Integration by Parts

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Take the integral of e to the power x, times sine x, d x. Integrate it by parts, then by parts again, and out comes the very same integral we started with! So are we going round in a circle forever? Hold on to that question; that circle is exactly what solves it. Today: integration by parts, for integrating a product of two functions. It comes up often in JEE problems. So, let's begin.

0:34The integration by parts formula

So, here is the formula. The long S is the integral sign, and d x means with respect to x. u and v are two functions of x, multiplied together, and u prime, with the little dash, is the derivative of u, okay? Read it with me. The integral of u times v, d x, equals u times the integral of v d x, minus the integral of, u prime times the integral of v d x, in round brackets, then d x. Got it? In plain words: u gets differentiated, and v gets integrated. Keep u, times the integral of v, then subtract the integral of u prime times that same integral of v. With me? Since these are indefinite integrals, add plus capital C, the constant of integration, once, at the very end, okay? Now here is the real question. In a product of two functions, which one should be u? Does it even matter?

1:34ILATE: which function should be u?

Yes, it matters a lot! Choose the wrong u, and the new integral can get harder. So we choose u in a fixed order, called ILATE, okay? Read the five letters with me. Inverse trigonometric, logarithmic, algebraic, like x or x squared, trigonometric, and finally exponential. Whichever function comes first in this list, call it u. The other one is v, the one we integrate. Clear? Quick test. Take x times e to the power x. Which one is u? Yes, exactly, u equals x! Algebraic comes before exponential, so v is e to the power x. See why it works: x differentiates to just one, and e to the power x integrates to itself. Simpler, right?

2:27Only one function? Take v=1

Now, a tricky case. Look at these three: the integral of log x, d x, where l n on screen means the natural log; then tan inverse x, and sine inverse x. The minus one written up top means inverse, not one over, okay? Each has only one function. So where is the product? What is v here? Simple trick: take v equal to one! Write log x as log x times one. Log x is u, since we know its derivative, one over x, and one is easy to integrate. Same idea for tan inverse x and sine inverse x. Clever, isn't it?

3:10Two results worth remembering

Now two results that save time. First: e to the power x, times a square bracket, f of x plus f prime of x, where f prime is the derivative of f. So the bracket is a function plus its own derivative. Then the integral is just e to the power x times f of x, plus capital C, okay? Second: e to the power a x, times sine b x, with a and b constants. The integral is e to the power a x, times the bracket, a sine b x minus b cos b x, all over a squared plus b squared, plus capital C, the constant. Clear? And if it is cos b x instead of sine b x? What do you think changes? The bracket becomes a cos b x plus b sine b x. Minus with sine, plus with cos: watch that sign. In Example two, we will check the sine result. Got it?

4:12Example 1: integrals of ln x and x ln x by parts

Now let's use all this. Example one. Find the integral of log x, d x, and the integral of x times log x, d x. Hint: remember the v equals one trick. Pause the video and try it first. Done? Okay, let's check. Log x is alone, so u is log x and v is one. Then u times the integral of v gives log x times x, minus the integral of u prime, one over x, times x, d x. Okay? Now one over x, times x, is just one, and its integral is x. So the answer is x log x, minus x, plus capital C, the constant. Really quick! Now the second one, x times log x. Many students take u equals x, since x is so easy to differentiate. What do you say: u equals x, or u equals log x? No, no, careful here! With u equals x, v would be log x, and we would first have to integrate log x itself, the longer way round. Logarithmic comes before algebraic, so u is log x and v is x. Clear? The integral of v is x squared over two. So: log x times x squared over two, minus the integral of one over x, times x squared over two, d x. With me? One over x, times x squared over two, is x over two, which integrates to x squared over four. Answer: x squared over two, times log x, minus x squared over four, plus capital C, as always.

5:56Example 2: e to the x sin x, the integral that comes back

Now, our puzzle from the start, as Example two. Find the integral of e to the power x, times sine x, d x. Hint: call it capital I, and go by parts twice. Pause the video and try it first. Done? Okay, let's check. We call the integral capital I, okay? Trigonometric comes before exponential, so u is sine x, and v is e to the power x. Right? Now e to the power x integrates to itself. So capital I equals e to the power x sine x, minus the integral of e to the power x cos x, d x. That cos x is u prime, okay? We need by parts once more. This time, which one is u: cos x, or e to the power x? Same order as before, so u is cos x. Its derivative is minus sine x, and minus times minus gives plus. So we get e to the power x cos x, plus the integral of e to the power x sine x. Look, that last integral is capital I, back again! Switching to u equals e to the power x would only undo the first step. Got it? So here is the circle from our puzzle. Many students stop right here. Are we really stuck? Not at all, and this is the beautiful part. Put it back into the first line, minding the minus sign on the whole bracket. Capital I equals e to the power x sine x, minus e to the power x cos x, minus capital I, okay? Now treat it like an equation. Move capital I to the left: two capital I equals e to the power x, times sine x minus cos x. Divide by two. So capital I is one half e to the power x, times sine x minus cos x, plus capital C, the constant. The circle itself gave us the answer! Quick check with our formula: a equals one, and b equals one. We get e to the power x, times sine x minus cos x, over one plus one, that is two. Same answer. Perfect!

8:14Quick revision: ILATE, v = 1, e to the x rule, solve for I

So, quick revision. Differentiate u, integrate v, and choose u by ILATE, okay? Log x or inverse trig alone: take v equal to one. And e to the power x, times f plus f prime, gives e to the power x times f. If the integral comes back, solve for it like an equation. And always add plus capital C, the constant. Now try the practice problems on this page. All the best!