Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take the integral from zero to pi of x times cos squared x, d x. That x in front is a real nuisance, right? Integration by parts with cos squared gets long. So, can we get rid of that x without ever integrating it? Sounds impossible, doesn't it? Today, you will do it in a few lines. This is Set two of our mixed problems, and questions like these come up often in JEE Main. So, let's begin.

0:34Set 2: what is inside

So, Set two holds nine solved problems, all in the JEE Main pattern. Okay? They mix the whole unit: substitution, integration by parts, partial fractions, the King property, Leibniz's rule, and more. Right? Today: two tools, then two problems from this set, the easier one first. With me?

0:55Tool 1: the King property

Tool one, from Chapter four: the King property. The long S is the integral sign, a and b are the limits, and d x means we integrate with respect to x. The integral from a to b of f of x, d x, equals the integral from a to b of f of a plus b minus x, d x. Replace x by a plus b minus x, and the value does not change. Okay? Now, a quick question. If the limits are zero and pi, the Greek letter, what does x become? Pi minus x, yes, exactly! Here a is zero and b is pi, so a plus b minus x is pi minus x. Right? Now, why is this useful? Call the integral capital I, with an awkward integrand, capital F of x. Write I both ways and add. Two I equals the integral from a to b of, in square brackets, F of x plus F of a plus b minus x, d x. If that bracket is simple, we are done. Clear? And a standard result from the course, built this way. The integral from zero to pi of x times f of sine x, d x, equals pi over two, times the integral from zero to pi of f of sine x, d x, for any function f of sine x. The x in front just disappears! But only for the limits zero to pi. Got it?

2:16Tool 2: t = tan x/2, then change the limits

Tool two, from Chapter two. For the integral of one over, a plus b sine x, d x, put t equals tan of x over two. Okay? Then sine x becomes two t over one plus t squared, and cos x becomes one minus t squared, over one plus t squared. Right? And d x? Differentiate t equals tan of x over two. What is d x in terms of d t? Very good! d t equals one half, sec squared of x over two, d x, and sec squared is one plus t squared. The double arrow means, so it follows: d x equals two d t, over one plus t squared. Clear? Now, the trap. In a definite integral, change the limits along with the variable. Once you are in t, finish in t. Never go back to x. Don't put the old x limits into a t answer! With me? And if the new denominator splits into distinct linear factors, use partial fractions, with the cover-up rule. Got it?

3:21Example 1: x cos squared x from 0 to pi by King property

Now, remember our puzzle from the start? Let's finish it. Evaluate the integral from zero to pi of x cos squared x, d x. Cos squared x means cos x, the whole squared. Hint: use tool one. Pause the video, and try it first. Done? Okay, let's check. Call the integral capital I, okay? The limits are zero and pi, so we replace x by pi minus x. Now, cos of pi minus x is minus cos x. So the integrand picks up a minus sign, right? No, no, careful here! The cos is squared. Minus cos x, the whole squared, is just cos squared x. So it does not change at all. Got it? So only the x in front changes, to pi minus x. I equals the integral from zero to pi of, pi minus x, times cos squared x, d x. Clear? Now we have I in two forms. What do we do with them? Add them, exactly! x plus pi minus x is just pi, a constant. So two I equals the integral from zero to pi of pi cos squared x, d x. Take pi outside and halve it: I equals pi over two, times the integral from zero to pi of cos squared x. Look, the x is gone! With me? Cos squared x is one plus cos two x, over two. The half gives pi over two, and cos two x runs through a full cycle from zero to pi, so it gives zero. So this integral is pi over two. Okay? So I equals pi over two, times pi over two; the dot means times. The answer is pi squared over four. And the x in front vanished without any integration by parts, just as we promised at the start. Perfect!

5:13Example 2: 1/(4 + 5 sin x) from 0 to pi/2 is (1/3) ln 2

Now a harder one. Evaluate the integral from zero to pi over two of one over, four plus five sine x, d x. Hint: use tool two, and watch the limits. Pause the video, and try it first. Done? Okay, let's check. Put t equals tan of x over two, and first, the limits. When x is zero, and when x is pi over two, what is t? When x is zero, t is tan zero, which is zero. When x is pi over two, t is tan of pi over four, which is one. Okay? Now the denominator. Five sine x becomes ten t, over one plus t squared. Over a common denominator: four t squared plus ten t plus four, over one plus t squared. Right? Now bring in d x, two d t over one plus t squared. The one plus t squared cancels completely! So capital I is the integral from zero to one of two d t over, four t squared plus ten t plus four. Take two common, and this is one over, two t plus one, times t plus two. Clear? Now partial fractions. The numerator is a constant, so the fraction is proper, and no division is needed. Cover up two t plus one, and put t equals minus one half in the rest: two thirds. Cover up t plus two, and put t equals minus two: minus one third. With me? Now integrate. On screen, l n means the natural log. So this gives two thirds natural log of two t plus one, right? Oh no, careful with that one! The coefficient of t is two, so divide by two: one third natural log of two t plus one, minus one third natural log of t plus two. Both are positive for t from zero to one, so no modulus is needed. Combine the logs. The square brackets with zero and one mean: value at one, minus value at zero. Got it? At t equals one: natural log of three over three, which is zero. At t equals zero: natural log of one half. Zero minus natural log of one half is natural log two. So the answer is one third natural log two. Perfect!

7:32Quick revision: King property, tan x/2 limits, divide by 2

So, quick revision. One: the King property. Replace x by a plus b minus x, then add the two forms. Two: for one over, a plus b sine x, put t equals tan of x over two, and change the limits at once. Three: when you integrate one over two t plus one, divide by two, the coefficient of t. Now go and try the practice problems on this page. All the best!