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Integrals of Trigonometric Expressions Help

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Introduction to Integrals of Trigonometric Expressions

Trigonometric expressions arise frequently in our work, especially as a result of substitutions. In this section we develop a few examples of trigonometric integrals.

The following trigonometric identities will be particularly useful for us.

I. We have

Methods of Integration 7.4 Integrals of Trigonometric Expressions

The reason is that

cos 2x = cos2x − sin2x = [1 − sin2x ] − sin2x = 1 − 2 sin2x .

II. We have

Methods of Integration 7.4 Integrals of Trigonometric Expressions

The reason is that

cos 2x = cos2x − sin2x

sin2x = cos2x − [1 − cos2x ] = 2 cos2x − 1.

Now we can turn to some examples.

Examples

Example 1

Calculate the integral

∫ cos2x dx .

Solution 1

Of course we will use formula II . We write

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Example 2

Calculate the integral

∫ sin3x  cos2x dx .

Solution 2

When sines and cosines occur together, we always focus on the odd power (when one occurs). We write

sin3x  cos2x = sinx sin2x cos2x = sin x (1 − cos2x ) cos2x

= [cos2x − cos4x ] sinx .

Then

∫ sin3x  cos2 dx = ∫ [cos2x − cos4x ] sin x dx .

A u -substitution is suggested: We let u = cos x , du = − sin x dx . Then the integral becomes

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Resubstituting for the u variable, we obtain the final solution of

Methods of Integration 7.4 Integrals of Trigonometric Expressions

You Try It : Calculate the integral

∫ sin23x cos53x dx .

Example 3

Calculate

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Solution 3

Substituting

Methods of Integration 7.4 Integrals of Trigonometric Expressions

into the integrand yields

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Again using formula II, we find that our integral becomes

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Applying formula II one last time yields

Methods of Integration 7.4 Integrals of Trigonometric Expressions

You Try It : Calculate the integral

Methods of Integration 7.4 Integrals of Trigonometric Expressions

You Try It : Calculate the integral

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Integrals involving the other trigonometric functions can also be handled with suitable trigonometric identities. We illustrate the idea with some examples that are handled with the identity

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Examples

Example 4

Calculate

∫ tan3x sec3x dx .

Solution 4

Using the same philosophy about odd exponents as we did with sines and cosines, we substitute sec 2 x − 1 for tan 2 x . The result is

∫ tan x (sec2x − l) sec3 x dx .

We may regroup the terms in the integrand to obtain

∫ [sec4 x − sec2 x ] sec x tan x dx .

A u -substitution suggests itself: We let u = sec x and therefore du = sec x tan x dx . Thus our integral becomes

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Resubstituting the value of u gives

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Example 5

Calculate

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Solution 5

We write

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Letting u = tan x and du = sec2 x dx then gives the integral

Methods of Integration 7.4 Integrals of Trigonometric Expressions

You Try It : Calculate the integral

Methods of Integration 7.4 Integrals of Trigonometric Expressions

Further techniques in the evaluation of trigonometric integrals will be explored in the exercises.

Find practice problems and solutions for these concepts at: Methods Of Integration Practice Test.

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