# Question Video: Determining the Definite Integral of a Trigonometric Function Mathematics • Higher Education

Determine β«_(βπ/4)^(βπ/6) (4 + π cos 9π₯) dπ₯.

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### Video Transcript

Determine the definite integral from negative π by four to negative π by six of four plus π times the cos of nine π₯ with respect to π₯.

Weβre asked to evaluate the definite integral of a constant plus a trigonometric function. And we know how to evaluate definite integrals by using the fundamental theorem of calculus. So weβll start by recalling the part of the fundamental theorem of calculus which tells us how to evaluate definite integrals.

We have if lowercase π is continuous on the closed interval from π to π and capital πΉ prime of π₯ is equal to lowercase π of π₯, then the definite integral from π to π of lowercase π of π₯ with respect to π₯ is equal to capital πΉ evaluated at π minus capital πΉ evaluated at π. In other words, if our integrand is continuous on the interval of integration and we have an antiderivative of our integrand, then we can evaluate our definite integral by using our antiderivative. Itβs equal to capital πΉ evaluated at π, the upper limit of integration, minus capital πΉ evaluated at π, the lower limit of integration.

So letβs start by finding the values of these given to us in the question. We see our upper limit of integration is negative π by six, and the lower limit of integration is negative π by four. So π is negative π by four, and π is negative π by six. Next, we set lowercase π of π₯ to be our integrand, four plus π times the cos of nine π₯.

Now, remember, to use the fundamental theorem of calculus, we need to check that our integrand is continuous on the interval of integration. In this case, we can see that our integrand is the sum of two continuous functions. Itβs a constant plus a trigonometric function. In particular, we know the cos of nine π₯ is continuous for all real values of π₯. So our integrand is the sum of two functions continuous for all real values of π₯. In particular, this means that π will be continuous on our interval of integration. So weβve shown that the first part is true.

We now need to find an antiderivative of our integrand. And there are several different ways of doing this. However, the easiest is to use what we know about indefinite integrals. Remember, indefinite integrals is a way of finding antiderivatives. And we know how to do this term by term.

First, the integral of the constant four with respect to π₯ is equal to four π₯. Next, to integrate our second term, we need to recall one of our standard integral results for trigonometric functions. For any real constant π not equal to zero, the integral of the cos of ππ₯ with respect to π₯ is equal to the sin of ππ₯ divided by π plus a constant of integration πΆ. In our case, the value of π is equal to nine. So we get π times the sin of nine π₯ all divided by nine. And we add a constant of integration πΆ. And this gives us antiderivatives of our integrand.

In fact, this is an antiderivative for any value of the constant πΆ. But we only need one integrand. So weβll just set the value of πΆ equal to zero. This gives us capital πΉ of π₯ is equal to four π₯ plus π times the sin of nine π₯ divided by nine.

Now that weβve found our antiderivative and shown that our integral is continuous across the entire interval of integration, weβre ready to use the fundamental theorem of calculus to evaluate our definite integral. We get that this definite integral is equal to four π₯ plus π times the sin of nine π₯ divided by nine evaluated at the limits of integration, negative π by four and negative π by six.

Now, all thatβs left to do is evaluate this at the limits of integration. Weβll start with the upper limits of integration, when π₯ is equal to negative π by six. We substitute π₯ is equal to negative π by six into our antiderivative. This gives us four times negative π by six plus π times the sin of nine times negative π by six all divided by nine. And if we evaluate this expression, we get negative five π by nine.

Weβll now do the same with our lower limits of integration. Remember, we need to subtract this value. So we substitute π₯ is equal to negative π by four into our antiderivative. This gives us four times negative π by four plus π times the sin of nine times negative π by four divided by nine. And now we can start simplifying once again. First, four times negative π by four is equal to negative π. Next, if we evaluate π times the sin of negative nine π by four divided by nine, we get negative root two π divided by 18. So this entire expression simplified to give us negative five π by nine minus negative π minus root two π divided by 18.

And finally, if we evaluate this expression, we get root two π divided by 18 plus four π by nine, which is our final answer. Therefore, by using the fundamental theorem of calculus and what we know about indefinite integrals of trigonometric functions, we were able to show the definite integral from negative π by four to negative π by six of four plus π times the cos of nine π₯ with respect to π₯ is equal to root two π divided by 18 plus four π by nine.