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Lesson Explainer: Trigonometric Integrals

In this explainer, we will learn how to evaluate integrals of the products of trigonometric terms using trigonometric identities.

There are many trigonometric functions we cannot easily integrate. For example, imagine we were asked to determine ο„Έπ‘₯π‘₯sind. This is not a standard trigonometric function whose integral we know, so we would need to find a function whose derivative is sinπ‘₯. This is quite a difficult task, so instead, we can try rewriting the integrand into an easier form to integrate.

We can recall two identities involving the square of the sine function. We have the Pythagorean identity, which says that for any angle π‘₯, sincosπ‘₯+π‘₯≑1.

If we used this identity to rewrite the integrand, we would need to integrate the square of the cosine function and this is not any easier.

Instead, we can recall the double-angle formula for the cosine function, which tells us that for any angle π‘₯, cossin2π‘₯≑1βˆ’2π‘₯.

We can rearrange this identity to find an expression for sinπ‘₯ as follows: cossincossincossin2π‘₯βˆ’1β‰‘βˆ’2π‘₯βˆ’12(2π‘₯βˆ’1)≑π‘₯12(1βˆ’2π‘₯)≑π‘₯.

We can integrate the left-hand side of this equation, so this will allow us to integrate the square of the sine function. We have ο„Έπ‘₯π‘₯=ο„Έ12(1βˆ’2π‘₯)π‘₯=12ο„Έ(1βˆ’2π‘₯)π‘₯.sindcosdcosd

We then recall that ο„Έ1π‘₯=π‘₯+dC and ο„Έπ‘Žπ‘₯π‘₯=1π‘Žπ‘Žπ‘₯+cosdsinC for any nonzero constant π‘Ž. Thus, 12ο„Έ(1βˆ’2π‘₯)π‘₯=12ο„Έ1π‘₯βˆ’12ο„Έ2π‘₯π‘₯=π‘₯2βˆ’12ο€Ό122π‘₯+=π‘₯2βˆ’142π‘₯+,cosddcosdsinCsinC where we combine the constants of integration into a single constant called C.

Hence, ο„Έπ‘₯π‘₯=π‘₯2βˆ’142π‘₯+.sindsinC

There are many different identities we can use to rewrite integrands into a form that we may be able to integrate. These include the Pythagorean identity, angle sum and difference identities, double-angle identities, and half-angle identities.

Let’s now see an example of integrating the square of the tangent function by rewriting it using the Pythagorean identity.

Example 1: Integrating Tangent Squared

Determine ο„Έ39π‘₯π‘₯tand.

Answer

It is not immediately obvious how to directly integrate the square of the tangent function; however, we can manipulate the integrand into a form that we can integrate more easily.

We can recall that we can find an expression for the square of the tangent function by using the Pythagorean identity, which tells us that for any angle πœƒ, we have sincosοŠ¨οŠ¨πœƒ+πœƒβ‰‘1.

Dividing both sides of the identity by cosοŠ¨πœƒ and using the fact that 1πœƒβ‰‘πœƒcossec, we get sincoscoscoscostansecοŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨πœƒπœƒ+πœƒπœƒβ‰‘1πœƒπœƒ+1β‰‘πœƒ.

We can rearrange this identity to make tanοŠ¨πœƒ the subject: tansecοŠ¨οŠ¨πœƒβ‰‘πœƒβˆ’1.

We now rewrite this identity by substituting πœƒ=9π‘₯ to get tansec9π‘₯≑9π‘₯βˆ’1.

We can now substitute this expression into our integral. This gives us ο„Έ39π‘₯π‘₯=ο„Έ3ο€Ή9π‘₯βˆ’1π‘₯.tandsecd

We can take the constant factor of 3 outside of the integral to get ο„Έ3ο€Ή9π‘₯βˆ’1π‘₯=3ο„Έο€Ή9π‘₯βˆ’1π‘₯.secdsecd

We can now integrate this term by term by recalling that ο„Έπ‘Žπ‘₯π‘₯=1π‘Žπ‘Žπ‘₯+secdtanC for any nonzero constant π‘Ž and ο„Έ1π‘₯=π‘₯+dC.

Instead of adding separate constants of integration, we can add a single constant of integration at the end of the expression to get 3ο„Έο€Ή9π‘₯βˆ’1π‘₯=3ο€Ό199π‘₯βˆ’π‘₯+.secdtanC

We can leave our answer like this; however, we can rewrite this by taking out a factor of βˆ’19 to get 3ο€Ό199π‘₯βˆ’π‘₯+=3ο€Όβˆ’19(βˆ’9π‘₯+9π‘₯)+=βˆ’13(9π‘₯βˆ’9π‘₯)+.tanCtanCtanC

Hence, we have shown that ο„Έ39π‘₯π‘₯=βˆ’13(9π‘₯βˆ’9π‘₯)+.tandtanC

In our next example, we will integrate the square of the cosine function by rewriting it using a double-angle identity.

Example 2: Integrating Cosine Squared

Determine ο„Έ8π‘₯π‘₯cosd.

Answer

It is not immediately obvious how to directly integrate the square of the cosine function; however, we can manipulate the integrand into a form that we can integrate. We might be tempted to do this by using the Pythagorean identity; however, this would give us cossinπ‘₯≑1βˆ’π‘₯, and this expression is no easier to integrate than what we started with.

Instead, we can recall that the double-angle identity for the cosine states that for any value of π‘₯, we have that coscos2π‘₯≑2π‘₯βˆ’1.

We can rearrange this identity to make cosπ‘₯ the subject. We have coscoscoscos2π‘₯+1≑2π‘₯12(2π‘₯+1)≑π‘₯.

We can substitute this into the integral to get ο„Έ8π‘₯π‘₯=ο„Έ8ο€Ό12(2π‘₯+1)π‘₯=ο„Έ4(2π‘₯+1)π‘₯.cosdcosdcosd

We can also take out the constant factor of 4 to simplify the integral as shown: ο„Έ4(2π‘₯+1)π‘₯=4ο„Έ(2π‘₯+1)π‘₯.cosdcosd

We can now integrate each term separately by recalling that ο„Έπ‘Žπ‘₯π‘₯=1π‘Žπ‘Žπ‘₯+cosdsinC for any nonzero constant π‘Ž and ο„Έ1π‘₯=π‘₯+dC.

We now evaluate the integral and instead of adding separate constants of integration, we can add a single constant of integration at the end of the expression to get 4ο„Έ(2π‘₯+1)π‘₯=4ο€Όπ‘₯+122π‘₯+=4π‘₯+22π‘₯+.cosdsinCsinC

In our next example, we will integrate a trigonometric expression by rewriting it using various trigonometric identities and integral results.

Example 3: Integrating a Trigonometric Function Using Identities

Determine ο„Έ((π‘₯)+(π‘₯))π‘₯cotcscd.

Answer

This integral is hard to integrate directly, so let’s instead rewrite the integrand. We can start by expanding the brackets to obtain ο„Έ((π‘₯)+(π‘₯))π‘₯=ο„Έο€Ήπ‘₯+2π‘₯π‘₯+π‘₯π‘₯.cotcscdcotcotcsccscd

We can integrate two of these terms using standard integral results. We recall the following: ο„Έπ‘₯π‘₯π‘₯=βˆ’π‘₯+,ο„Έπ‘₯π‘₯=βˆ’π‘₯+.cotcscdcscCcscdcotC

To integrate the term cotπ‘₯, we can rewrite it using the Pythagorean identity. We have for any angle π‘₯, cotcscπ‘₯≑π‘₯βˆ’1.

We can use this to rewrite the integral as ο„Έπ‘₯+2π‘₯π‘₯+π‘₯π‘₯=ο„Έο€Ήπ‘₯βˆ’1+2π‘₯π‘₯+π‘₯π‘₯=ο„Έο€Ή2π‘₯βˆ’1+2π‘₯π‘₯π‘₯.cotcotcsccscdcsccotcsccscdcsccotcscd

We can now integrate each term separately. We have ο„Έ2π‘₯π‘₯=2ο„Έπ‘₯π‘₯=βˆ’2π‘₯+,ο„Έβˆ’1π‘₯=βˆ’π‘₯+,ο„Έ2π‘₯π‘₯π‘₯=2ο„Έπ‘₯π‘₯π‘₯=βˆ’2π‘₯+.cscdcscdcotCdCcotcscdcotcscdcscC

Therefore, we can add these results together (with a single constant of integration) to evaluate the integral. We have ο„Έο€Ήπ‘₯βˆ’1+2π‘₯π‘₯+π‘₯π‘₯=βˆ’2π‘₯βˆ’π‘₯βˆ’2π‘₯+.csccotcsccscdcotcscC

Hence, ο„Έ((π‘₯)+(π‘₯))π‘₯=βˆ’2π‘₯βˆ’π‘₯βˆ’2π‘₯+.cotcscdcotcscC

In the previous example, we can simplify our answer by multiplying through by βˆ’1. βˆ’C is still a constant, so we can just call this new constant C. We get ο„Έ((π‘₯)+(π‘₯))π‘₯=2π‘₯+π‘₯+2π‘₯+.cotcscdcotcscC

In our next example, we will integrate a trigonometric function by rewriting it using trigonometric identities.

Example 4: Integrating a Trigonometric Function Using Identities

Determine ο„Έ(1βˆ’π‘₯)π‘₯π‘₯cossind.

Answer

It is hard to evaluate this integral directly, so let’s instead rewrite the integrand. We can start by multiplying out the parentheses to get ο„Έ(1βˆ’π‘₯)π‘₯π‘₯=ο„Έ1βˆ’2π‘₯+π‘₯π‘₯π‘₯.cossindcoscossind

We can now divide each term in the numerator by the denominator separately, giving us ο„Έ1βˆ’2π‘₯+π‘₯π‘₯π‘₯=ο„Έο€Ύ1π‘₯βˆ’2π‘₯π‘₯+π‘₯π‘₯π‘₯.coscossindsincossincossind

We can then rewrite the integrand using trigonometric and reciprocal trigonometric identities: ο„Έο€Ύ1π‘₯βˆ’2π‘₯π‘₯+π‘₯π‘₯π‘₯=ο„Έο€Ήπ‘₯βˆ’2π‘₯π‘₯+π‘₯π‘₯.sincossincossindcsccotcsccotd

We cannot integrate each term in the integrand directly, we will rewrite cotπ‘₯ using the Pythagorean identity. We recall that for any angle π‘₯, we have cotcscπ‘₯≑π‘₯βˆ’1.

Substituting this into the integrand and simplifying gives us ο„Έο€Ήπ‘₯βˆ’2π‘₯π‘₯+π‘₯π‘₯=ο„Έο€Ήπ‘₯βˆ’2π‘₯π‘₯+π‘₯βˆ’1π‘₯=ο„Έο€Ή2π‘₯βˆ’2π‘₯π‘₯βˆ’1π‘₯.csccotcsccotdcsccotcsccscdcsccotcscd

We can now integrate each term separately by recalling the following integral results: ο„Έπ‘₯π‘₯=βˆ’π‘₯+,ο„Έβˆ’1π‘₯=βˆ’π‘₯+,ο„Έπ‘₯π‘₯π‘₯=βˆ’π‘₯+.cscdcotCdCcotcscdcscC

We can now integrate each term separately, adding a single constant of integration to get ο„Έο€Ή2π‘₯βˆ’2π‘₯π‘₯βˆ’1π‘₯=βˆ’2π‘₯+2π‘₯βˆ’π‘₯+.csccotcscdcotcscC

Hence, ο„Έ(1βˆ’π‘₯)π‘₯π‘₯=2π‘₯βˆ’2π‘₯βˆ’π‘₯+.cossindcsccotC

In our final example, we evaluate a definite integral by rewriting it using a half-angle identity.

Example 5: Evaluating a Definite Integral of a Trigonometric Function Using Identities

Evaluate ο„Έπœƒ(1+πœƒ)πœƒο‘½οŽ‘ο‘½οŽ‘οŠ±οŠ¨οŠ¨sincosd.

Answer

There are many different ways we could rewrite the integrand in order to evaluate this integral. The easiest way is to note the similarity between the integrand and the half-angle identity for the tangent function:: tansincosο€½πœƒ2ο‰β‰‘πœƒ1+πœƒ.

We can use this to rewrite the integrand as shown: ο„Έπœƒ(1+πœƒ)πœƒ=ο„Έο€½πœƒ1+πœƒο‰πœƒ=ο„Έο€½ο€½πœƒ2ο‰ο‰πœƒ=ο„Έο€½πœƒ2ο‰πœƒ.ο‘½οŽ‘ο‘½οŽ‘ο‘½οŽ‘ο‘½οŽ‘ο‘½οŽ‘ο‘½οŽ‘ο‘½οŽ‘ο‘½οŽ‘οŠ±οŠ¨οŠ¨οŠ±οŠ¨οŠ±οŠ¨οŠ±οŠ¨sincosdsincosdtandtand

We now rewrite tanοŠ¨ο€½πœƒ2 using the Pythagorean identity. We recall that for any angle π‘₯, we have tansecπ‘₯≑π‘₯βˆ’1.

Substituting π‘₯=πœƒ2 into the identity gives us tansecοŠ¨οŠ¨ο€½πœƒ2=ο€½πœƒ2ο‰βˆ’1.

Substituting this into our integrand gives us ο„Έο€½πœƒ2ο‰πœƒ=ο„Έο€½ο€½πœƒ2ο‰βˆ’1ο‰πœƒ.ο‘½οŽ‘ο‘½οŽ‘ο‘½οŽ‘ο‘½οŽ‘οŠ±οŠ¨οŠ±οŠ¨tandsecd

We can now integrate this expression term by term by recalling that ο„Έ(π‘Žπœƒ)πœƒ=1π‘Ž(π‘Žπœƒ)+secdtanC for any nonzero constant π‘Ž and ο„Έ1πœƒ=πœƒ+dC.

Substituting π‘Ž=12 into this integral result yields ο„Έο€½πœƒ2ο‰πœƒ=1ο€»ο‡ο€½πœƒ2+=2ο€½πœƒ2+.secdtanCtanC

We can use this to evaluate the integral. We have ο„Έο€½ο€½πœƒ2ο‰βˆ’1ο‰πœƒ=2ο€½πœƒ2ο‰βˆ’πœƒο‘.ο‘½οŽ‘ο‘½οŽ‘ο‘½οŽ‘ο‘½οŽ‘οŠ±οŠ¨οŠ±secdtan

Now, all that is left to do is to evaluate the limits of integration. We get 2ο€½πœƒ2ο‰βˆ’πœƒο‘=22οŽβˆ’πœ‹2βˆ’ο‚2ο‚ο€»βˆ’ο‡2οŽβˆ’ο€»βˆ’πœ‹2ο‡οŽ=2ο€»πœ‹4ο‡βˆ’πœ‹2βˆ’ο€»2ο€»βˆ’πœ‹4ο‡βˆ’ο€»βˆ’πœ‹2=2βˆ’πœ‹2+2βˆ’πœ‹2=4βˆ’πœ‹.tantantantantanο‘½οŽ‘ο‘½οŽ‘οŠ±οŽ„οŠ¨οŽ„οŠ¨

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We can evaluate some integrals of trigonometric functions by rewriting them using the Pythagorean identity, angle sum and difference identities, double-angle identities, and half-angle identities.
  • There are often many ways to rewrite the integrands, so it is a good idea to check different options to see which gives an easy-to-evaluate integral.

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