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Calculus Cameron Geisler

Section 6.3 Integrals of Trigonometric Functions

In this section, we explore the integrals (antiderivatives) of various trigonometric functions. First, recall the antiderivative of sine and cosine,
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\begin{align*} \int \sin{x} \,dx \amp= -\cos{x} + C\\ \text{and} \qquad \int \cos{x} \,dx \amp= \sin{x} + C \end{align*}
The integrals of the other trigonometric functions are somewhat more complicated, and require using substitution. First, here are all of them together,
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\begin{equation*} \boxed{\begin{array}{ll} \int \sin{x} \,dx = -\cos{x} + C \amp \int \cos{x} \,dx = \sin{x} + C \\ \int \sec{x} \,dx = \ln{\abs{\sec{x} + \tan{x}}} + C \amp \int \csc{x} \,dx = -\ln{\abs{\csc{x} + \cot{x}}} + C \\ \int \tan{x} \,dx = -\ln{\abs{\cos{x}}} + C \amp \int \cot{x} \,dx = \ln{\abs{\sin{x}}} + C \end{array}} \end{equation*}
You may get a formula sheet on your final exam with these formulas, or you may be expected to memorize them (or derive them if needed).
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Remark 6.3.1.

A pattern that can help you memorize these antiderivatives, is that the antiderivative of the “co” functions (cosine, cosecant, cotangent) are similar to their corresponding non-co functions (sine, secant, tangent), except,
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  • They have the opposite sign (so if the non-co function has a positive sign, the co function has a negative sign, and vice versa).
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  • They have similar function parts, but “flipped”, in that non-co functions become co functions, and co becomes non-co. In particular,
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    integral of... non-co-function co-function
    \(\cos{x}\) \(\longleftrightarrow\) \(\sin{x}\)
    \(\ln{\abs{\sec{x}+\tan{x}}}\) \(\longleftrightarrow\) \(\ln{\abs{\csc{x}+\cot{x}}}\)
    \(\ln{\abs{\cos{x}}}\) \(\longleftrightarrow\) \(\ln{\abs{\sin{x}}}\)
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Subsection 6.3.1 Integrals of Tangent, Secant, Cosecant, and Cotangent

Remark 6.3.3.

Note that these two antiderivatives are equivalent, because,
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\begin{align*} -\ln{\abs{\cos{x}}} \amp= \ln{\brac{\abs{\cos{x}}^{-1}}} \amp\amp\text{by log laws}\\ \amp= \ln{\abs{\frac{1}{\cos{x}}}}\\ \amp= \ln{\abs{\sec{x}}} \end{align*}
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Example 6.3.4. Deriving the Integral of Tangent.

To integrate \(\tan{x}\text{,}\) recall that \(\tan{x} = \frac{\sin{x}}{\cos{x}}\text{.}\) We can rewrite it as \(\frac{\sin{x}}{\cos{x}}\text{,}\) and use substitution,
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\begin{equation*} \int \tan{x} \,dx = \int \frac{\sin{x}}{\cos{x}} \,dx \end{equation*}
Notice that there is \(\cos{x}\) in the denominator, and its derivative \(-\sin{x}\) is in the numerator (except with a negative sign), so we can use substitution. Then,
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\begin{align*} u = \cos{x} \qquad du \amp= -\sin{x} \,dx\\ -du \amp= \sin{x} \,dx \end{align*}
Then,
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\begin{align*} \amp= -\int \frac{1}{u} \,du\\ \amp= -\ln{\abs{u}} + C\\ \amp= -\ln{\abs{\cos{x}}} + C\\ \amp= \ln{\abs{(\cos{x})^{-1}}} + C\\ \amp= \ln{\abs{\sec{x}}} + C \end{align*}
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Example 6.3.6. Deriving the Integral of Secant.

There are multiple ways to integrate \(\sec{x}\text{,}\) but the most common way is to use a very special trick, of multiplying the numerator and denominator by \(\sec{x} + \tan{x}\text{,}\)
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\begin{equation*} \int \sec{x} \,dx = \int \frac{\sec{x}(\sec{x} + \tan{x})}{\sec{x} + \tan{x}} \,dx \end{equation*}
It turns out that this allows us to use substitution, because there is \(\sec{x}+\tan{x}\) in the denominator, and its derivative is \(\sec{x} \tan{x} + \sec^2{x}\text{,}\) which is exactly what is in the numerator. Then,
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\begin{align*} u = \sec{x} + \tan{x} \qquad du \amp= (\sec{x} \tan{x} + \sec^2{x}) \,dx\\ \amp= \sec{x}(\sec{x} + \tan{x}) \,dx \end{align*}
Then,
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\begin{align*} \amp= \int \frac{1}{u} \,du\\ \amp= \ln{\abs{u}} + C\\ \amp= \ln{\abs{\sec{x} + \tan{x}}} + C \end{align*}
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Example 6.3.8. Deriving the Integral of Cosecant.

To integrate \(\csc{x}\text{,}\) we can use the same trick as secant, but this time multiplying the numerator and denominator by \(\csc{x} + \cot{x}\text{,}\)
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\begin{equation*} \int \csc{x} \,dx = \int \frac{\csc{x}(\csc{x} + \cot{x})}{\csc{x} + \cot{x}} \,dx \end{equation*}
Again, this allows us to use substitution, because there is \(\csc{x} + \cot{x}\) in the denominator, and its derivative is \(-\csc{x} \cot{x} - \csc^2{x}\text{,}\) which is what is in the numerator (except with a negative sign). Then,
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\begin{align*} u = \csc{x} + \cot{x} \qquad du \amp= -\csc{x} \cot{x} - \csc^2{x} \,dx\\ -du \amp= \csc{x} \cot{x} + \csc^2{x} \,dx \end{align*}
Then,
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\begin{align*} \amp= -\int \frac{1}{u} \,du\\ \amp= -\ln{\abs{u}} + C\\ \amp= -\ln{\abs{\csc{x} + \cot{x}}} + C \end{align*}
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Remark 6.3.10.

Note that these two antiderivatives are equivalent, because,
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\begin{align*} \ln{\abs{\csc{x}}} \amp= \ln{\abs{\frac{1}{\sin{x}}}} \amp\amp\text{by definition of cosecant}\\ \amp= \ln{\abs{\sin{x}}^{-1}}\\ \amp= -\ln{\abs{\sin{x}}} \amp\amp\text{by log laws} \end{align*}
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Example 6.3.11. Deriving the Integral of Cotangent.

To integrate \(\cot{x}\text{,}\) it is similar to \(\tan{x}\text{.}\) We can rewrite it as \(\frac{\cos{x}}{\sin{x}}\text{,}\) and use substitution,
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\begin{equation*} \int \cot{x} \,dx = \int \frac{\cos{x}}{\sin{x}} \,dx \end{equation*}
Then,
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\begin{align*} u = \sin{x} \qquad du \amp= \cos{x} \,dx \end{align*}
Then,
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\begin{align*} \amp= \int \frac{1}{u} \,du\\ \amp= \ln{\abs{u}} + C\\ \amp= \ln{\abs{\sin{x}}} + C \end{align*}
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Subsection 6.3.2 Integral of \(\sin^2{x}\) and \(\cos^2{x}\)

Two important trig integrals that come up often are \(\int \sin^2{x} \,dx\) and \(\int \cos^2{x} \,dx\text{.}\)
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These integrals show up often enough that they are helpful to memorize, or at least memorize the method of deriving them.
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Example 6.3.13. Integral of \(\cos^2{x}\).

Consider,
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\begin{equation*} \int \cos^2{x} \,dx \end{equation*}
Solution.
Unfortunately, there is no simple substitution that can be used to evaluate this integral (like, for example, \(u=\cos{x}\)). Also, you can’t just integrate \(\cos^2{x}\) as if it were 2 \(\cos{x}\) functions multiplied together,,
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\begin{align*} \int \cos^2{x} \,dx \amp\neq \int \cos{x} \,dx \cdot \int \cos{x} \,dx\\ \int \cos^2{x} \,dx \amp\neq \sin^2{x} + C \end{align*}
Instead, we have to use a particular trigonometric identity, called the power-reducing identities. There are 2 of them, one for cosine and one for sine,
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\begin{equation*} \cos^2{x} = \frac{1+\cos{(2x)}}{2} \qquad \text{and} \qquad \sin^2{x} = \frac{1-\cos{(2x)}}{2} \end{equation*}
As their name suggests, these identities express \(\cos^2{x}\) and \(\sin^2{x}\) (cosine and sine to the 2nd power) in terms of \(\cos{(2x)}\text{,}\) which is to the 1st power.
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These identities can be derived from the double-angle identities for cosine,
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\begin{align*} \cos{(2x)} \amp= 2\cos^2{x} - 1\\ \cos{(2x)} \amp= 1 - 2\sin^2{x} \end{align*}
If you solve the first equation to isolate for \(\cos^2{x}\text{,}\) and the second equation for \(\sin^2{x}\text{,}\) you get the power-reducing identities.
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Using these identities, we can evaluate the integral of \(\cos^2{x}\text{,}\)
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\begin{align*} \int \cos^2{x} \,dx \amp= \int \frac{1 + \cos{(2x)}}{2} \,dx\\ \amp= \frac{1}{2} \int (1 + \cos{(2x)}) \,dx\\ \amp= \frac{1}{2} \brac{x + \frac{1}{2} \sin{(2x)}} + C \end{align*}
Further, the double-angle identity \(\sin{(2x)} = 2 \sin{x} \cos{x}\) can be used to rewrite the antiderivative in terms of only single angles \(x\text{,}\)
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\begin{align*} \amp= \frac{1}{2} x + \frac{1}{2} \sin{x} \cos{x} + C\\ \amp= \frac{1}{2}(x + \sin{x} \cos{x}) + C \end{align*}
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The integral of \(\sin^2{x}\) can be evaluated in the same way, using the other power-reducing identity,
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Example 6.3.14. Integral of \(\sin^2{x}\).

Consider the integral,
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\begin{equation*} \int \sin^2{x} \,dx \end{equation*}
Solution.
Use the power-reducing identity for sine \(\sin^2{x} = \frac{1-\cos{(2x)}}{2}\text{,}\)
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\begin{align*} \int \sin^2{x} \,dx \amp= \int \frac{1 - \cos{(2x)}}{2} \,dx\\ \amp= \frac{1}{2} \int (1 - \cos{(2x)}) \,dx\\ \amp= \frac{1}{2} \brac{x - \frac{1}{2} \sin{(2x)}} + C \end{align*}
Again, we can use the double-angle identity \(\sin{(2x)} = 2 \sin{x} \cos{x}\) to rewrite the antiderivative in terms of only single angles \(x\text{,}\)
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\begin{align*} \amp= \frac{1}{2} x - \frac{1}{2} \sin{x} \cos{x} + C\\ \amp= \frac{1}{2}(x - \sin{x} \cos{x}) + C \end{align*}
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For reference, here are the power-reducing identities again,
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\begin{equation*} \boxed{\begin{array}{rl} \cos^2{x} \amp= \frac{1+\cos{(2x)}}{2} \\[5pt] \sin^2{x} \amp= \frac{1-\cos{(2x)}}{2} \end{array}} \end{equation*}
You may get these identities on your formula sheet for your exams, or you may need to memorize them.
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Subsection 6.3.3 Examples

Exercise Group 6.3.1. Basic Trigonometric Integrals Practice.

Evaluate each integral.
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(a)
\(\int (\sec{x}+1)^2 \,dx\)
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Hint.
Expand, and use the formula for \(\int \sec^2{x} \,dx\text{.}\)
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Answer.
\(\tan{x} + 2\ln{\abs{\sec{x} + \tan{x}}} + x + C\)
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(b)
\(\int \frac{1}{1-\sin{x}} \,dx\)
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Hint.
Multiply the numerator and denominator by \(1+\sin{x}\text{,}\) and split the fraction.
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Answer.
\(\tan{x} + \sec{x} + C\)
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(c)
\(\int \frac{1+\sin{x}}{\cos^2{x}} \,dx\)
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Hint.
Split the fraction into \(\sec^2{x} + \sec{x}\tan{x}\text{.}\)
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Answer.
\(\tan{x} + \sec{x} + C\)
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(d)
\(\int (\sec{x} - \tan{x})^2 \,dx\)
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Hint.
Expand, and use \(\tan^2{x} = \sec^2{x} - 1\text{.}\)
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Answer.
\(2\tan{x} - 2\sec{x} - x + C\)
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(e)
\(\int (\csc{x} - \sec{x})(\sin{x} + \cos{x}) \,dx\)
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Hint.
Distribute; use \(\csc{x} \sin{x} = 1\) and \(\sec{x} \cos{x} = 1\text{.}\)
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Answer.
\(\ln{\abs{\sin{x}}} + \ln{\abs{\cos{x}}} + C\)
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(f)
\(\int_{\pi/4}^{\pi/3} \frac{dx}{\cos^2{x}\tan{x}}\)
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Hint.
Rewrite as \(\int \frac{\sec^2{x}}{\tan{x}} \,dx\) and use \(u = \tan{x}\text{.}\)
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Answer.
\(\frac{1}{2}\ln{3}\)
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Exercise Group 6.3.2. More Trigonometric Integrals Practice.

Evaluate each integral.
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(a)
\(\int \frac{dx}{\cos{x} - 1}\)
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Hint.
Multiply numerator and denominator by \(\cos{x} + 1\text{,}\) and use \(\cos^2{x} - 1 = -\sin^2{x}\text{.}\)
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Answer.
\(\csc{x} + \cot{x} + C\)
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(b)
\(\int \frac{dx}{\sec{x} + \tan{x}}\)
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Hint.
Multiply numerator and denominator by \(\sec{x} - \tan{x}\text{,}\) and use \(\sec^2{x} - \tan^2{x} = 1\text{.}\)
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Answer.
\(\ln{\abs{1 + \sin{x}}} + C\)
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(c)
\(\int (\cos{x}\sin{2x} + \sin{x}\cos{2x}) \,dx\)
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Hint.
Use the sine addition formula \(\sin{(A+B)} = \sin{A}\cos{B} + \cos{A}\sin{B}\text{.}\)
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Answer.
\(-\frac{1}{3}\cos{3x} + C\)
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(d)
\(\int \frac{dx}{1 - \sec{x}}\)
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Hint.
Rewrite as \(\int \frac{\cos{x}}{\cos{x} - 1} \,dx\text{,}\) then multiply by the conjugate \(\frac{\cos{x}+1}{\cos{x}+1}\text{.}\)
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Answer.
\(\cot{x} + \csc{x} + x + C\)
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(e)
\(\int (\sec{x} + \cot{x})^2 \,dx\)
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Hint.
Expand, and use \(\cot^2{x} = \csc^2{x} - 1\) and \(\sec{x}\cot{x} = \csc{x}\text{.}\)
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Answer.
\(\tan{x} - 2\ln{\abs{\csc{x} + \cot{x}}} - \cot{x} - x + C\)
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