Lesson Explainer: Trigonometric Integrals | Nagwa Lesson Explainer: Trigonometric Integrals | Nagwa

Lesson Explainer: Trigonometric Integrals Mathematics

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 𝑥𝑥sind. This is not a standard trigonometric function whose integral we know, so we would need to find a function whose derivative is sin𝑥. 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 𝑥, sincos𝑥+𝑥1.

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 𝑥, cossin2𝑥12𝑥.

We can rearrange this identity to find an expression for sin𝑥 as follows: cossincossincossin2𝑥12𝑥12(2𝑥1)𝑥12(12𝑥)𝑥.

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 𝑥𝑥=12(12𝑥)𝑥=12(12𝑥)𝑥.sindcosdcosd

We then recall that 1𝑥=𝑥+dC and 𝑎𝑥𝑥=1𝑎𝑎𝑥+cosdsinC for any nonzero constant 𝑎. Thus, 12(12𝑥)𝑥=121𝑥122𝑥𝑥=𝑥212122𝑥+=𝑥2142𝑥+,cosddcosdsinCsinC where we combine the constants of integration into a single constant called C.

Hence, 𝑥𝑥=𝑥2142𝑥+.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 39𝑥𝑥tand.

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 𝜃, we have sincos𝜃+𝜃1.

Dividing both sides of the identity by cos𝜃 and using the fact that 1𝜃𝜃cossec, we get sincoscoscoscostansec𝜃𝜃+𝜃𝜃1𝜃𝜃+1𝜃.

We can rearrange this identity to make tan𝜃 the subject: tansec𝜃𝜃1.

We now rewrite this identity by substituting 𝜃=9𝑥 to get tansec9𝑥9𝑥1.

We can now substitute this expression into our integral. This gives us 39𝑥𝑥=39𝑥1𝑥.tandsecd

We can take the constant factor of 3 outside of the integral to get 39𝑥1𝑥=39𝑥1𝑥.secdsecd

We can now integrate this term by term by recalling that 𝑎𝑥𝑥=1𝑎𝑎𝑥+secdtanC for any nonzero constant 𝑎 and 1𝑥=𝑥+dC.

Instead of adding separate constants of integration, we can add a single constant of integration at the end of the expression to get 39𝑥1𝑥=3199𝑥𝑥+.secdtanC

We can leave our answer like this; however, we can rewrite this by taking out a factor of 19 to get 3199𝑥𝑥+=319(9𝑥+9𝑥)+=13(9𝑥9𝑥)+.tanCtanCtanC

Hence, we have shown that 39𝑥𝑥=13(9𝑥9𝑥)+.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 8𝑥𝑥cosd.

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 cossin𝑥1𝑥, 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 𝑥, we have that coscos2𝑥2𝑥1.

We can rearrange this identity to make cos𝑥 the subject. We have coscoscoscos2𝑥+12𝑥12(2𝑥+1)𝑥.

We can substitute this into the integral to get 8𝑥𝑥=812(2𝑥+1)𝑥=4(2𝑥+1)𝑥.cosdcosdcosd

We can also take out the constant factor of 4 to simplify the integral as shown: 4(2𝑥+1)𝑥=4(2𝑥+1)𝑥.cosdcosd

We can now integrate each term separately by recalling that 𝑎𝑥𝑥=1𝑎𝑎𝑥+cosdsinC for any nonzero constant 𝑎 and 1𝑥=𝑥+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 4(2𝑥+1)𝑥=4𝑥+122𝑥+=4𝑥+22𝑥+.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 ((𝑥)+(𝑥))𝑥cotcscd.

Answer

This integral is hard to integrate directly, so let’s instead rewrite the integrand. We can start by expanding the brackets to obtain ((𝑥)+(𝑥))𝑥=𝑥+2𝑥𝑥+𝑥𝑥.cotcscdcotcotcsccscd

We can integrate two of these terms using standard integral results. We recall the following: 𝑥𝑥𝑥=𝑥+,𝑥𝑥=𝑥+.cotcscdcscCcscdcotC

To integrate the term cot𝑥, we can rewrite it using the Pythagorean identity. We have for any angle 𝑥, cotcsc𝑥𝑥1.

We can use this to rewrite the integral as 𝑥+2𝑥𝑥+𝑥𝑥=𝑥1+2𝑥𝑥+𝑥𝑥=2𝑥1+2𝑥𝑥𝑥.cotcotcsccscdcsccotcsccscdcsccotcscd

We can now integrate each term separately. We have 2𝑥𝑥=2𝑥𝑥=2𝑥+,1𝑥=𝑥+,2𝑥𝑥𝑥=2𝑥𝑥𝑥=2𝑥+.cscdcscdcotCdCcotcscdcotcscdcscC

Therefore, we can add these results together (with a single constant of integration) to evaluate the integral. We have 𝑥1+2𝑥𝑥+𝑥𝑥=2𝑥𝑥2𝑥+.csccotcsccscdcotcscC

Hence, ((𝑥)+(𝑥))𝑥=2𝑥𝑥2𝑥+.cotcscdcotcscC

In the previous example, we can simplify our answer by multiplying through by 1. C is still a constant, so we can just call this new constant C. We get ((𝑥)+(𝑥))𝑥=2𝑥+𝑥+2𝑥+.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 (1𝑥)𝑥𝑥cossind.

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 (1𝑥)𝑥𝑥=12𝑥+𝑥𝑥𝑥.cossindcoscossind

We can now divide each term in the numerator by the denominator separately, giving us 12𝑥+𝑥𝑥𝑥=1𝑥2𝑥𝑥+𝑥𝑥𝑥.coscossindsincossincossind

We can then rewrite the integrand using trigonometric and reciprocal trigonometric identities: 1𝑥2𝑥𝑥+𝑥𝑥𝑥=𝑥2𝑥𝑥+𝑥𝑥.sincossincossindcsccotcsccotd

We cannot integrate each term in the integrand directly, we will rewrite cot𝑥 using the Pythagorean identity. We recall that for any angle 𝑥, we have cotcsc𝑥𝑥1.

Substituting this into the integrand and simplifying gives us 𝑥2𝑥𝑥+𝑥𝑥=𝑥2𝑥𝑥+𝑥1𝑥=2𝑥2𝑥𝑥1𝑥.csccotcsccotdcsccotcsccscdcsccotcscd

We can now integrate each term separately by recalling the following integral results: 𝑥𝑥=𝑥+,1𝑥=𝑥+,𝑥𝑥𝑥=𝑥+.cscdcotCdCcotcscdcscC

We can now integrate each term separately, adding a single constant of integration to get 2𝑥2𝑥𝑥1𝑥=2𝑥+2𝑥𝑥+.csccotcscdcotcscC

Hence, (1𝑥)𝑥𝑥=2𝑥2𝑥𝑥+.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 𝜃(1+𝜃)𝜃sincosd.

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:: tansincos𝜃2𝜃1+𝜃.

We can use this to rewrite the integrand as shown: 𝜃(1+𝜃)𝜃=𝜃1+𝜃𝜃=𝜃2𝜃=𝜃2𝜃.sincosdsincosdtandtand

We now rewrite tan𝜃2 using the Pythagorean identity. We recall that for any angle 𝑥, we have tansec𝑥𝑥1.

Substituting 𝑥=𝜃2 into the identity gives us tansec𝜃2=𝜃21.

Substituting this into our integrand gives us 𝜃2𝜃=𝜃21𝜃.tandsecd

We can now integrate this expression term by term by recalling that (𝑎𝜃)𝜃=1𝑎(𝑎𝜃)+secdtanC for any nonzero constant 𝑎 and 1𝜃=𝜃+dC.

Substituting 𝑎=12 into this integral result yields 𝜃2𝜃=1𝜃2+=2𝜃2+.secdtanCtanC

We can use this to evaluate the integral. We have 𝜃21𝜃=2𝜃2𝜃.secdtan

Now, all that is left to do is to evaluate the limits of integration. We get 2𝜃2𝜃=22𝜋222𝜋2=2𝜋4𝜋22𝜋4𝜋2=2𝜋2+2𝜋2=4𝜋.tantantantantan

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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