Integration techniques | Integration by parts

5 Oct 2019 2 Comments

U-substitution is a very useful technique for integration. You can apply it in every case, but it doesn’t always make things easier. But here comes this new technique in help! Integration by parts is used to calculate the integral of products. Well, everything is actually a product, (x² for example is x*x but also x²*1) so you can apply it every time you see that u-sub is not paying off.

This is how you calculate the integral of the product of two functions:

$\int f(x)g(x)dx=f(x)\int g(x)dx-\int \left ( \frac{\mathrm{d} }{\mathrm{d} x}f(x)\int g(x)dx \right )dx=$ $f(x)G(x)-\int f'(x)G(x)dx$

to make things look even more simple, we will let u = f(x) and dv = g(x)dx. This means that du = f(x)dx (if you take the derivative of u you get du and for f(x) you get f(x)dx) and v = g(x) (because if you take the integral of dv you get v and the integral of g(x)dx is g(x) ). Therefore, we can write

$\int f(x)g(x)dx=uv-\int udv$

Example:

$\int x\sin(x)dx$

U-sub won’t help here. We can use integration by parts though. With this method you can choose what to differentiate and what to integrate. If we differentiate the x we get 1, which is great. But if we integrate it, we add another x (we get x²), which we don’t want. So,

$u = x\; \; \; \; \; \; \; \; dv = \sin(x)dx$ $du = dx\; \; \; \; \, v = -\cos(x)$

I suggest you write everything like this every time you perform this technique, so you can imagine that u and v are crossed and go together (uv) and that du and v are aligned (vdu). We get:

$-x\cos(x)-\int -\cos(x)dx=-x\cos(x)+\int \cos(x)dx=$ $-x\cos(x)+\sin(x)+C$

Another one:

$\int \ln(x)dx$

Whooa! Where’s the product? Well, it is ln(x)*1. And obviously, we are going to differentiate ln(x) and integrate 1, right? It would make no sense to do the other way around, since the integral of ln(x) is what we are trying to solve. If you don’t remember the fundamental integrals and derivatives and their rules, click these links:

https://addjustabitofpi.wordpress.com/2019/10/01/fundamental-derivatives/

https://addjustabitofpi.wordpress.com/2019/10/03/fundamental-integrals/

https://addjustabitofpi.wordpress.com/2019/09/30/the-rules-of-differentiation/

https://addjustabitofpi.wordpress.com/2019/10/03/rules-and-properties-of-integrals/

$u=\ln(x)\; \; \; \; \; \; \; dv=dx$ $du=\frac{1}{x}\; \; \; \; \; \; \; \; \; v=x$ $x\ln(x)-\int x\frac{1}{x}dx=x\ln(x)-\int dx=x\ln(x)-x = x\left ( \ln(x) +1\right )+C$

Sometimes you may need to perform an u-substitution first and then integration by parts, or the other way around:

$\int 4xe^{4x}dx$

We first let u = 4x, so du = 4 dx and dx = du/4;

$u=4x\; \; \; \; du=4\, dx\rightarrow dx=\frac{du}{4}$ $\int ue^u\frac{du}{4}=\frac{1}{4}\int ue^udu$

At this point we can perform integration by parts; since we have already used u, we will change letter. Usually w, t, z, q are used.

$t=u\; \; \; \; \; \; \; dv=e^udu$ $dt=du\; \; \; \; \; \; \; v=e^u$ $\frac{1}{4}\left ( ue^u-\int e^udu \right )=\frac{1}{4}\left ( ue^u-e^u \right )=\frac{1}{4}e^u(u-1)=\frac{1}{4}e^{4x}(4x-1)+C$

Like u-substitution, you can use this technique as many times as you want (well, as many times as needed is better).

$\int \ln^2(x)dx$ $u=\ln^2(x)\; \; \; \; \; dv=dx$ $du=\frac{2\ln(x)}{x}dx\; \; \; \; \; v=x$ $x\ln^2(x)-2\int \ln(x)dx$

You know the integral of ln(x) because we solved it before, but if we hadn’t already, you would have needed to perform another integration by parts to solve it. The result is

$x\left ( \ln^2(x)-2\ln(x)+2 \right )+C$

Now test your ability! Give these integrals a try and leave the answer in the comments! If you need help, I’ll be happy to clear things up.

$\int x^2\cos(\varphi x+\pi^2)dx$ $\int \sin(\ln(x))dx$

Congratulations! You can now solve most integrals! If you have any doubt or questions leave a comment, I will be happy to help!

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Integration, Integration techniques Calculus, Indefinite integrals, integration by parts

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Integration techniques | Integration by parts

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