The King Property

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take sine x to the power one hundred, over sine x to the power one hundred plus cos x to the power one hundred, and integrate it from zero to pi over two. That looks scary, right? Yet the answer is simply pi over four, for any power at all. The trick behind it is the King property, and it comes up often in JEE, so let's begin.

0:27The King property: replace x by a+b-x

Here it is: the King property. The long S is the integral sign, a and b are its limits, and d x says we integrate in x. The integral from a to b of f of x equals the integral from a to b of f of a plus b minus x. And a and b must be the limits of this same integral, okay? But why should that be true? Think about it. As x moves from a to b, where does a plus b minus x move? From b back to a, exactly! The same interval, walked in reverse. To prove it, put t equals a plus b minus x, so d x equals minus d t. The double arrow means gives: x equals a gives t equals b, and x equals b gives t equals a. The limits flip, the minus sign flips them back, and we rename t as x. With me? When the lower limit is zero and the upper limit is a, we simply replace x by a minus x. This is the form we use most often. Clear?

1:33The 2I trick: add the two forms

Now, how do we use it? Call the integral capital I, and its integrand capital F of x. So capital I has two forms: one with F of x, and one with F of a plus b minus x. Okay? Add the two forms. Two I equals the integral from a to b of F of x plus F of a plus b minus x, d x. The square brackets just group the sum. This helps when that sum is something simple. Right? Here is the most useful case: capital F of x equals f of x, over f of x plus f of a plus b minus x. Replace x by a plus b minus x. The top becomes f of a plus b minus x, and the bottom just swaps its two terms. So what do the two fractions add up to? One, exactly! Same bottom, and the two tops add up to it. So two I is the integral of one from a to b, which is b minus a, and capital I is b minus a, over two. Beautiful, isn't it? From the solved examples on this page: the integral from two to eight of root x, over root x plus root of ten minus x. The tick sign is the square root. Here a plus b is ten, so the answer is eight minus two, over two, that is three, with no integration at all. Got it?

2:57Two standard results

Next, two standard results. First: the integral from zero to pi over two of natural log of sine x equals minus pi over two, times natural log of two. Pi is the Greek letter, and l n is the natural log. Its proof is in the solved examples on this page, okay? Second: the integral from zero to pi of x times f of sine x equals pi over two, times the integral from zero to pi of f of sine x. The x in front disappears, and pi over two comes out. Clear? But why does it work? The limits are zero and pi, so a plus b minus x is pi minus x. Tell me: what is sine of pi minus x? Sine x itself, correct! The arrow on screen means replace x by pi minus x. So f of sine x stays the same, and only the x in front changes. Okay? Now add the two forms: x plus pi minus x is just pi, a constant, which comes outside. Divide by two, and there is our result. With me?

4:07Example 1: root sin x/(root sin x + root cos x) is pi/4

Now, example one. Evaluate the integral from zero to pi over two of root sine x, over root sine x plus root cos x, d x. The graph on the right is a hint. Pause the video and try it first. Done? Okay, let's check. With limits zero and pi over two, we replace x by pi over two minus x. Sine x becomes cos x, and cos x becomes sine x. They simply swap places, as the two-headed arrow shows. Right? So capital I also equals the integral of root cos x, over root cos x plus root sine x, with the same bottom. Add this to the original form. What is the new integrand? Just one, yes! The two tops add up to the bottom. So two I is the integral from zero to pi over two of one, d x, which is pi over two. Okay? So capital I equals pi over four. On the graph, the solid orange curve is f of x, and the dashed blue curve is f of pi over two minus x. They add up to one everywhere, so the shaded area under the orange curve is exactly half of the rectangle of height one and width pi over two. Lovely, isn't it? Notice, the square root played no special part. For any power n, sine to the power n x, over sine to the power n x plus cos to the power n x, still gives pi over four. So power one hundred also gives pi over four, and that solves our puzzle!

5:44Example 2: x sin x/(1 + cos squared x) is pi squared/4

Now a harder one. Evaluate the integral from zero to pi of x sine x, over one plus cos squared x, d x. Hint: an x in front, and limits zero and pi. Pause the video and try it first. Done? Okay, let's check. Replace x by pi minus x. Sine of pi minus x is sine x, no change. But cos of pi minus x is minus cos x. So does the bottom, one plus cos squared x, change? No, it does not! Minus cos x, squared, is just cos squared x. So the bottom stays the same. Right? So only the x in front changes, to pi minus x, and capital I takes this new form. With me? Add this to the original form: x plus pi minus x is pi, so it comes outside. Two I equals pi times the integral from zero to pi of sine x, over one plus cos squared x, d x. That is our x times f of sine x result in action. Okay? The top, sine x, is minus the derivative of cos x, so the antiderivative is minus tan inverse of cos x. Tan to the power minus one means tan inverse, not one over tan. The square brackets with zero and pi mean upper limit minus lower limit. At pi we get pi over four, and at zero, minus pi over four. Subtracting gives pi over two, and pi times pi over two is pi squared over two. So the answer is pi squared over two, right? No, no, careful here! That was two I, not capital I itself. Forgetting to divide by two is a very common slip. Divide by two, and capital I equals pi squared over four. That's it!

7:38Quick revision: x to a+b-x, add for 2I, then divide by 2

So, quick revision. Replace x by a plus b minus x: the integral does not change. If F of x plus F of a plus b minus x is simple, add the two forms to get two I, okay? Then divide by two: the answer is capital I, not two I, right? Now go and try the practice problems on this page. All the best!