Trigonometric Integrals

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Think about the integral of d x, over sine x times cos cubed x. No formula in our table fits it, and no substitution jumps out, right? Yet one substitution finishes it in just a few lines. Which one? Keep thinking. By the end of this video, you will spot it at a glance. Today we do trigonometric integrals: powers and products of sine and cos. They come up often in JEE problems. So, let's begin.

0:34One odd power: save one for dt

So, here is today's family. The long S is the integral sign. Inside it, sine to the power m of x, times cos to the power n of x, and then d x, which means we integrate with respect to x. The small m and n written up on sine and cos are the powers. Everything today depends on whether these powers are odd or even, okay? Now suppose m, the power of sine, is odd. Many students say: sine has the odd power, so put t equals sine x. What do you think: t equals sine x, or t equals cos x? It is t equals cos x, the other function! If you picked sine x, careful here, that is the trap many students fall into. Since d t equals minus sine x d x, one sine x from the odd power goes with d x to make d t. That is the whole trick: save one for d t. Right? What is left is an even power of sine, so it comes in squares. Sine squared x equals one minus cos squared x, that is, one minus t squared. Now everything is a polynomial in t. Clear? Now the mirror case: n, the power of cos, is odd. Put t equals sine x, so d t equals cos x d x. Save one cos x for d t, and the rest turns into sine. Got it?

1:57Both powers even, or a negative even sum

Alright, now both powers are even. Which substitution works now: t equals sine x, or t equals cos x? Honestly, neither of them works well here. With both powers even, there is no odd power to save for d t, okay? So we lower the powers instead. Sine squared x equals one minus cos two x, over two, and cos squared x equals one plus cos two x, over two. If a square appears again after you expand, use it once more. Right? Now the most useful trick. When sine or cos sits in the denominator, its power is negative. If m plus n is a negative even integer, put t equals tan x. Our opening puzzle is exactly this case; we solve it in Example two.

2:47Products: turn them into sums

Next, products of two different angles. Here we do not substitute; we turn the product into a sum. Two sine A cos B equals sine of A plus B, plus sine of A minus B, okay? Capital A and capital B are just two angles, and the brackets keep each angle together. Two cos A cos B equals cos of A plus B, plus cos of A minus B, the same pattern. Each piece is then easy to integrate. With me? Now the third one, two sine A sine B, also gives two cos terms. But which one comes first, and with what sign? It is cos of A minus B, minus cos of A plus B, in that order. The minus angle comes first. Students often flip it, and the whole answer changes sign. So write this one down.

3:38Powers of tan, even powers of sec

Now powers of tan. Write tan to the power n of x as tan to the power n minus two of x, times the bracket, sec squared x minus one. Sec x is the secant, one over cos x, okay? We split it this way because d of tan x equals sec squared x d x. So the part with sec squared is a straight substitution, t equals tan x, and the other part is a lower power of tan. Just repeat. Right? And for sec to the power two k, an even power of sec, keep one sec squared x for d t. Change the rest using sec squared x equals one plus tan squared x, and put t equals tan x. Clear?

4:18Example 1: sin cubed x, then sin squared x cos cubed x

Now let's use all this. Example one. Find the integral of sine cubed x, d x, and the integral of sine squared x times cos cubed x, d x. Decide which power is odd, then choose t. Pause the video and try it first. Done? Okay, let's check. In the first one, sine has power three, which is odd. So t equals cos x, and d t equals minus sine x d x. Save one sine x, and write sine squared x as one minus cos squared x. So we get minus the integral of one minus t squared, d t. Watch that minus sign from d t. That gives minus t, plus t cubed over three. Back to x: minus cos x, plus one third cos cubed x, plus capital C, the constant of integration. Never forget it, okay? Now the second one. Sine has power two, and cos has power three. So which one becomes t? Yes, exactly! Cos has the odd power, so t equals sine x, and d t equals cos x d x. The cos squared x left over becomes one minus sine squared x. If you said t equals cos x, that was the trap. Don't fall for it! Multiply out: t squared minus t to the power four. Integrate, and put t back as sine x: one third sine cubed x, minus one fifth sine to the power five x, plus capital C, the constant. Really quick, once t is chosen correctly.

5:55Example 2: 1 over sin x cos cubed x, put t = tan x

Now, our puzzle from the start, as Example two. Find the integral of d x, over sine x times cos cubed x. Hint: find m and n first, then look at their sum. Pause the video and try it first. Done? Okay, let's check. Sine x in the denominator means power minus one, so m equals minus one, and cos cubed x gives n equals minus three. So m plus n is minus four, a negative even integer. That means t equals tan x. To make tan x appear, we divide the top and the bottom by a power of cos x. Which power should it be? Yes, exactly! Cos to the power four, the total power in the denominator, one plus three. The top becomes sec to the power four x, and the bottom becomes sine x over cos x, which is tan x. Clear? Now write sec to the power four x as sec squared x times sec squared x. One sec squared x, with d x, is d t. The other is one plus tan squared x, that is, one plus t squared. Got it? So we get the integral of one plus t squared, over t, d t, which splits into one over t, plus t. One over t gives natural log of mod t. These two straight bars mean the modulus, okay? And t gives t squared over two, then plus capital C, as always. Put t back as tan x: natural log of modulus of tan x, plus one half tan squared x, plus capital C, the constant. And there it is: the puzzle needed just one substitution. That is the power of counting m plus n.

7:33Quick revision: odd power, both even, products, t = tan x

So, quick revision. One odd power: save one for d t, and put t equal to the other function. Both powers even: lower them with the cos two x formulas, and turn products into sums. If m plus n is a negative even integer, put t equals tan x. And always add plus capital C, okay? Now try the practice problems on this page. All the best!