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Trigonometric Integrals

Solve integration problems involving products and powers of

Integrating Products and Powers of sin

A key idea behind the strategy used to integrate combinations of products and powers of \(\text{sin}\ x\) and \(\text{cos}\ x\) involves rewriting these expressions as sums and differences of integrals of the form \(\int {\text{sin}}^{j}x\ \text{cos}\ x\ dx\) or \(\int {\text{cos}}^{j}x\ \text{sin}\ x\ dx.\) After rewriting these integrals, we evaluate them using u-substitution. Before describing the general process in detail, let’s take a look at the following examples.

Example

Try it.

Evaluate \(\int {\text{cos}}^{3}x\ \text{sin}\ x\ dx.\)

Solution

Use \(u\)-substitution and let \(u=\text{cos}\ x.\) In this case, \(du=\text{-}\text{sin}\ x\ dx.\) Thus,

\[\begin{array}{ll}\int {\text{cos}}^{3}\ x\ \text{sin}\ x\ dx & =\text{-}\int {u}^{3}\ du \\ & =-\frac{1}{4}{u}^{4}+C \\ & =-\frac{1}{4}{\text{cos}}^{4}\ x+C.\end{array}\]

Example

Try it.

Evaluate \(\int {\text{cos}}^{2}x\ {\text{sin}}^{3}x\ dx.\)

Solution

To convert this integral to integrals of the form \(\int {\text{cos}}^{j}x\ \text{sin}\ x\ dx,\) rewrite \({\text{sin}}^{3}x={\text{sin}}^{2}x\ \text{sin}\ x\) and make the substitution \({\text{sin}}^{2}x=1-{\text{cos}}^{2}x.\) Thus,

\[\begin{array}{lll}\int {\text{cos}}^{2}x\ {\text{sin}}^{3}x\ dx & =\int {\text{cos}}^{2}x(1-{\text{cos}}^{2}x)\text{sin}\ x\ dx & \text{Let}\ u=\text{cos}\ x;\ \text{then}\ du=\text{-}\text{sin}\ x\ dx. \\ & =\text{-}\int {u}^{2}(1-{u}^{2})du & \\ & =\int ({u}^{4}-{u}^{2})du & \\ & =\frac{1}{5}{u}^{5}-\frac{1}{3}{u}^{3}+C & \\ & =\frac{1}{5}{\text{cos}}^{5}x-\frac{1}{3}{\text{cos}}^{3}x+C. & \end{array}\]

In the next example, we see the strategy that must be applied when there are only even powers of \(\text{sin}\ x\) and \(\text{cos}\ x.\) For integrals of this type, the identities

\[{\text{sin}}^{2}x=\frac{1}{2}-\frac{1}{2}\text{cos}(2x)=\frac{1-\text{cos}(2x)}{2}\]

and

\[{\text{cos}}^{2}x=\frac{1}{2}+\frac{1}{2}\text{cos}(2x)=\frac{1+\text{cos}(2x)}{2}\]

are invaluable. These identities are sometimes known as power-reducing identities and they may be derived from the double-angle identity \(\text{cos}(2x)={\text{cos}}^{2}x-{\text{sin}}^{2}x\) and the Pythagorean identity \({\text{cos}}^{2}x+{\text{sin}}^{2}x=1.\)

The general process for integrating products of powers of \(\text{sin}\ x\) and \(\text{cos}\ x\) is summarized in the following set of guidelines.

Condensed — the full section is in OpenStax Calculus Volume 2.

Integrating Products and Powers of tan

Before discussing the integration of products and powers of \(\text{tan}\ x\) and \(\text{sec}\ x,\) it is useful to recall the integrals involving \(\text{tan}\ x\) and \(\text{sec}\ x\) we have already learned:

\(\int {\text{sec}}^{2}x\ dx=\text{tan}\ x+C\)
\(\int \text{sec}\ x\ \text{tan}\ x\ dx=\text{sec}\ x+C\)
\(\int \text{tan}\ x\ dx=\text{ln}|\text{sec}\ x|+C\)
\(\int \text{sec}\ x\ dx=\text{ln}|\text{sec}\ x+\text{tan}\ x|+C.\)

For most integrals of products and powers of \(\text{tan}\ x\) and \(\text{sec}\ x,\) we rewrite the expression we wish to integrate as the sum or difference of integrals of the form \(\int {\text{tan}}^{j}x\ {\text{sec}}^{2}x\ dx\) or \(\int {\text{sec}}^{j}x\ \text{tan}\ x\ dx.\) As we see in the following example, we can evaluate these new integrals by using u-substitution.

Example

Try it.

Evaluate \(\int {\text{sec}}^{5}x\ \text{tan}\ x\ dx.\)

Solution

Start by rewriting \({\text{sec}}^{5}x\ \text{tan}\ x\) as \({\text{sec}}^{4}x\ \text{sec}\ x\ \text{tan}\ x.\)

\[\begin{array}{llll}\int {\text{sec}}^{5}x\ \text{tan}\ x\ dx & =\int {\text{sec}}^{4}x\ \text{sec}\ x\ \text{tan}\ x\ dx & & \text{Let}\ u=\text{sec}\ x;\ \text{then},\ du=\text{sec}\ x\ \text{tan}\ x\ dx. \\ & =\int {u}^{4}du & & \text{Evaluate the integral}. \\ & =\frac{1}{5}{u}^{5}+C & & \text{Substitute}\ \text{sec}\ x=u. \\ & =\frac{1}{5}{\text{sec}}^{5}x+C & & \end{array}\]

We now take a look at the various strategies for integrating products and powers of \(\text{sec}\ x\) and \(\text{tan}\ x.\)

Example

Try it.

Evaluate \(\int {\text{tan}}^{6}x\ {\text{sec}}^{4}x\ dx.\)

Solution

Since the power on \(\text{sec}\ x\) is even, rewrite \({\text{sec}}^{4}x={\text{sec}}^{2}x\ {\text{sec}}^{2}x\) and use \({\text{sec}}^{2}x={\text{tan}}^{2}x+1\) to rewrite the first \({\text{sec}}^{2}x\) in terms of \(\text{tan}\ x.\) Thus,

\[\begin{array}{llll}\int {\text{tan}}^{6}x\ {\text{sec}}^{4}x\ dx & =\int {\text{tan}}^{6}x({\text{tan}}^{2}x+1){\text{sec}}^{2}x\ dx & & \text{Let}\ u=\text{tan}\ x\ \text{and}\ du={\text{sec}}^{2}x\text{dx}. \\ & =\int {u}^{6}({u}^{2}+1)du & & \text{Expand}. \\ & =\int ({u}^{8}+{u}^{6})du & & \text{Evaluate the integral}. \\ & =\frac{1}{9}{u}^{9}+\frac{1}{7}{u}^{7}+C & & \text{Substitute}\ \text{tan}\ x=u. \\ & =\frac{1}{9}{\text{tan}}^{9}x+\frac{1}{7}{\text{tan}}^{7}x+C. & & \end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Reduction Formulas

Evaluating \(\int {\text{sec}}^{n}x\ dx\) for values of \(n\) where \(n\) is odd requires integration by parts. In addition, we must also know the value of \(\int {\text{sec}}^{n-2}x\ dx\) to evaluate \(\int {\text{sec}}^{n}x\ dx.\) The evaluation of \(\int {\text{tan}}^{n}x\ dx\) also requires being able to integrate \(\int {\text{tan}}^{n-2}x\ dx.\) To make the process easier, we can derive and apply the following power reduction formulas. These rules allow us to replace the integral of a power of \(\text{sec}\ x\) or \(\text{tan}\ x\) with the integral of a lower power of \(\text{sec}\ x\) or \(\text{tan}\ x.\)

Example

Try it.

Apply a reduction formula to evaluate \(\int {\text{sec}}^{3}x\ dx.\)

Solution

By applying the first reduction formula, we obtain

\[\begin{array}{ll}\int {\text{sec}}^{3}x\ dx & =\frac{1}{2}\text{sec}\ x\ \text{tan}\ x+\frac{1}{2}\int \text{sec}\ x\ dx \\ & =\frac{1}{2}\text{sec}\ x\ \text{tan}\ x+\frac{1}{2}\text{ln}|\text{sec}\ x+\text{tan}\ x|+C.\end{array}\]

Example

Try it.

Evaluate \(\int {\text{tan}}^{4}x\ dx.\)

Solution

Applying the reduction formula for \(\int {\text{tan}}^{4}x\ dx\) we have

\[\begin{array}{llll}\int {\text{tan}}^{4}x\ dx & =\frac{1}{3}{\text{tan}}^{3}x-\int {\text{tan}}^{2}x\ dx & & \\ & =\frac{1}{3}{\text{tan}}^{3}x-(\text{tan}\ x-\int {\text{tan}}^{0}x\ dx) & & \text{Apply the reduction formula to}\int {\text{tan}}^{2}x\ dx. \\ & =\frac{1}{3}{\text{tan}}^{3}x-\text{tan}\ x+\int 1\ dx & & \text{Simplify}. \\ & =\frac{1}{3}{\text{tan}}^{3}x-\text{tan}\ x+x+C. & & \text{Evaluate}\int 1dx.\end{array}\]

Key Concepts

Integrals of trigonometric functions can be evaluated by the use of various strategies. These strategies include
1. Applying trigonometric identities to rewrite the integral so that it may be evaluated by u-substitution
2. Using integration by parts
3. Applying trigonometric identities to rewrite products of sines and cosines with different arguments as the sum of individual sine and cosine functions
4. Applying reduction formulas

Key Equations

To integrate products involving \(\text{sin}(ax),\) \(\text{sin}(bx),\) \(\text{cos}(ax),\) and \(\text{cos}(bx),\) use the substitutions.

Sine Products	\(\text{sin}(ax)\text{sin}(bx)=\frac{1}{2}\text{cos}((a-b)x)-\frac{1}{2}\text{cos}((a+b)x)\)
Sine and Cosine Products	\(\text{sin}(ax)\text{cos}(bx)=\frac{1}{2}\text{sin}((a-b)x)+\frac{1}{2}\text{sin}((a+b)x)\)
Cosine Products	\(\text{cos}(ax)\text{cos}(bx)=\frac{1}{2}\text{cos}((a-b)x)+\frac{1}{2}\text{cos}((a+b)x)\)
Power Reduction Formula	\(\int {\text{sec}}^{n}xdx=\frac{{\text{sec}}^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int {\text{sec}}^{n-2}x\ dx;n\ne 1\)
Power Reduction Formula	\(\int {\text{tan}}^{n}x\ dx=\frac{1}{n-1}{\text{tan}}^{n-1}x-\int {\text{tan}}^{n-2}x\ dx\)

Trigonometric Integrals

Fill in the blank to make a true statement.

Use an identity to reduce the power of the trigonometric function to a trigonometric function raised to the first power.

Evaluate each of the following integrals by u-substitution.

Compute the following integrals using the guidelines for integrating powers of trigonometric functions. Use a CAS to check the solutions. (Note: Some of the problems may be done using techniques of integration learned previously.)

For the following exercises, find a general formula for the integrals.

Use the double-angle formulas to evaluate the following integrals.

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible.

Condensed — the full section is in OpenStax Calculus Volume 2.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Evaluate \(\int {\text{cos}}^{3}x\ \text{sin}\ x\ dx.\)

Atskleisti atsakymą

Use \(u\)-substitution and let \(u=\text{cos}\ x.\) In this case, \(du=\text{-}\text{sin}\ x\ dx.\) Thus,
\[\begin{array}{ll}\int {\text{cos}}^{3}\ x\ \text{sin}\ x\ dx & =\text{-}\int {u}^{3}\ du \\ & =-\frac{1}{4}{u}^{4}+C \\ & =-\frac{1}{4}{\text{cos}}^{4}\ x+C.\end{array}\]
Evaluate \(\int {\text{sin}}^{4}x\ \text{cos}\ x\ dx.\)

Atskleisti atsakymą

\(\frac{1}{5}{\text{sin}}^{5}x+C\)
Evaluate \(\int {\text{cos}}^{2}x\ {\text{sin}}^{3}x\ dx.\)

Atskleisti atsakymą

To convert this integral to integrals of the form \(\int {\text{cos}}^{j}x\ \text{sin}\ x\ dx,\) rewrite \({\text{sin}}^{3}x={\text{sin}}^{2}x\ \text{sin}\ x\) and make the substitution \({\text{sin}}^{2}x=1-{\text{cos}}^{2}x.\) Thus,
\[\begin{array}{lll}\int {\text{cos}}^{2}x\ {\text{sin}}^{3}x\ dx & =\int {\text{cos}}^{2}x(1-{\text{cos}}^{2}x)\text{sin}\ x\ dx & \text{Let}\ u=\text{cos}\ x;\ \text{then}\ du=\text{-}\text{sin}\ x\ dx. \\ & =\text{-}\int {u}^{2}(1-{u}^{2})du & \\ & =\int ({u}^{4}-{u}^{2})du & \\ & =\frac{1}{5}{u}^{5}-\frac{1}{3}{u}^{3}+C & \\ & =\frac{1}{5}{\text{cos}}^{5}x-\frac{1}{3}{\text{cos}}^{3}x+C. & \end{array}\]
Evaluate \(\int {\text{cos}}^{3}x\ {\text{sin}}^{2}x\ dx.\)

Atskleisti atsakymą

\(\frac{1}{3}{\text{sin}}^{3}x-\frac{1}{5}{\text{sin}}^{5}x+C\)
Evaluate \(\int {\text{sin}}^{2}x\ dx.\)

Atskleisti atsakymą

To evaluate this integral, let’s use the trigonometric identity \({\text{sin}}^{2}x=\frac{1}{2}-\frac{1}{2}\text{cos}(2x).\) Thus,
\[\begin{array}{ll}\int {\text{sin}}^{2}x\ dx & =\int (\frac{1}{2}-\frac{1}{2}\text{cos}(2x))dx \\ & =\frac{1}{2}x-\frac{1}{4}\text{sin}(2x)+C.\end{array}\]
Evaluate \(\int {\text{cos}}^{2}x\ dx.\)

Atskleisti atsakymą

\(\frac{1}{2}x+\frac{1}{4}\text{sin}(2x)+C\)
Evaluate \(\int {\text{cos}}^{8}x\ {\text{sin}}^{5}x\ dx.\)

Atskleisti atsakymą

Since the power on \(\text{sin}\ x\) is odd, use strategy 1. Thus,
\[\begin{array}{llll}\int {\text{cos}}^{8}x\ {\text{sin}}^{5}x\ dx & =\int {\text{cos}}^{8}x\ {\text{sin}}^{4}x\ \text{sin}\ x\ dx & & \text{Break off}\ \text{sin}\ x. \\ & =\int {\text{cos}}^{8}x{({\text{sin}}^{2}x)}^{2}\text{sin}\ x\ dx & & \text{Rewrite}\ {\text{sin}}^{4}x={({\text{sin}}^{2}x)}^{2}. \\ & =\int {\text{cos}}^{8}x{(1-{\text{cos}}^{2}x)}^{2}\text{sin}\ x\ dx & & \text{Substitute}\ {\text{sin}}^{2}x=1-{\text{cos}}^{2}x. \\ & =\int {u}^{8}{(1-{u}^{2})}^{2}(\text{-}du) & & \text{Let}\ u=\text{cos}\ x\ \text{and}\ du=\text{-}\text{sin}\ x\ dx. \\ & =\int (\text{-}{u}^{8}+2{u}^{10}-{u}^{12})du & & \text{Expand}. \\ & =-\frac{1}{9}{u}^{9}+\frac{2}{11}{u}^{11}-\frac{1}{13}{u}^{13}+C & & \text{Evaluate the integral}. \\ & =-\frac{1}{9}{\text{cos}}^{9}x+\frac{2}{11}{\text{cos}}^{11}x-\frac{1}{13}{\text{cos}}^{13}x+C. & & \text{Substitute}\ u=\text{cos}\ x.\end{array}\]
Evaluate \(\int {\text{sin}}^{4}x\ dx.\)

Atskleisti atsakymą

Since the power on \(\text{sin}\ x\) is even \((k=4)\) and the power on \(\text{cos}\ x\) is even \((j=0),\) we must use strategy 3. Thus,
\[\begin{array}{llll}\int {\text{sin}}^{4}x\ dx & =\int {({\text{sin}}^{2}x)}^{2}dx & & \text{Rewrite}\ {\text{sin}}^{4}x={({\text{sin}}^{2}x)}^{2}. \\ & =\int {(\frac{1}{2}-\frac{1}{2}\text{cos}(2x))}^{2}dx & & \text{Substitute}\ {\text{sin}}^{2}x=\frac{1}{2}-\frac{1}{2}\text{cos}(2x). \\ & =\int (\frac{1}{4}-\frac{1}{2}\text{cos}(2x)+\frac{1}{4}{\text{cos}}^{2}(2x))dx & & \text{Expand}\ {(\frac{1}{2}-\frac{1}{2}\text{cos}(2x))}^{2}. \\ & =\int (\frac{1}{4}-\frac{1}{2}\text{cos}(2x)+\frac{1}{4}(\frac{1}{2}+\frac{1}{2}\text{cos}(4x))dx. & & \end{array}\]
Since \({\text{cos}}^{2}(2x)\) has an even power, substitute \({\text{cos}}^{2}(2x)=\frac{1}{2}+\frac{1}{2}\text{cos}(4x)\text{:}\)
\[\begin{array}{ll}=\int (\frac{3}{8}-\frac{1}{2}\text{cos}(2x)+\frac{1}{8}\text{cos}(4x))dx & \text{Simplify}. \\ =\frac{3}{8}x-\frac{1}{4}\text{sin}(2x)+\frac{1}{32}\ \text{sin}(4x)+C & \text{Evaluate the integral}.\end{array}\]
Evaluate \(\int {\text{cos}}^{3}x\ dx.\)

Atskleisti atsakymą

\(\text{sin}\ x-\frac{1}{3}{\text{sin}}^{3}x+C\)
Evaluate \(\int {\text{cos}}^{2}(3x)dx.\)

Atskleisti atsakymą

\(\frac{1}{2}x+\frac{1}{12}\ \text{sin}(6x)+C\)
Evaluate \(\int \text{sin}(5x)\text{cos}(3x)dx.\)

Atskleisti atsakymą

Apply the identity \(\text{sin}(5x)\text{cos}(3x)=\frac{1}{2}\text{sin}(2x)+\frac{1}{2}\text{sin}(8x).\) Thus,
\[\begin{array}{ll}\int \text{sin}(5x)\text{cos}(3x)dx & =\int \frac{1}{2}\text{sin}(2x)\text{dx}+\int \frac{1}{2}\text{sin}(8x)\text{dx} \\ & =-\frac{1}{4}\text{cos}(2x)-\frac{1}{16}\ \text{cos}(8x)+C.\end{array}\]
Evaluate \(\int \text{cos}(6x)\text{cos}(5x)dx.\)

Atskleisti atsakymą

\(\frac{1}{2}\text{sin}\ x+\frac{1}{22}\ \text{sin}(11x)+C\)
Evaluate \(\int {\text{sec}}^{5}x\ \text{tan}\ x\ dx.\)

Atskleisti atsakymą

Start by rewriting \({\text{sec}}^{5}x\ \text{tan}\ x\) as \({\text{sec}}^{4}x\ \text{sec}\ x\ \text{tan}\ x.\)
\[\begin{array}{llll}\int {\text{sec}}^{5}x\ \text{tan}\ x\ dx & =\int {\text{sec}}^{4}x\ \text{sec}\ x\ \text{tan}\ x\ dx & & \text{Let}\ u=\text{sec}\ x;\ \text{then},\ du=\text{sec}\ x\ \text{tan}\ x\ dx. \\ & =\int {u}^{4}du & & \text{Evaluate the integral}. \\ & =\frac{1}{5}{u}^{5}+C & & \text{Substitute}\ \text{sec}\ x=u. \\ & =\frac{1}{5}{\text{sec}}^{5}x+C & & \end{array}\]
Evaluate \(\int {\text{tan}}^{5}x\ {\text{sec}}^{2}x\ dx.\)

Atskleisti atsakymą

\(\frac{1}{6}{\text{tan}}^{6}x+C\)
Evaluate \(\int {\text{tan}}^{6}x\ {\text{sec}}^{4}x\ dx.\)

Atskleisti atsakymą

Since the power on \(\text{sec}\ x\) is even, rewrite \({\text{sec}}^{4}x={\text{sec}}^{2}x\ {\text{sec}}^{2}x\) and use \({\text{sec}}^{2}x={\text{tan}}^{2}x+1\) to rewrite the first \({\text{sec}}^{2}x\) in terms of \(\text{tan}\ x.\) Thus,
\[\begin{array}{llll}\int {\text{tan}}^{6}x\ {\text{sec}}^{4}x\ dx & =\int {\text{tan}}^{6}x({\text{tan}}^{2}x+1){\text{sec}}^{2}x\ dx & & \text{Let}\ u=\text{tan}\ x\ \text{and}\ du={\text{sec}}^{2}x\text{dx}. \\ & =\int {u}^{6}({u}^{2}+1)du & & \text{Expand}. \\ & =\int ({u}^{8}+{u}^{6})du & & \text{Evaluate the integral}. \\ & =\frac{1}{9}{u}^{9}+\frac{1}{7}{u}^{7}+C & & \text{Substitute}\ \text{tan}\ x=u. \\ & =\frac{1}{9}{\text{tan}}^{9}x+\frac{1}{7}{\text{tan}}^{7}x+C. & & \end{array}\]
Evaluate \(\int {\text{tan}}^{5}x\ {\text{sec}}^{3}x\ dx.\)

Atskleisti atsakymą

Since the power on \(\text{tan}\ x\) is odd, begin by rewriting \({\text{tan}}^{5}x\ {\text{sec}}^{3}x={\text{tan}}^{4}x\ {\text{sec}}^{2}x\ \text{sec}\ x\ \text{tan}\ x.\) Thus,
\[\begin{array}{llllll}{\text{tan}}^{5}x\ {\text{sec}}^{3}x & = & {\text{tan}}^{4}x\ {\text{sec}}^{2}x\ \text{sec}\ x\ \text{tan}\ x. & & & \text{Write}\ {\text{tan}}^{4}x={({\text{tan}}^{2}x)}^{2}. \\ \int {\text{tan}}^{5}x\ {\text{sec}}^{3}x\ dx & = & \int {({\text{tan}}^{2}x)}^{2}{\text{sec}}^{2}x\ \text{sec}\ x\ \text{tan}\ x\ dx & & & \text{Use}\ {\text{tan}}^{2}x={\text{sec}}^{2}x-1. \\ & = & \int {({\text{sec}}^{2}x-1)}^{2}{\text{sec}}^{2}x\ \text{sec}\ x\ \text{tan}\ x\ dx & & & \text{Let}\ u=\text{sec}\ x\ \text{and}\ du=\text{sec}\ x\ \text{tan}\ x\ dx. \\ & = & \int {({u}^{2}-1)}^{2}{u}^{2}du & & & \text{Expand}. \\ & = & \int ({u}^{6}-2{u}^{4}+{u}^{2})du & & & \text{Integrate}. \\ & = & \frac{1}{7}{u}^{7}-\frac{2}{5}{u}^{5}+\frac{1}{3}{u}^{3}+C & & & \text{Substitute}\ \text{sec}\ x=u. \\ & = & \frac{1}{7}{\text{sec}}^{7}x-\frac{2}{5}{\text{sec}}^{5}x+\frac{1}{3}{\text{sec}}^{3}x+C. & & & \end{array}\]
Evaluate \(\int {\text{tan}}^{3}x\ dx.\)

Atskleisti atsakymą

Begin by rewriting \({\text{tan}}^{3}x=\text{tan}\ x\ {\text{tan}}^{2}x=\text{tan}\ x({\text{sec}}^{2}x-1)=\text{tan}\ x\ {\text{sec}}^{2}x-\text{tan}\ x.\) Thus,
\[\begin{array}{ll}\int {\text{tan}}^{3}x\ dx & =\int (\text{tan}\ x\ {\text{sec}}^{2}x-\text{tan}\ x)dx \\ & =\int \text{tan}\ x\ {\text{sec}}^{2}x\ dx-\int \text{tan}\ x\ dx \\ & =\frac{1}{2}{\text{tan}}^{2}x-\text{ln}|\text{sec}\ x|+C.\end{array}\]
For the first integral, use the substitution \(u=\text{tan}\ x.\) For the second integral, use the formula.
Integrate \(\int {\text{sec}}^{3}x\ dx.\)

Atskleisti atsakymą

This integral requires integration by parts. To begin, let \(u=\text{sec}\ x\) and \(dv={\text{sec}}^{2}x\text{dx}.\) These choices make \(du=\text{sec}\ x\ \text{tan}\ x\) and \(v=\text{tan}\ x.\) Thus,
\[\begin{array}{llll}\int {\text{sec}}^{3}x\ dx & =\text{sec}\ x\ \text{tan}\ x-\int \text{tan}\ x\ \text{sec}\ x\ \text{tan}\ x\ dx & & \\ & =\text{sec}\ x\ \text{tan}\ x-\int {\text{tan}}^{2}x\ \text{sec}\ x\ dx & & \text{Simplify}. \\ & =\text{sec}\ x\ \text{tan}\ x-\int ({\text{sec}}^{2}x-1)\text{sec}\ x\ dx & & \text{Substitute}\ {\text{tan}}^{2}x={\text{sec}}^{2}x-1. \\ & =\text{sec}\ x\ \text{tan}\ x+\int \text{sec}\ x\ dx-\int {\text{sec}}^{3}x\ dx & & \text{Rewrite}. \\ & =\text{sec}\ x\ \text{tan}\ x+\text{ln}|\text{sec}\ x+\text{tan}\ x|-\int {\text{sec}}^{3}x\ dx. & & \text{Evaluate}\int \text{sec}\ x\ dx.\end{array}\]
We now have
\[\int {\text{sec}}^{3}x\ dx=\text{sec}\ x\ \text{tan}\ x+\text{ln}|\text{sec}\ x+\text{tan}\ x|-\int {\text{sec}}^{3}x\ dx.\]
Since the integral \(\int {\text{sec}}^{3}x\ dx\) has reappeared on the right-hand side, we can solve for \(\int {\text{sec}}^{3}x\ dx\) by adding it to both sides. In doing so, we obtain
\[2\int {\text{sec}}^{3}x\ dx=\text{sec}\ x\ \text{tan}\ x+\text{ln}|\text{sec}\ x+\text{tan}\ x|.\]
Dividing by 2, we arrive at
\[\int {\text{sec}}^{3}x\ dx=\frac{1}{2}\text{sec}\ x\ \text{tan}\ x+\frac{1}{2}\text{ln}|\text{sec}\ x+\text{tan}\ x|+C.\]
Evaluate \(\int {\text{tan}}^{3}x\ {\text{sec}}^{7}x\ dx.\)

Atskleisti atsakymą

\(\frac{1}{9}{\text{sec}}^{9}x-\frac{1}{7}{\text{sec}}^{7}x+C\)
Apply a reduction formula to evaluate \(\int {\text{sec}}^{3}x\ dx.\)

Atskleisti atsakymą

By applying the first reduction formula, we obtain
\[\begin{array}{ll}\int {\text{sec}}^{3}x\ dx & =\frac{1}{2}\text{sec}\ x\ \text{tan}\ x+\frac{1}{2}\int \text{sec}\ x\ dx \\ & =\frac{1}{2}\text{sec}\ x\ \text{tan}\ x+\frac{1}{2}\text{ln}|\text{sec}\ x+\text{tan}\ x|+C.\end{array}\]
Evaluate \(\int {\text{tan}}^{4}x\ dx.\)

Atskleisti atsakymą

Applying the reduction formula for \(\int {\text{tan}}^{4}x\ dx\) we have
\[\begin{array}{llll}\int {\text{tan}}^{4}x\ dx & =\frac{1}{3}{\text{tan}}^{3}x-\int {\text{tan}}^{2}x\ dx & & \\ & =\frac{1}{3}{\text{tan}}^{3}x-(\text{tan}\ x-\int {\text{tan}}^{0}x\ dx) & & \text{Apply the reduction formula to}\int {\text{tan}}^{2}x\ dx. \\ & =\frac{1}{3}{\text{tan}}^{3}x-\text{tan}\ x+\int 1\ dx & & \text{Simplify}. \\ & =\frac{1}{3}{\text{tan}}^{3}x-\text{tan}\ x+x+C. & & \text{Evaluate}\int 1dx.\end{array}\]
Apply the reduction formula to \(\int {\text{sec}}^{5}x\ dx.\)

Atskleisti atsakymą

\(\int {\text{sec}}^{5}x\ dx=\frac{1}{4}{\text{sec}}^{3}x\ \text{tan}\ x+\frac{3}{4}\int {\text{sec}}^{3}x\)
\({\text{sin}}^{2}x+_______=1\)

Atskleisti atsakymą

\({\text{cos}}^{2}x\)
\({\text{sec}}^{2}x-1=_______\)
\({\text{sin}}^{2}x=_______\)

Atskleisti atsakymą

\(\frac{1-\text{cos}(2x)}{2}\)
\({\text{cos}}^{2}x=_______\)
\(\int {\text{sin}}^{3}x\ \text{cos}\ x\ dx\)

Atskleisti atsakymą

\(\frac{{\text{sin}}^{4}x}{4}+C\)
\(\int \sqrt{\text{cos}\ x}\ \text{sin}\ x\ dx\)
\(\int {\text{tan}}^{5}(2x){\text{sec}}^{2}(2x)dx\)

Atskleisti atsakymą

\(\frac{1}{12}{\text{tan}}^{6}(2x)+C\)
\(\int {\text{sin}}^{7}(2x)\text{cos}(2x)dx\)
\(\int \text{tan}(\frac{x}{2}){\text{sec}}^{2}(\frac{x}{2})dx\)

Atskleisti atsakymą

\((\frac{x}{2})\left(\frac{x}{2}\right)+{C}_{1}{\text{or sec}}^{2}\left(\frac{x}{2}\right)+{C}_{2}\)
\(\int {\text{tan}}^{2}x\ {\text{sec}}^{2}x\ dx\)
\(\int {\text{sin}}^{3}x\ dx\)

Atskleisti atsakymą

\(-\text{cos}x+\frac{1}{3}{\text{cos}}^{3}x+C\)
\(\int {\text{cos}}^{3}x\ dx\)
\(\int \text{sin}\ x\ \text{cos}\ x\ dx\)

Atskleisti atsakymą

\(-\frac{1}{2}{\text{cos}}^{2}x+C\) or \(\frac{1}{2}{\text{sin}}^{2}x+C\)
\(\int {\text{cos}}^{5}x\ dx\)
\(\int {\text{sin}}^{5}x\ {\text{cos}}^{2}x\ dx\)

Atskleisti atsakymą

\(-\frac{1}{3}{\text{cos}}^{3}x+\frac{2}{5}{\text{cos}}^{5}x-\frac{1}{7}{\text{cos}}^{7}x+C\)
\(\int {\text{sin}}^{3}x\ {\text{cos}}^{3}x\ dx\)
\(\int \sqrt{\text{sin}\ x}\ \text{cos}\ x\ dx\)

Atskleisti atsakymą

\(\frac{2}{3}{(\text{sin}\ x)}^{\frac{3}{2}}+C\)
\(\int \sqrt{\text{sin}\ x}\ {\text{cos}}^{3}x\ dx\)

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\neq

not equal
The two sides are different.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Trigonometric Integrals

Solve integration problems involving products and powers of
Solve integration problems involving products and powers of
Use reduction formulas to solve trigonometric integrals.
If
If
If both
If
If

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Pabandyk savo pačių

Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Daugiau informacijos Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests