Harder Indefinite Integrals

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Take e to the power x, times one minus x whole squared, over one plus x squared whole squared, and integrate it. Integration by parts? Partial fractions? What if the full answer fits in one short line? Hold that thought; Example one shows it. Today: harder indefinite integrals. Questions like these come up often in JEE, and the key is not a new formula, but a hidden pattern. So, let's begin.

0:33The secret: make f and f' appear

So, the secret of harder integrals: make a function and its derivative appear. Three results we already know. First: the integral of f dash of x over f of x, d x, equals natural log of the modulus of f of x, plus capital C, okay? The long S is the integral sign, the little dash on f means its derivative, the two straight bars mean the modulus, and capital C is the constant of integration. Clear? Second: the integral of f dash of x, times f of x in square brackets, to the power n, d x. It equals f of x to the power n plus one, over n plus one, plus capital C, the constant. This needs n not equal to minus one; for minus one, use the log result above. Got it? Third, from integration by parts: the integral of e to the power x, times f of x plus f dash of x, in square brackets, d x. It equals e to the power x times f of x, plus capital C, as before. Right? But why is the third one true? Hint: differentiate the right side. Yes, exactly, the product rule! d over d x, the derivative with respect to x, of e to the power x times f of x, equals e to the power x f of x, plus e to the power x f dash of x. That is exactly the integrand. And keep this habit: to check any integral, differentiate the answer. Okay? So, the plan for every hard integral: rewrite it until f and f dash appear. With me?

2:17Four ways to make the derivative appear

Now, how do we make the derivative appear? Four tricks, all from the solved examples on this page. Trick one: split the top. x to the power four plus one, over x to the power six plus one, breaks into one over x squared plus one, plus x squared over x to the power six plus one. In the second piece, x squared is one third of the derivative of x cubed. Okay? Trick two: take the highest power out of a bracket. Under this square root, take x to the power four out; it comes out of the root as x squared. Inside we get two, minus two x to the power minus two, plus x to the power minus four. Now differentiate the new inside, and compare it with what is left outside. Clear? Trick three: multiply and divide by a constant, such as sine of a minus b. Multiplying and dividing by the same non-zero constant changes nothing, but it can build the shape we need. Example two uses it. Right? And trick four: spot a hidden derivative. Look at x sine x plus cos x. Quickly, what is its derivative? Product rule on x sine x gives sine x, plus x cos x. Then cos x gives minus sine x. The sines cancel, and we get just x cos x! So when x cos x shows up next to this block, that is your hint: the derivative of the block is hiding in the question. Got it?

3:50Example 1: e^x (1 - x)^2 over (1 + x^2)^2 is e^x (f + f')

Now, remember our puzzle from the start? Here it is as Example one. Find the integral of e to the power x, times one minus x whole squared, over one plus x squared whole squared, d x. Pause the video and try it first. Done? Okay, let's check. See the e to the power x sitting outside the fraction. Which of our three results does that remind you of? Yes, exactly, the pattern e to the power x, times f plus f dash! So we must break the fraction into some f, plus its derivative. Expand the top: one minus x whole squared is one minus two x plus x squared, which we group as one plus x squared, minus two x. Okay? Now divide by one plus x squared, whole squared. The first part cancels once, leaving one over one plus x squared. The second part is minus two x, over one plus x squared whole squared. With me? Now, many students see this and jump into partial fractions. Oh no, careful, that is the long road! Instead, look at the two pieces. Is the second one the derivative of the first? Yes! Take f equal to one over one plus x squared. Its derivative is minus two x, over one plus x squared whole squared; even the minus sign matches. So the fraction is exactly f plus f dash. Right? So the answer is e to the power x, over one plus x squared, plus capital C, the constant. One short line, as promised: puzzle solved! Differentiate it, and the integrand comes back.

5:38Example 2: 1/(sin(x - a) sin(x - b)), multiply by sin(a - b)

Now a harder one. Find the integral of d x over, sine of x minus a, times sine of x minus b, where a and b are constants, and sine of a minus b is not zero; so a minus b is not a multiple of pi. Pause the video and try it first. Done? Okay, let's check. Expanding both sines directly gets really messy. So look at the two angles, x minus a, and x minus b. What is x minus b, minus, x minus a? Yes, exactly, the x cancels, and we get a minus b, a constant! So multiply and divide by sine of a minus b. It has no x in it, and it is non-zero, as given. On the top, write a minus b as x minus b, minus, x minus a. Clear? Now expand the top with the formula for the sine of a difference; the square bracket just holds the angle. We get sine of x minus b, times cos of x minus a, minus cos of x minus b, times sine of x minus a. Okay? Divide each term by the bottom, sine of x minus a, times sine of x minus b. The first term leaves cos of x minus a, over sine of x minus a: that is cot of x minus a. The second leaves cot of x minus b. Beautiful, right? Now, before we integrate, think. We used sine of a minus b on the top. What happens to the sine of a minus b we divided by? It stays outside, as one over sine of a minus b. So capital I, our integral, is one over sine of a minus b, times the integral of cot of x minus a, minus cot of x minus b, d x. Forgetting this constant in front is a very common mistake. Got it? The integral of cot is natural log of the modulus of sine, and the two logs combine into one. Answer: one over sine of a minus b, times natural log of the modulus of, sine of x minus a over sine of x minus b, plus capital C, the constant. One smart multiplication, and a scary integral became two cots!

8:02Quick revision: make f and f' appear, e^x (f + f'), 4 tricks

So, quick revision. Hard integrals: rewrite until a function and its derivative appear. e to the power x outside: look for e to the power x, times f plus f dash; the answer is e to the power x times f, plus capital C, the constant. And the four tricks: split the top, take a power out, multiply and divide by a constant, and spot the hidden derivative. Now go and try the practice problems on this page. All the best!