Antiderivatives and the Standard Integrals

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle. I differentiated a function and got two x as the answer. Which function did I start with? x squared, right? But x squared plus five also gives two x, and so does x squared minus one hundred. So which one was it? You cannot tell! Today we run differentiation backwards. That is integration, and the table we build today is used all through integral calculus for JEE, okay? So, let's begin.

0:32Going backwards: the antiderivative

So, first the new word. F prime of x, the capital F with a little dash, is the derivative of capital F, okay? If F prime of x equals small f of x on an interval, then capital F is an antiderivative, or primitive, of small f on that interval. Clear? Back to our puzzle. d over d x means the derivative. The derivative of x squared is two x, and of x squared plus five, also two x, since a constant differentiates to zero. So, how many antiderivatives does two x have? Infinitely many, yes! And the course tells us more: on an interval, any two antiderivatives differ only by a constant. So find one, and you have them all. With me? So we write it like this. The tall stretched S is the integral sign, and d x says that x is the variable. The integral of f of x, d x, equals capital F of x, plus capital C, the constant of integration. Forgetting plus capital C is the classic mistake, so never drop it, okay?

1:38Powers, logs and exponentials

Now the standard integrals, each a derivative read backwards. Powers first: raise the power by one, and divide by the new power. The integral of x to the power n, d x, is x to the power n plus one, over n plus one, plus capital C, right? But look at the condition: n not equal to minus one. The slashed equals sign means, not equal to. What goes wrong when n is minus one? Yes, exactly! Then n plus one is zero, and we would divide by zero. So d x over x has its own formula: the natural log of mod x, plus capital C, okay? On screen, the letters l n mean natural log, and the two straight bars are the modulus, read as mod x. Right? Next, e to the power x, where e is the base of natural logs. It is its own derivative, so also its own integral. Similarly, for a positive constant base small a, other than one, small a to the power x integrates to small a to the power x, over the natural log of small a, plus capital C, of course. From here on, every answer on screen ends in plus capital C, even if I skip saying it. Got it?

2:52Trigonometric integrals: derivatives backwards

Now trigonometry. Quick question: is the integral of sine x equal to cos x? No, no, careful here! The derivative of cos x is minus sine x, so sine x integrates to minus cos x. And cos x integrates to sine x, with no minus sign. Right? Next, tan x differentiates to sec squared x, so sec squared x integrates to tan x. The c s c you see is cosec, okay? Cosec squared x integrates to minus cot x. With me? Then sec x tan x gives sec x, and cosec x cot x gives minus cosec x. Here is a small trick: in these six, every answer that is a co function carries a minus sign: minus cos, minus cot, minus cosec. Easy to remember!

3:41Four integrals that give logarithms

Now four integrals that give logarithms. Tan x integrates to the natural log of the modulus of sec x, and cot x to the natural log of the modulus of sine x. Okay? The integral of sec x is the natural log of the modulus of, sec x plus tan x. Learn this one by heart; it comes up often. Clear? And cosec x integrates to the natural log of the modulus of, cosec x minus cot x. The course also writes this as the natural log of the modulus of tan of x over two; both forms are equal. So do not panic if an option shows the tan form!

4:19Inverse trigonometric forms

Now the inverse trigonometric forms. The long tick is the square root sign. d x over the square root of one minus x squared integrates to sine inverse x. The little minus one means inverse, not one over sine. Right? Then, d x over one plus x squared gives tan inverse x; we need it in Example two. And d x over x times the square root of x squared minus one gives sec inverse x. Derivatives you know, read backwards. Got it?

4:50Two rules that do the rest

Now two rules. First, linearity: for constants a and b, the integral of, a f plus b g, is a times the integral of f, plus b times the integral of g. So, integrate term by term. Okay? Second, the linear change. If the integral of f of x, d x, is capital F of x plus capital C, then f of, a x plus b, integrates to one over a, times capital F of, a x plus b, plus capital C again. The one over a cancels the factor a from the chain rule. Clear? Let's try one from the course examples: the integral of one half times sec squared of x over two, d x. What is it? Here the constant a is one half, so one over a is two. The dot means times: one half, times two, times tan of x over two, which is just tan of x over two, plus capital C, okay?

5:42Example 1: (root x + 1/root x) squared, expand and integrate

Alright, let's use the table. Example one. Find the integral of, root x plus one over root x, the whole bracket squared, d x. Pause the video, and try it first. Done? Okay, let's check. No formula fits a bracket squared, so expand it. Root x, squared, is x; twice root x times one over root x is two; and one over root x, squared, is one over x. So we get x plus two plus one over x. Clear? Now term by term. The power rule handles x and the constant two easily. But one over x? Can we use the power rule there? No, we cannot! One over x is x to the power minus one, exactly the case the power rule cannot handle, so it gives the natural log of the modulus of x. And x gives x squared over two, two gives two x. So the answer is x squared over two, plus two x, plus the natural log of the modulus of x, plus capital C, okay?

6:45Example 2: (x4 + 1)/(x2 + 1), divide to get tan inverse x

Now a slightly harder one. Example two. Find the integral of, x to the power four plus one, over x squared plus one, d x. Pause the video, and try it first. Okay, look at the fraction. The top has degree four, and the bottom has degree two. So what do we do first? Divide, yes, just like with numbers! x to the power four plus one equals, x squared plus one, times, x squared minus one, plus a remainder of two. Check: that product is x to the power four minus one, and adding two gives it back. Perfect! Now divide both sides by x squared plus one. The integrand becomes x squared minus one, plus two over, x squared plus one. Okay? Look at the last term: two over, x squared plus one. Which standard integral is this? Tan inverse, correct! Term by term, the answer is x cubed over three, minus x, plus two tan inverse x, plus capital C, okay? And the trap? Forgetting the remainder two, which loses the whole tan inverse part. With me?

7:59Quick revision: plus C, power rule n not -1, rewrite first

So, quick revision. Integration reverses differentiation, and every indefinite integral gets plus capital C, okay? The power rule works for every n except minus one; one over x gives the natural log of the modulus of x. And if the integrand is not in the table, rewrite it first: expand, or divide, then integrate. Now go and try the practice problems on this page. All the best!