How to Do Integration by Parts | Easiest Method With Example

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How to Do Integration by Parts

Integration is a mathematical technique used in Calculus for integrating a function partially or by parts. You can use integration by parts while dealing with the multiplication of continuously differentiable functions. It can be done for performing both – definite and indefinite integration.

A definite integral usually has an interval by which it is bound i.e., we calculate the integral value for the interval (a, b). Whereas, in an indefinite integral, also known as a primary integral, results in a differentiable function after integration is complete. If you differentiate the indefinite integral, you will get the previous function that you integrated originally.

Hence, indefinite integration is the reverse of differentiation and is termed as anti-differentiation.

Must Learn:

In this article, you will learn to derive the formula for doing integration by parts, the method of integrating, and when to use such a technique.

The Formula for Integration by Parts

It can be derived using the product rule that you might have learned during differentiating functions that are the multiplication of two or more functions.

Just to refresh your memory, the product rule is given as follows:

(d(u*v))/dx=u*dv/dx+ v*du/dx

To get the partial integration formula, we shall integrate the product rule equation as follows:
∫▒(d(u*v))/dx=∫▒〖u*dv/dx〗+ ∫▒〖v*du/dx〗

Taking it a step further, you can guess by looking at the L.H.S. that you will be left with simply (u*v).
(u*v)=∫▒〖u*dv/dx〗+ ∫▒〖v*du/dx〗

Now, we will rearrange the above formula by taking L.H.S. to R.H.S. and taking the first term of R.H.S to L.H.S., and we will have the formula for integration by parts as given below:
∫▒〖u*dv/dx〗=(u*v)- ∫▒〖v*du/dx〗

Above formula is the one that is to be used for performing integration by parts. The formula for definite integration over the interval [a, b] as follows:
∫_a^b▒〖u*dv/dx〗=(u*v)- ∫_a^b▒〖v*du/dx〗

Steps for Integration by Parts

Looking at the formula, you must be thinking that all there is to integration by parts is substituting values in the formula and then your problem will be solved. But integration by parts is rather trickier than it appears. Why?

To provide the reason, you must remember first that integration by parts is a solution to integrate a function which is the product of two functions. Now if you see the formula once more, you will notice that the two functions being multiplied are u and dv/dx.

This means that both of the given functions must be continuously differentiable and you will have to choose which is u and which is dv/dx. If you choose these correctly, then you will be able to integrate without any difficulty.

Follow the steps given below to perform integration by parts:

  1. Given a problem of integration which has two functions in multiplication, select one of them as u and the other one as dv/dx.
  2. Integrate dv/dx and you will have v.
  3. Differentiate u and you will have du/dx.
  4. Now you have all the terms you need to move forward.
  5. You simply have to substitute u, dv/dx, v, and du/dx in the equation for integration by parts viz.
    ∫▒〖u*dv/dx〗=(u*v)- ∫▒〖v*du/dx〗
Integration
Integration

Now just solve the remaining equation and simplify to get the result.

  • For understanding it better, let’s take an example and solve it using the above steps.
  • Let’s say we are required to find ∫▒〖x*sin⁡(x)dx〗.
  • So we will select x as u and sin⁡(x) as dv/dx.
  • Moving on, we know that the integration of sin⁡(x) (which is our dv/dx) is -cos(x)
  • Differentiation of x (which is our u) with respect to x (i.e. dx/dx ) is simply 1.
  • Now we just have to substitute these terms in the equation, and the following is what we’ll get after doing so:
    ∫▒〖x*〗 sin⁡(x)dx=(x* -cos(x))- ∫▒〖-cos⁡(x)*1 dx〗

This can be simplified further as follows:
∫▒〖x*〗 sin⁡(x)dx=-x cos(x) -(-sin⁡(x) )+C
∫▒〖x*〗 sin⁡(x)dx=-x cos(x)+sin⁡(x)+C

Hence, we have the solution for the given problem using integration by parts.

When to Use Integration by Parts

It may not strike immediately by looking at a problem that you should do integration by parts to solve it. What you need to make sure is simply that when you see a function that is the multiplication of two other functions, then check that one of those functions must be continuously differentiable and the other function must be integrable.

See:

If you find that it is true, then you can proceed with using partial integration. Sometimes, people use this method to solve problems that don’t generally seem to be a product of two functions.

The most common example of this is the partial integration of ln⁡(x). It is supposed that the function is 1*ln⁡(x), and then it is easily solvable.

When does Integration by Parts Fail

You must know that integration by parts has its own limitations, and you cannot use it to solve every integral that comes your way. There are cases when applying integration by parts leads to an infinite loop. We can observe it in the subsequent example:
∫▒〖e^x*cos⁡(x)dx〗= e^x*cos⁡(x)- ∫▒〖e^x*-sin⁡(x)dx〗
= e^x*cos⁡(x)+ ∫▒〖e^x*sin⁡(x)dx〗

At this point, you may be capable of noticing that you need to apply integration by parts again for ∫▒〖e^x*sin⁡(x)dx〗. If you do so, what you’ll get is:
= e^x*cos⁡(x)+e^x*sin⁡(x)- ∫▒〖e^x*cos⁡(x)dx〗

As you can see, we reached to ∫▒〖e^x*cos⁡(x)dx〗 again, which was our original problem. So it will continue infinitely, and you will not reach a solution using integration by parts in this case.

Simple Trick to Select u and dv/dx

There is a mnemonic that can be used to decide which function to select as u from the given problem – LIATE. Here, L stands for Logarithmic function, I stands for Inverse Trigonometric function, A stands for Algebraic function, T stands for Trigonometric function, and E stands for Exponential function. LIATE provides you with a priority for selecting u.

So, whenever you have a Logarithmic function, select it as u and select the rest of the term as dv/dx; second comes Inverse Trigonometric function and so on. Hence, start looking for LIATE one by one, and when you find a term that matches, you have found your u.

Conclusion

By now, you must have a good idea of how to do integration by parts. The procedure is not that difficult to understand, but the trick is in choosing the right u and dv/dx. If you make a mistake in this step, the chances are that you will have to restart solving your problem by changing your selection.

But after frequent practice, you’ll be able to see the next few steps in your head before writing them on paper, and then you will find this step to be quite straightforward.

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