A lecture on trigonometric integrals and trigonometric substitution

1 Trigonometric Integrals

First, we must recall a few trigonometric identities:

sin2x+cos2x = 1 (1)
sec2x = 1+tan2x (2)
sin2x = 1-cos(2x)2 (3)
cos2x = 1+cos(2x)2 (4)
sin(2x) = 2sinxcosx (5)
cos(2x) = cos2x-sin2x. (6)

The most usual integralsDlmfPlanetmath which involve trigonometric functionsDlmfMathworldPlanetmath can be solved using the identities above.

Example 1.1.

sinxdx=-cosx+C and cosxdx=sinx+C are immediate integrals.

Example 1.2.

For sin2xdx,cos2xdx we use formulas (3) and (4) respectively, e.g.

Example 1.3.

For integrals of the form cosmxsinxdx or sinmxcosxdx we use substitution with u=cosx or u=sinx respectively, e.g.


In the following examples, we use equations (1) in the forms sin2x=1-cos2x or cos2x=1-sin2x to transform the integral into one of the type described in Example 1.3.

Example 1.4.
sin3xdx = sin2xsinxdx=(1-cos2x)sinxdx
= sinxdx-cos2xsinxdx
= -cosx+cos3x3+C.

Similarly one can solve cos3xdx.

Example 1.5.
cos3xsin2xdx = cos2xcosxsin2xdx=(1-sin2x)cosxsin2xdx
= cosxsin2xdx-cosxsin4xdx
= sin3x3-sin5x5+C.
Example 1.6.

In order to solve cos5xsin3xdx we express it first as cos5xsin2xsinx=cos5x(1-cos2x)sinxdx and then proceed as in the previous example.

One can use similar tricks to solve integrals which involve products of powers of secx and tanx, by using Equation (2). Also, recall that the derivative of tanx is sec2x while the derivative of secx is secxtanx.

Example 1.7.
tan5xsec4xdx = tan5xsec2xsec2xdx=tan5x(1+tan2x)sec2xdx
= tan5xsec2xdx+tan7xsec2xdx
= tan6x6+tan8x8+C.
Example 1.8.
tan3xsec4xdx = tanxtan2xsec4xdx=tanx(sec2x-1)sec4xdx
= tanxsecxsec5xdx-tanxsecxsec3xdx
= sec6x6-sec4x4+C.

2 Trigonometric Substitutions

One can easily deduce that 011-x2𝑑x has value π4. Why? Simply because the graph of the functionMathworldPlanetmath y=1-x2 is half a circumference of radius r=1 (because if you square both sides of y=1-x2 you obtain x2+y2=1 which is the equation of a circle or radius r=1). Therefore, the area under the graph is a quarter of the area of a circle.

How does one compute 011-x2𝑑x without using the geometry of the problem? This is the prototype of integral where a trigonometric substitution will work very nicely. Notice that neither substitution nor integration by parts will work appropriately.

Example 2.1.

Suppose we want to solve 011-x2𝑑x with analytic methods. We will use a substitution x=sinθ (so θ will be our new variable of integration), because, as we know from Equation (1), 1-x2=1-sin2θ=cosθ, thus getting rid of the pesky square root. Notice that dx=cosθdθ. We also need to find the new limits of integration with respect to the new variable of integration, namely θ. When x=0=sinθ we must have θ=0. Similarly, when x=1=sinθ one has θ=π/2. We are now ready to integrate:

011-x2𝑑x = 0π/2(cosθ)cosθdθ=0π/2cos2θdθ
= 0π/21+cos(2θ)2𝑑θ=12(θ+sin(2θ)2)0π/2=π/4.

Notice that we made use of Equation (4) in the second line.

Example 2.2.

Similarly, one can solve 0rr2-x2𝑑x by using a substitution x=rsinθ. Indeed, r2-x2=r2-r2sin2θ=rcosθ and dx=rcosθdθ. The limits of integration with respect to θ are again θ=0 to θ=π/2 (check this!). Thus:

0rr2-x2𝑑x = 0π/2r2(cosθ)cosθdθ=r20π/2cos2θdθ
= r20π/21+cos(2θ)2𝑑θ=r22(θ+sin(2θ)2)0π/2=r2π/4.

Thus, we have proved that a quarter of a circle of radius r has area r2π/4 which implies that the area of such a circle is πr2, as usual.

The trigonometric substitutions usually work when expressions like r2-x2, r2+x2, x2-r2 appear in the integral at hand, for some real number r. Here is a table of the suggested change of variables in each particular case:

If you see… try this… because…
1-x2 x=sinθ 1-sin2θ=cosθ
r2-x2 x=rsinθ r2-sin2θ=rcosθ
1+x2 x=tanθ 1+tan2θ=secθ
r2+x2 x=rtanθ r2+tan2θ=rsecθ
x2-1 x=secθ sec2θ-1=tanθ
x2-r2 x=rsinθ sec2θ-1=rtanθ
Remark 2.3.

The above are “suggested” substitutions, they may not be the most ideal choice! For example, for the integral 2x1-x2𝑑x, the change u=1-x2 will work much better than x=sinθ.

Example 2.4.

We would like to find the value of


Since neither a u-substitution nor integration by parts seem appropriate, we try x=secθ, dx=secθtanθdθ. When x=2=secθ one has θ=π/4 while x=2 implies θ=π/3. Hence:

221x3x2-1𝑑x = π/4π/3secθtanθsec3θtanθ𝑑θ=π/4π/31sec2θ𝑑θ=π/4π/3cos2θdθ

and the last integral is easy to compute using Equation (4).

Title A lecture on trigonometric integrals and trigonometric substitution
Canonical name ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution
Date of creation 2013-03-22 15:38:39
Last modified on 2013-03-22 15:38:39
Owner alozano (2414)
Last modified by alozano (2414)
Numerical id 4
Author alozano (2414)
Entry type Feature
Classification msc 26A36
Related topic ALectureOnIntegrationByParts
Related topic ALectureOnIntegrationBySubstitution
Related topic ALectureOnThePartialFractionDecompositionMethod