Integration by parts practice problems

In a previous lesson, I explained the integration by parts formula and how to use it. Sometimes though, finding an integral using integration by parts isn’t as simple as the problem I did in that lesson. So I’d like to show some other more complex cases and how to work through them.

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Example 1

Evaluate the integral $$\int 4x^2 \ sin(5x) \ dx.$$

I will proceed through this problem following the same steps that I used in the integration by parts lesson.

1. Picking u and dv

Remember, we want to pick the piece of our function that we’d rather differentiate to be u, and the piece we’d rather integrate to be dv. The integral we need to evaluate can clearly be thought of as the product of and . Therefore, we need to take the derivative of one of these and the anti-derivative of the other.

First consider the piece. If we take the derivative of this function we’ll end up with an x term, but if we take the anti-derivative we’ll end up with an term. This is due to the power rule. Generally it’s best to choose whichever will result in the simplest option. When it comes to polynomials, the simpler one is whichever has the lower power. Therefore, it is preferred to take the derivative of the piece because the x term resulting from the derivative is simpler than the term resulting from taking the anti-derivative.

But before we say that we definitely want to assign the to be u so that we can take its derivative, we want to think about what this will mean for the sin(5x) piece.

What we need to consider is the difference between taking the derivative and the anti-derivative of sin(5x). In either case, we will end up with some constant multiplied by cos(5x) (we can find this using the chain rule for the derivative or u-substitution for the anti-derivative). But you can see that the derivative and the anti-derivative of this piece are equally complex, so it doesn’t make much of a difference whether we say the sin(5x) piece is assigned to u or dv.

Since the derivative of is much simpler than its anti-derivative, we would rather call it u than dv. And it doesn’t make a difference if sin(5x) is considered to be u or dv. So we will say $$u=4x^2$$ $$dv=sin(5x) \ dx.$$

2. Finding v and du

Now to find v we simply need to take the anti-derivative of the dv piece from the previous section. And to find du we need to take the derivative of u.

You can find the anti-derivative of sin(5x) by using u-substitution. I’m not going to show all the steps for this, but we will need to use the fact that the anti-derivative of sin(x) is -cos(x). Knowing this, we can find that the anti-derivative of sin(5x) would give us $$v=- \ \frac{1}{5} cos(5x).$$

Now that we have found v, let’s move onto finding du. This can simply be done by finding the derivative of u from part 1. Finding the derivative of can simply be found using the power rule. Doing this gives us $$du=8x \ dx.$$

3. Plugging it all into the formula

Now that we have found all 4 of the pieces we need, we just have to plug them into the integration by parts formula. To summarize, the 4 pieces we have up to this point are $$u=4x^2$$ $$dv=sin(5x) \ dx$$ $$v=- \ \frac{1}{5} cos(5x)$$ $$du=8x \ dx.$$

So we just need to use the integration by parts formula with these. $$\int u \ dv = uv – \int v \ du$$ $$\int 4x^2 \ sin(5x) \ dx = \Big(4x^2\Big) \bigg(- \ \frac{1}{5} cos(5x) \bigg) – \int – \ \frac{1}{5} cos(5x) \ 8x \ dx$$

Now we can simplify and evaluate the integral on the right side of our equation. $$(1): \ \int 4x^2 \ sin(5x) \ dx = \ – \ \frac{4}{5} x^2 \ cos(5x) \ + \ \frac{8}{5}\int x \ cos(5x) \ dx$$

But notice, the integral on the right side of our equation is still fairly complex. We still have an integral which is the product of two simpler functions, x and cos(5x). In order to evaluate this integral we’ll actually need to use integration by parts again. So now we need to use integration by parts to evaluate $$\int x \ cos(5x) \ dx.$$ We’ll go ahead and follow the same steps as we did before, but now we have a new integral and will need to reassign our u and dv.

4. Picking u and dv

Now that we are going through this process a second time, we don’t really have much of a choice when we pick which piece will be u and dv. The reason for this is that we will need to make this determination based on what we did the first time through. Consider where each of our pieces came from. One of our pieces is x, which came from the in our original integral. And the other piece is cos(5x) which came from the sin(5x) in the original integral.

Since the x piece in our current integral came from the piece in the original integral, and we decided that the piece would be u earlier, we need to follow up by doing the same here. Therefore, we will say $$u=x$$ this time around.

By the same reasoning, we will need to say that $$dv = cos(5x) \ dx$$ this time around since we said assigned the sin(5x) to dv the first time through.

5. Finding v and du

Now we just need to take the u and dv from the previous step and use them to find v and du.

To find v we just need to find the anti-derivative of dv. We previously decided that . Just like before, we can find this anti-derivative using u-substitution. Doing this tells us that $$v = \frac{1}{5} sin(5x).$$

And now we just need to find du by taking the derivative of u. Since we know , we know that $$du=dx.$$

6. Plugging it all into the formula

And finally we just need to plug the 4 pieces we have found into the integration by parts formula. So far we have found $$u=x$$ $$dv = cos(5x) \ dx$$ $$v = \frac{1}{5} sin(5x)$$ $$du=dx.$$

Now putting these into the integration by parts formula we find $$\int u \ dv = uv – \int v \ du$$ $$\int x \ cos(5x) \ dx \ = \ x \bigg( \frac{1}{5} sin(5x) \bigg) \ – \int \frac{1}{5} sin(5x) \ dx$$ $$\int x \ cos(5x) \ dx \ = \ \frac{1}{5} x \ sin(5x) \ – \ \frac{1}{5} \int sin(5x) \ dx$$

And now the integral we need to evaluate is much simpler than what we started with. $$\int x \ cos(5x) \ dx \ = \ \frac{1}{5} x \ sin(5x) \ – \ \frac{1}{5} \bigg( – \frac{1}{5} cos(5x) \bigg)$$ $$\int x \ cos(5x) \ dx \ = \ \frac{1}{5} x \ sin(5x) \ + \ \frac{1}{25} cos(5x)$$

Wrapping it all together

Now that we have found $$\int x \ cos(5x) \ dx \ = \ \frac{1}{5} x \ sin(5x) \ + \ \frac{1}{25} cos(5x)$$ we can bring this back to our equation (1) back in step 3. And all we need to do is replace the with . Doing this tells us that $$\int 4x^2 \ sin(5x) \ dx = \ – \ \frac{4}{5} x^2 \ cos(5x) \ + \ \frac{8}{5}\int x \ cos(5x) \ dx$$ $$= \ – \ \frac{4}{5} x^2 \ cos(5x) \ + \ \frac{8}{5} \bigg( \frac{1}{5} x \ sin(5x) \ + \ \frac{1}{25} cos(5x) \bigg)$$ $$= \ – \ \frac{4}{5} x^2 \ cos(5x) \ + \ \frac{8}{25} x \ sin(5x) \ + \ \frac{8}{125} cos(5x)$$

And that’s our answer! Clearly a bit more complicated than the first integration by parts example I did, but it isn’t too bad. You essentially just need to apply the same process two time in a row. As long as you stay consistent in your designations of u and dv each time, it should all work out in the end.

Example 2

$$\int xe^{-2x} \ dx$$

We’ll start this by deciding which piece we’ll call u and which piece is dv.

$$u=x$$ $$dv=e^{-2x} \ dx$$

Then we need to use these to figure out du and v.

$$du=1 \cdot dx=dx$$ $$v=-\frac{1}{2}e^{-2x}$$

Now we can plug all 4 of these pieces into the integration by parts formula.

$$\int xe^{-2x} \ dx \ = \ x \bigg( -\frac{1}{2}e^{-2x} \bigg) – \int -\frac{1}{2}e^{-2x} \ dx$$

At this point we are left with a simpler integral to evaluate.

$$= -\frac{1}{2}xe^{-2x} + \frac{1}{2} \int e^{-2x} \ dx$$ $$= -\frac{1}{2}xe^{-2x} + \frac{1}{2} \bigg( -\frac{1}{2} e^{-2x} \bigg)$$ $$= -\frac{1}{2}xe^{-2x} – \frac{1}{4} e^{-2x}$$ $$= -\frac{1}{2}e^{-2x} \bigg( x + \frac{1}{2} \bigg)$$

Example 3

$$\int x^{\frac{3}{2}}ln(x) \ dx$$

We’ll start this by deciding which piece we’ll call u and which piece is dv.

$$u=ln(x)$$ $$dv=x^{\frac{3}{2}} \ dx$$

Then we need to use these to figure out du and v.

$$du=\frac{1}{x} \ dx$$ $$v=\frac{2}{5}x^{\frac{5}{2}}$$

Now we can plug all 4 of these pieces into the integration by parts formula.

$$\int x^{\frac{3}{2}}ln(x) \ dx \ = \ \frac{2}{5}x^{\frac{5}{2}} ln(x) \ – \int \frac{1}{x} \cdot \frac{2}{5}x^{\frac{5}{2}} \ dx$$

At this point we are left with a simpler integral to evaluate.

$$= \ \frac{2}{5}x^{\frac{5}{2}} ln(x) \ – \ \frac{2}{5} \int x^{\frac{3}{2}} \ dx$$ $$= \ \frac{2}{5}x^{\frac{5}{2}} ln(x) \ – \ \frac{2}{5} \cdot \frac{2}{5} x^{\frac{5}{2}}$$ $$= \ \frac{2}{5}x^{\frac{5}{2}} \bigg( ln(x) \ – \ \frac{2}{5} \bigg)$$

As always, let me know if you have any questions. If anything was confusing here leave a comment or send me an email at jakesmathlessons@gmail.com and I’ll get back to you with an answer. You can also use the contact form below to reach out and I’ll send you my FREE calculus 1 study guide as a bonus! Also check out my other lessons about integrals!