text{,}\) so that we can write it in terms of \(u\text{?}\)
Hint.
We have a trigonometric identity (the Pythagorean Identity):
\begin{equation*} \sin^2(x)+\cos^2(x) = 1\text{.} \end{equation*}

(d)

Why would this strategy not have worked if we were looking at the integrals \(\displaystyle \int\sin^4(x)\cos^2(x)\;dx\) or\(\displaystyle\int\sin^4(x)\cos^4(x)\;dx\text{?}\) What, specifically, did we need in order to use this combination of substitution and trigonometric identity to solve the integral?

Integrating Powers of Sine and Cosine.

For integrals in the form\(\displaystyle\int\sin^p(x)\cos^q(x)\;dx\) where \(p\) and \(q\) are real number exponents:
  • If \(q\text{,}\) the exponent on \(\cos(x)\) is odd, we should use \(u=\sin(x)\) and \(du=\cos(x)\;dx\text{.}\) Then, we can apply the Pythagorean Identity \(\cos^2(x)=1-\sin^2(x)\text{.}\)
  • If \(p\text{,}\) the exponent on \(\sin(x)\) is odd, we should use \(u=\cos(x)\) and \(du=-\sin(x)\;dx\text{.}\) Then, we can apply the Pythagorean Identity \(\sin^2(x)=1-\cos^2(x)\text{.}\)
  • If both \(p\) and \(q\) are even, we can either use Integration by Parts or use the following power-reducing trigonometric identities:
    \begin{align*} \sin^2(x) \amp=\dfrac{1-\cos(2x)}{2} = \frac{1}{2}-\frac{\cos(2x)}{2} \\ \cos^2(x) \amp=\dfrac{1+\cos(2x)}{2} = \frac{1}{2}+\frac{\cos(2x)}{2} \end{align*}
A strange note, here, is that we typically pick our \(u\)-substitution based on looking to see a suitable candidate for \(u\text{:}\) we look for functions that are composed “inside” of other functions or we look for a function whose derivative is in the integral (the “function-derivative pair” that we talk about in Section 5.6). Here, though, we’re selecting our substitution based on \(du\text{:}\) we’re looking to see which function we can set aside one copy of for the differential, and then have an even power left over so that we can apply the Pythagorean Identity to translate the rest.

Example 7.5.2.

For each of the following, identify an appropriate substitution, make a note of which trigonometric identity you’ll use, and then integrate.

(a)

\(\displaystyle \int \sin^5(x)\cos^6(x)\;dx\)
Hint.
Notice that the exponent on \(\sin(x)\) is odd: if you let \(u=\cos(x)\text{,}\) you’ll end up with \(\sin^4(x)\) left over in your integral, and you can write it as \((\underbrace{1-\cos^2(x)}_{\sin^2(x)})^2\text{.}\)

(b)

\(\displaystyle \int \sin^{4/3}(x)\cos^5(x)\;dx\)
Hint.
This one seems scary at first, because of the fraction exponent. Notice, though, that you have no hope of converting fractional exponents of sine functions into cosine functions easily. So, pick \(u=\sin(x)\) and try to convert any remaining cosines using \(\cos^2(x)=1-\sin^2(x)\text{.}\)

(c)

\(\displaystyle \int \sin^3(x)\cos^9(x)\;dx\)
Hint.
You get a choice here! Both exponents are odd, so picking either function as \(u\) will leave you with an even exponent on the other function on which to use the Pythagorean Identity. Is there a choice of \(u\) that will be easier than the other choice?

Subsection Building a Strategy for Powers of Secants and Tangents

Activity 7.5.3 Compare and Contrast (Again).

We’re going to do another Compare and Contrast, but this time we’re only going to consider one integral:
\begin{equation*} \int \sec^4(x)\tan^3(x)\;dx\text{.} \end{equation*}
We’re going to employ another strategy, similar to the one for Integrating Powers of Sine and Cosine.

(a)

Before you start thinking about this integral, let’s build the relevant version of the Pythagorean Identity that we’ll use. Our standard version of this is:
\begin{equation*} \sin^2(x)+\cos^2(x)=1\text{.} \end{equation*}
Since we want a version that connects \(\tan(x)\text{,}\) which also is written as \(\frac{\sin(x)}{\cos(x)}\text{,}\) with \(\sec(x)\text{,}\) or \(\frac{1}{\cos(x)}\text{,}\) let’s divide everything in the Pythagorean Identity by \(\cos^2(x)\text{:}\)
\begin{align*} \frac{\sin^2(x)}{\fillinmath{XXXXX}} + \frac{\cos^2(x)}{\fillinmath{XXXXX}} \amp = \frac{1}{\fillinmath{XXXXX}}\\ \fillinmath{XXXXX} + \fillinmath{XXXXX} \amp = \fillinmath{XXXXX} \end{align*}
Solution.
\begin{equation*} \tan^2(x)+1 = \sec^2(x) \end{equation*}

(b)

Now start with the integral. We’re going to use two different processes here, two different \(u\)-substitutions. First, set \(u=\tan(x)\text{.}\) Complete the substitution and solve the integral.
Hint.
Here, \(du = \sec^2(x)\;dx\text{.}\) We’ll also use \(\sec^2(x) = \tan^2(x)+1\text{.}\)
Solution.
\begin{align*} \int \sec^4(x)\tan^3(x)\;dx \amp = \int \underbrace{\sec^2(x)}_{\tan^2(x)+1} \tan^3(x) \sec^2(x)\;dx \\ \amp =\int (u^2+1)u^3\;du\\ \amp =\int u^5+u^3\;du\\ \amp = \frac{u^6}{6}+\frac{u^4}{4}+C\\ \amp = \frac{\tan^6(x)}{6}+\frac{\tan^4(x)}{4}+C \end{align*}

(c)

Now try the integral again, this time using \(u=\sec(x)\) as your substitution.
Hint.
Now \(du=\sec(x)\tan(x)\;dx\text{,}\) and we’ll use the same Pythagorean Identity, just rewritten as \(\tan^2(x) = \sec^2(x)-1\text{.}\)
Solution.
\begin{align*} \int \sec^4(x)\tan^3(x)\;dx \amp = \int \sec^3(x)\underbrace{\tan^2(x)}_{\sec^2(x)-1}\sec(x)\tan(x)\;dx \\ \amp = \int u^3(u^2-1)\;du\\ \amp = \int u^5-u^3\;du\\ \amp = \frac{u^6}{6}-\frac{u^4}{4}+C\\ \amp = \frac{\sec^6(x)}{6} - \frac{\sec^4(x)}{4} + C \end{align*}

(d)

For each of these integrals, why were the exponents set up just right for \(u\)-substitution each time? How does the structure of the derivatives of each function play into this?
Hint.
Notice we had an even exponent on the \(\sec(x)\) function, but an odd exponent on the \(\tan(x)\) function.

(e)

Which substitution would be best for the integral\(\displaystyle\int\sec^4(x)\tan^4(x)\;dx\text{.}\) Why?

(f)

Which substitution would be best for the integral\(\displaystyle\int\sec^3(x)\tan^3(x)\;dx\text{.}\) Why?

Integrating Powers of Secant and Tangent.

For integrals in the form\(\displaystyle\int\sec^p(x)\tan^q(x)\;dx\) where \(p\) and \(q\) are real number exponents:
  • If \(q\text{,}\) the exponent on \(\tan(x)\text{,}\) is odd, we can use \(u=\sec(x)\) and \(du=\sec(x)\tan(x)\;dx\text{.}\) Then we can apply the Pythagorean Identity \(\tan^2(x)=\sec^2(x)-1\text{.}\)
  • If \(p\text{,}\) the exponent on \(\sec(x)\text{,}\) is even, we can use \(u=\tan(x)\) and \(du=\sec^2(x)\;dx\text{.}\) Then we can apply the Pythagorean Identity \(\sec^2(x)=\tan^2(x)+1\text{.}\)
  • If \(p\) is odd and \(q\) is even, we can use Integration by Parts.

Example 7.5.4.

For each of the following, identify an appropriate substitution, make a note of which trigonometric identity you’ll use, and then integrate.

(a)

\(\displaystyle \int \tan^3(x)\sec^8(x)\;dx\)
Hint.
You have some choices here! If you use \(u=\sec(x)\text{,}\) then you’ll end up with a remaining \(tan^2(x)\) to convert using a Pythagorean Identity. Alternatively, if you use \(u=\tan(x)\text{,}\) you’ll end up needing to convert the remaining \(\sec^6(x) = (\sec^2(x))^3\text{.}\)

(b)

\(\displaystyle \int \tan^{-5/2}(x)\sec^6(x)\;dx\)
Hint.
Another fraction exponent that is pushing us towards using \(u=\tan(x)\text{.}\) There will still be a \(\sec^4(x)\) to convert after substituting in \(du\text{.}\)

(c)

\(\displaystyle \int \tan^5(x)\sec^3(x)\;dx\)
Hint.
The odd exponent on tangent is fine, since we can use \(u=\sec(x)\) to leave us with a \(\tan^4(x)\) to convert.

Practice Problems Practice Problems

1.

For an integral \(\dint \sin^a(x)\cos^b(x)\;dx\text{,}\) how do you know whether to use \(u=\sin(x)\) or \(u=\cos(x)\) as the substitution?

2.

For an integral \(\dint \tan^a(x)\sec^b(x)\;dx\text{,}\) how do you know whether to use \(u=\tan(x)\) or \(u=\sec(x)\) as the substitution?

3.

Integrate the following.

(a)

\(\dint \sin^3(x)\cos^2(x)\;dx\)
Solution.
\begin{align*} u \amp = \cos(x)\\ du \amp = -\sin(x)\;dx\\ \int \sin^3(x)\cos^2(x)\;dx \amp = -\int(1-\cos^2(x))\cos^2(x)(-\sin(x))\;dx\\ \amp = -\int (1-u^2)u^2\;du\\ \amp = - \int u^2-u^4\;du\\ \amp = - \left(\frac{u^3}{3} - \frac{u^5}{5}\right)+C\\ \amp = \frac{\cos^5(x)}{5} - \frac{\cos^3(x)}{3} + C \end{align*}

(b)

\(\dint \sin^2(x)\cos^3(x)\;dx\)
Solution.
\begin{align*} u \amp = \sin(x)\\ du \amp = \cos(x)\;dx\\ \int \sin^2(x)\cos^3(x)\;dx \amp = \int \sin^2(x)(1-\sin^2(x))\cos(x)\;dx\\ \amp = \int u^2(1-u^2)\;du\\ \amp = \int u^2 - u^4\;du\\ \amp = \frac{u^3}{3} - \frac{u^5}{5} + C\\ \amp = \frac{\sin^3(x)}{3} - \frac{\sin^5(x)}{5} + C \end{align*}

(c)

\(\dint \sin^3(x)\cos^3(x)\;dx\)
Solution 1.
\begin{align*} u \amp = \sin(x)\\ du \amp = \cos(x)\;dx\\ \int \sin^3\cos^3(x)\;dx \amp = \int u^3(1-u^2)\;du\\ \amp = \frac{u^4}{4} - \frac{u^6}{6} + C\\ \amp = \frac{\sin^4(x)}{4} - \frac{\sin^6(x)}{6} + C \end{align*}
Solution 2.
\begin{align*} u \amp = \cos(x)\\ du \amp = -\sin(x)\;dx\\ \int \sin^3\cos^3(x)\;dx \amp = -\int u^3(1-u^2)\;du\\ \amp = -\frac{u^4}{4} + \frac{u^6}{6} + C\\ \amp = \frac{\cos^6(x)}{6} - \frac{\cos^4(x)}{4} + C \end{align*}

(d)

\(\dint \tan^4(x)\sec^4(x)\;dx\)
Solution.
\begin{align*} u \amp = \tan(x)\\ du \amp = \sec^2(x)\;dx\\ \int \tan^4(x) \sec^4(x)\;dx \amp = \int \tan^4(x) (\tan^2(x)+1)\sec^2(x)\;dx\\ \amp = \int u^4(u^2+1)\;du\\ \amp = \int u^6 + u^4 \;du\\ \amp = \frac{u^7}{7} + \frac{u^5}{5}+C\\ \amp = \frac{\tan^7(x)}{7} + \frac{\tan^5(x)}{5} + C \end{align*}

(e)

\(\dint \tan^3(x)\sec^3(x)\;dx\)
Solution.
\begin{align*} u \amp = \sec(x)\\ du \amp = \sec(x)\tan(x)\;dx\\ \int \tan^3(x) \sec^3(x)\;dx \amp = \int \sec^2(x)(\sec^2(x)-1)\sec(x)\tan(x)\;dx\\ \amp = \int u^2(u^2-1)\;du\\ \amp = \int u^4 - u^2\;du\\ \amp = \frac{u^5}{5} - \frac{u^3}{3} + C\\ \amp = \frac{\sec^5(x)}{5} - \frac{\sec^3(x)}{3} + C \end{align*}

(f)

\(\dint \tan^3(x)\sec^4(x)\;dx\)
Solution 1.
\begin{align*} u \amp = \tan(x)\\ du \amp = \sec^2(x)\;dx\\ \int \tan^3(x)\sec^4(x)\;dx \amp = \int u^3(u^2+1)\;du\\ \amp = \frac{u^6}{6} + \frac{u^4}{4} + C\\ \amp = \frac{\tan^6(x)}{6} + \frac{\tan^4(x)}{4} + C \end{align*}
Solution 2.
\begin{align*} u \amp = \sec(x)\\ du \amp = \sec(x)\tan(x)\;dx\\ \int \tan^3(x) \sec^4(x)\;du \amp = \int u^3(u^2-1)\;du\\ \amp = \frac{u^6}{6} - \frac{u^4}{4} + C\\ \amp = \frac{\sec^6(x)}{6} - \frac{\sec^4(x)}{4} + C \end{align*}

4.

Integrate the following.

(a)

\(\dint \sin^{3/4}(x)\cos^5(x)\;dx\)
Solution.
\begin{align*} u \amp = \sin(x)\\ du \amp = \cos(x)\;dx\\ \int \sin^{3/4}(x)\cos^5(x)\;dx \amp = \int u^{3/4}(1-u^2)^2\;du\\ \amp = \int u^{3/4} - 2u^{11/4} + u^{19/4}\;du \\ \amp = \frac{4u^{7/4}}{7} - \frac{8u^{15/4}}{15} + \frac{4u^{23/4}}{23} + C\\ \amp = \frac{4\sin^{7/4}(x)}{7} - \frac{8\sin^{15/4}(x)}{15} + \frac{4\sin^{23/4}(x)}{23} + C \end{align*}

(b)

\(\dint \tan^5(x)\sec^{-1/2}(x)\;dx\)
Solution.
\begin{align*} u \amp = \sec(x)\\ du \amp = \sec(x)\tan(x)\;dx\\ \int \tan^5(x)\sec^{-1/2}(x)\;dx \amp = \int \tan^4(x)\sec^{-3/2}(x)\sec(x)\tan(x)\;dx\\ \amp = \int (u^2-1)^2u^{-3/2}\;du\\ \amp = \int u^{1/2}-u^{-3/2} \;du\\ \amp = \frac{2u^{3/2}}{3} + \frac{2}{u^{1/2}} + C\\ \amp = \frac{2\sec^{3/2}(x)}{3} + \frac{2}{\sqrt{\sec(x)}} + C \end{align*}

5.

Consider the integral \(\dint \sin^2(x)\;dx\text{.}\)

(a)

Use the trigonometric identity:
\begin{equation*} \sin^2(x) = \frac{1-\cos(2x)}{2} \end{equation*}
to integrate.
Solution.
\begin{align*} \int \sin^2(x)\;dx \amp = \frac{1}{2}\int 1-\cos(2x)\;dx\\ \amp = \frac{x}{2} - \frac{\sin(2x)}{4} + C \end{align*}

(b)

Use integration by parts to integrate.
Hint.
Solution.
\begin{align*} u \amp = \sin(x) \amp v \amp = -\cos(x)\\ du \amp = \cos(x)\;dx \amp dv \amp = \sin(x)\;dx \end{align*}
\begin{align*} \int \sin^2(x)\;dx \amp = -\sin(x)\cos(x) + \int \cos^2(x)\;dx \\ \int \sin^2(x)\;dx \amp = -\sin(x)\cos(x) + \int 1 - \sin^2(x)\;dx\\ \int \sin^2(x)\;dx \amp = -\sin(x)\cos(x)+ x -\int \sin^2(x)\;dx\\ 2\int \sin^2(x)\;dx \amp = -\sin(x)\cos(x)+x\\ \int \sin^2(x)\;dx \amp = -\frac{\sin(x)\cos(x)}{2} + \frac{x}{2} + C \end{align*}

(c)

Which of these techniques do you think was easier to implement and use? Why is that?

6.

Consider the integral \(\dint \cos^4(x)\;dx\text{.}\)

(a)

Use the trigonometric identity:
\begin{equation*} \cos^2(x) = \frac{1+\cos(2x)}{2} \end{equation*}
to integrate.
Solution.
\begin{align*} \int \cos^4(x)\;dx \amp = \int \left(\frac{1 + \cos(2x)}{2}\right)^2\;dx\\ \amp = \frac{1}{4}\int 1 + 2\cos(2x) + \cos^2(2x)\;dx\\ \amp = \frac{1}{4}\int 1 + 2\cos(2x) + \frac{1}{2} + \frac{\cos(4x)}{2}\;dx\\ \amp = \frac{3x}{8} + \frac{\sin(2x)}{4} + \frac{\sin(4x)}{32} + C \end{align*}

(b)

Use integration by parts to integrate.
Hint.
Try picking \(u=\cos^3(x)\) and \(dv=\cos(x)\;dx\text{.}\)
Solution.
\begin{align*} u \amp = \cos^3(x) \amp v \amp = \sin(x)\\ du \amp = -3\cos^2(x)\sin(x)\;dx \amp dv \amp = \cos(x)\;dx \end{align*}
\begin{align*} \int \cos^4(x)\;dx \amp = \sin(x)\cos^3(x) + 3\int \sin^2(x)\cos^2(x)\;dx\\ \int \cos^4(x)\;dx \amp = \sin(x)\cos^3(x) + 3 \int(1-\cos^2(x))\cos^2(x)\;dx\\ \int \cos^4(x)\;dx \amp = \sin(x)\cos^3(x) + 3\int \cos^2(x)\;dx - 3\int\cos^4(x)\;dx\\ 4\int \cos^4(x)\;dx \amp = \sin(x)\cos^3(x) + 3\int \cos^2(x)\;dx\\ \int \cos^4(x)\;dx \amp = \frac{\sin(x)\cos^3(x)}{4} + \frac{3}{4} \int \cos^2(x)\;dx \end{align*}
We can handle \(\dint \cos^2(x)\;dx\) in two ways: either use the trig identity, or integration by parts. I’ll show integration by parts, since that’s what I started with.
\begin{align*} u \amp = \cos(x) \amp v \amp = \sin(x)\\ du \amp = -\sin(x)\;dx \amp dv \amp = \cos(x)\;dx \end{align*}
\begin{align*} \int \cos^2(x)\;dx \amp = \sin(x)\cos(x) + \int \sin^2(x)\;dx\\ \int \cos^2(x)\;dx \amp = \sin(x)\cos(x) + \int 1-\cos^2(x)\;dx\\ 2\int \cos^2(x)\;dx \amp = \sin(x)\cos(x) + x\\ \int \cos^2(x)\;dx \amp = \frac{\sin(x)\cos(x)}{2} + \frac{x}{2} + C \end{align*}
So we can return to our original integral:
\begin{align*} \int \cos^4(x)\;dx \amp = \frac{\sin(x)\cos^3(x)}{4} + \frac{3}{4} \int \cos^2(x)\;dx\\ \int\cos^4(x)\;dx \amp = \frac{\sin(x)\cos^3(x)}{4} + \frac{3\sin(x)\cos(x)}{8} + \frac{3x}{8} + C \end{align*}

(c)

Which of these techniques did you prefer? Why?

7.

Integrate the following integrals.

(a)

\(\dint \tan^2(x)\;dx\)
Hint.
Use a Pythagorean Identity to convert this to be written in terms of secant functions.
Solution.
\begin{align*} \int \tan^2(x)\;dx \amp = \int \sec^2(x) - 1\;dx\\ \amp = \tan(x)-x + C \end{align*}

(b)

\(\dint \sec^3(x)\;dx\)
Hint.
Integration by parts works well here, and it’s helpful to know the derivative of \(\sec(x)\) and an antiderivative of \(\sec^2(x)\text{.}\)
Solution.
\begin{align*} u \amp = \sec(x) \amp v \amp = \tan(x)\\ du \amp = \sec(x)\tan(x)\;dx \amp dv \amp = \sec^2(x)\;dx \end{align*}
\begin{align*} \int \sec^3(x)\;dx \amp = \sec(x)\tan(x) - \int \sec(x)\tan^2(x)\;dx\\ \int \sec^3(x)\;dx \amp = \sec(x)\tan(x) - \int \sec(x)(\sec^2(x) - 1)\;dx \\ \int \sec^3(x)\;dx \amp = \sec(x)\tan(x) + \int \sec(x)\;dx - \int \sec^3(x)\;dx\\ 2\int\sec^3(x)\;dx \amp = \sec(x)\tan(x) + \ln|\sec(x)+\tran(x)| + C\\ \int \sec^3(x)\;dx \amp = \frac{\sec(x)\tan(x)}{2} + \frac{1}{2}\ln|\sec(x)+\tan(x)|+ C \end{align*}

(c)

\(\dint \tan^5(x)\;dx\)
Hint.
You can technically use either \(u=\sec(x)\) or \(u=\tan(x)\) here.
Solution.
We’ll look at the solution when \(u=\tan(x)\) and \(du=\sec^2(x)\;dx\text{.}\)
\begin{align*} \int \tan^5(x) \;dx\amp = \int \frac{u^5}{u^2+1}\;du\\ \amp = \int u^3 - u + \frac{u}{u^2+1}\;du\\ \amp = \frac{u^4}{4} - \frac{u^2}{2} + \frac{1}{2}\ln(u^2+1) + C\\ \amp = \frac{\tan^4(x)}{4} - \frac{\tan^2(x)}{2} + \frac{1}{2}\ln(sec^2(x)) + C \end{align*}

(d)

\(\dint \sin^5(x)\;dx\)
Solution.
\begin{align*} u \amp = \cos(x)\\ du \amp = -\sin(x)\;dx\\ \int \sin^5(x)\;dx \amp = -\int (1-u^2)^2\;du\\ \amp = -\int u^4-2u^2+1\;du\\ \amp = -\frac{u^5}{5} + \frac{2u^3}{3} - u + C\\ \amp = - \frac{\cos^5(x)}{5} + \frac{2\cos^3(x)}{3} - \cos(x)+C \end{align*}
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