**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

The reason is that

cos 2*x* = cos^{2}*x* − sin^{2}*x* = [1 − sin^{2}*x* ] − sin^{2}*x* = 1 − 2 sin^{2}*x* .

**II**. We have

The reason is that

cos 2*x* = cos^{2}*x* − sin^{2}*x*

sin^{2}*x* = cos^{2}*x* − [1 − cos^{2}*x* ] = 2 cos^{2}*x* − 1.

Now we can turn to some examples.

**Examples**

**Example 1**

Calculate the integral

∫ cos^{2}*x dx* .

**Solution 1**

Of course we will use formula **II** . We write

**Example 2**

Calculate the integral

∫ sin^{3}*x* cos^{2}*x dx* .

**Solution 2**

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

sin^{3}*x* cos^{2}*x* = sin*x* sin^{2}*x* cos^{2}*x* = sin *x* (1 − cos^{2}*x* ) cos^{2}*x*

= [cos^{2}*x* − cos^{4}*x* ] sin*x* .

Then

∫ sin^{3}*x* cos^{2} *dx* = ∫ [cos^{2}*x* − cos^{4}*x* ] sin *x dx* .

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

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

**You Try It** : Calculate the integral

∫ sin^{2}3*x* cos^{5}3*x dx* .

**Example 3**

Calculate

**Solution 3**

Substituting

into the integrand yields

Again using formula **II**, we find that our integral becomes

Applying formula **II** one last time yields

**You Try It** : Calculate the integral

**You Try It** : Calculate the integral

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

**Examples**

**Example 4**

Calculate

∫ tan^{3}*x* sec^{3}*x 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* (sec^{2}*x* − l) sec^{3} *x dx* .

We may regroup the terms in the integrand to obtain

∫ [sec^{4} *x* − sec^{2} *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

Resubstituting the value of *u* gives

**Example 5**

Calculate

**Solution 5**

We write

Letting *u* = tan *x* and *du* = sec^{2} *x dx* then gives the integral

**You Try It** : Calculate the integral

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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