Question Video: Finding the Integration of an Polynomial Function Multiplied by a Trigonometric Function Using Integration by Parts Mathematics • Higher Education

Use integration by parts to evaluate β«π₯ sin π₯ dπ₯.

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Video Transcript

Use integration by parts to evaluate the indefinite integral of π₯ times the sin of π₯ with respect to π₯.

In this question, we are asked to evaluate the indefinite integral of the product of two functions: an algebraic function π₯ and a trigonometric function sin of π₯. And this should hint to us that we could try using integration by parts to evaluate this integral, even if we were not told to use this method in the question. So, letβs start by recalling the formula for integration by parts.

It tells us the indefinite integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯. To apply this to evaluate our integral, weβre going to need to choose which of the factors in our integrand needs to be π’ and which one should be dπ£ by dπ₯. So, we need to determine how weβre going to choose which factor will be π’ and which will be dπ£ by dπ₯.

To do this, we need to know when weβre using this formula, the most difficult part of this expression to evaluate is the integral. So, we want to choose our functions to make this integral as easy to evaluate as possible. And usually, we want to do this by choosing π’ which makes dπ’ by dπ₯ simpler. However, sometimes, we need to choose our function π’ such that it cancels when multiplied by part of π£. In either case, letβs look at our functions π₯ and sin of π₯.

We know when we differentiate π₯, its degree will decrease. However, if we were to differentiate the sin of π₯, we would just get another trigonometric function. This would not be any simpler. So, weβll set our function π’ to be equal to π₯ and dπ£ by dπ₯ to be equal to the sin of π₯. Now, to apply integration by parts, weβre going to need to find an expression for dπ’ by dπ₯ and π£.

So, letβs start by finding dπ’ by dπ₯. Thatβs the derivative of π₯ with respect to π₯, which is just equal to one. We also want to find an expression for π£ from dπ£ by dπ₯. And we can recall π£ will be an antiderivative of this function. So, we know that π£ is going to be equal to the indefinite integral of the sin of π₯ with respect to π₯, which we can recall is negative the cos of π₯ plus the constant of integration. However, we donβt need to add a constant of integration in this case.

To see this, consider what would happen if we replace π£ by π£ plus πΆ in our integration by parts formula. In the first term, we would get π’ multiplied by the constant of integration πΆ. However, in our second term, we get πΆ times dπ’ by dπ₯. Negative the integral of this is negative π’ time πΆ, so these two terms cancel. So, itβs another case of constants of integration canceling, so we donβt need to include this constant.

Letβs now substitute our expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯ into our integration by parts formula. We get the indefinite integral of π₯ times the sin of π₯ with respect to π₯ is equal to π₯ times negative the cos of π₯ minus the indefinite integral of negative cos of π₯ multiplied by one with respect to π₯. The first term simplifies to give us negative π₯ times the cos of π₯. And in our second term, we take out the factor of negative one. This means we add the indefinite integral of the cos of π₯ with respect to π₯.

Now, all thatβs left to do is evaluate this indefinite integral. We know that the sin of π₯ is an antiderivative of cos of π₯. So, the indefinite integral of cos of π₯ is sin of π₯ plus the constant of integration πΆ. Therefore, we were able to show the indefinite integral of π₯ times the sin of π₯ with respect to π₯ is negative π₯ cos of π₯ plus sin of π₯ plus πΆ. And it is worth noting we can check our answer by differentiating it with respect to π₯ and checking we get π₯ multiplied by the sin of π₯.