Rational Functions of sin x and cos x; Irrational Integrands

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Take the integral of one over, root x plus cube root x, d x. Put t equal to root x, and the cube root stays. Put t equal to cube root x, and the square root stays. Is there one substitution that removes both roots together? Hold that thought; Example two answers it. Today: fractions of sine and cos, and integrands with roots. They come up often in JEE, so let's begin.

0:33The half-angle substitution t = tan x/2

So, the first family. The long S is the integral sign, and d x means with respect to x. Look at these three: d x over, a plus b cos x; d x over, a plus b sine x; and d x over, a sine x plus b cos x plus c, with a, b and c constants, okay? Fractions like these are called rational functions of sine x and cos x. For all three, put t equals tan of x over two, the half angle. Then sine x becomes two t, over one plus t squared, and cos x becomes one minus t squared, over one plus t squared. With me? But what about d x? If t is tan of x over two, what does d x become? Differentiate: d t equals one half, sec squared of x over two, d x. Sec squared is one plus tan squared, so that is one half of, one plus t squared, d x. So d x equals two d t, over one plus t squared. Clear? See the beauty: sine, cosine and d x all become fractions in t. No trigonometry left, just algebra. Right?

1:51Three more trigonometric forms

Now, what if there is no constant at the bottom, only a sine x plus b cos x? Write it as r times sine of, x plus alpha, where alpha, the Greek letter, is a fixed angle, and r is a constant. Then one over a single sine is an integral we already know. Okay? Next, sine squared and cos squared at the bottom. Divide top and bottom by cos squared x, and put t equals tan x. Then d t is sec squared x, d x, which is exactly what the top becomes. So everything turns into t. Got it? Now sine and cos sit on the top as well. Tan of x over two works, but it gets really messy. Is there a smarter way? Here is the trick. Write the numerator as capital A times the denominator, plus capital B times the derivative of the denominator. Den is short for denominator, and the little dash after the bracket means derivative, okay? So the integral is capital A x, from the capital A part. The capital B part is a derivative over its function, giving capital B times natural log of the modulus of the denominator; the two straight bars mean the modulus. Then plus capital C, the constant. Compare the sine and cos terms to find capital A and capital B, okay? And one trap here. The derivative of the denominator is a cos x minus b sine x, because the derivative of cos x is minus sine x. Many students lose that minus sign, and then capital B comes out wrong. Right?

3:31Integrands with roots

Now, integrands with roots. First, d x over, p x plus q, times the square root of, a x plus b. Put t squared equals a x plus b. Why t squared, and not just t? Yes, exactly! Then the root is simply t, and differentiating, two t d t equals a d x. No root left anywhere. Clear? Next, the square root of, a minus x over a plus x. Rationalise: multiply above and below by root of a minus x. We get a minus x, over root of, a squared minus x squared. Split it into two standard forms. Okay? And finally, mixed roots: x to the power one over m, with x to the power one over n. Put x equals t to the power l c m of m and n, the least common multiple. Then every root becomes a whole power of t. That is the key to our puzzle! With me?

4:32Example 1: 1 over 3 + 5 cos x with t = tan x/2

Example one. Find the integral of d x over, three plus five cos x. Pause the video and try it first. Done? Okay, let's check. Put t equals tan of x over two. The bottom is three, plus five times, one minus t squared, over one plus t squared. That is eight minus two t squared, over one plus t squared. With me? Flip it, and multiply by d x, which is two d t over, one plus t squared. The one plus t squared cancels! We are left with d t over, four minus t squared. Right? Now, many students see t squared at the bottom and write tan inverse at once. Is it tan inverse here, or a log? Look at the sign: four minus t squared. There is a minus, so it is a log, not tan inverse! The standard form: d t over, a squared minus t squared, gives one over two a, times natural log of the modulus of, a plus t over a minus t. Here a is two, okay? So we get one quarter, natural log of the modulus of, two plus t over two minus t. Put back t equals tan of x over two, and add plus capital C, the constant. Trigonometry turned into simple algebra!

5:56Example 2: 1 over root x + cube root x, put x = t to the 6

Now, our puzzle from the start, as Example two. Find the integral of d x over, root x plus cube root x; the small three on the root sign means cube root. Pause the video and try it first. Done? Okay, let's check. Root x is x to the power one half, and cube root x is x to the power one third. Which power of t clears both? Yes, six, the l c m of two and three! Put x equals t to the power six, so d x is six t to the power five, d t. Root x is t cubed, and cube root x is t squared. Both roots gone in one go: puzzle solved! Got it? So we get six t to the power five, d t, over t cubed plus t squared. Cancel t squared from the bottom: six times the integral of t cubed over, t plus one, d t. Okay? Can we split t cubed over, t plus one, into partial fractions straight away? No, no, careful here! The top has degree three and the bottom degree one, so the fraction is improper. Partial fractions need a proper fraction, so divide first. We get t squared minus t plus one, minus one over, t plus one. Clear? Integrate term by term: t cubed over three, minus t squared over two, plus t, minus natural log of, one plus t, all times six. No modulus needed: t is the sixth root of x, never negative, so one plus t is positive. Right? Finally, put t back; the small six on the root sign means the sixth root. So: two root x, minus three cube root x, plus six times the sixth root of x, minus six natural log of, one plus the sixth root of x, plus capital C, the constant. One clever power of t, and both roots disappeared!

8:00Quick revision: tan x/2, tan x, A den + B den', lcm power

So, quick revision. With a plus b cos x, a plus b sine x, or a sine x plus b cos x plus c at the bottom: put t equals tan of x over two. Squares at the bottom: put t equals tan x. Trig on top as well: capital A times the denominator, plus capital B times its derivative. Roots: t squared equals a x plus b; mixed roots: x equals t to the power l c m. And always add plus capital C, okay? Now go and try the practice problems on this page. All the best!