Integration by Substitution

Transcript

0:00Introduction

Hello everyone! One small challenge first. The derivative of sine of x squared is cos of x squared, times two x, by the chain rule. Easy, right? Now turn it around. What is the integral of two x times cos of x squared, d x? Many students freeze here. Today's move makes it one line: integration by substitution. It comes up often in JEE, so let's begin.

0:27The idea: change the variable

So, here is the rule. This tall curved sign is the integral sign, and d x says that x is the variable. If x equals g of t, the integral of f of x, d x, becomes the integral of f of g of t, times g prime of t, d t. The small dash, called prime, means the derivative. So d x becomes g prime of t, d t. Okay? But in a real question, nobody hands you g. So spot a function whose derivative, or a multiple of it, is another factor, and call it t. Clear? Back to our challenge: two x times cos of x squared, d x. The small raised two means squared. Which function has its derivative sitting right beside it? x squared, yes, exactly! Put t equals x squared, so d t equals two x d x, and that is sitting right there. With me? That gives the integral of cos t, d t, which is sine t, plus capital C, the constant of integration. But careful, the question was in x, so we go back to x: sine of x squared, plus C, and our challenge is solved!

1:33Three forms to spot at sight

Three forms are worth knowing at sight. First: the integral of f prime of x over f of x, d x, equals the natural log of mod f of x, plus C, okay? The letters l n mean natural log, and the two vertical bars mean mod, the modulus. Quick check: the integral of d x over x times natural log x. Which form is it? The first form, correct! f is natural log x, and one over x is its derivative. So the answer is the natural log of the modulus of natural log x, plus C, right? Second form: f prime of x, times f of x in square brackets to the power n, gives f of x to the power n plus one, over n plus one, plus C, for n not minus one; the equals sign with a slash means not equal to. Now you try: natural log x, the whole squared, over x, d x. Again f is natural log x, and n equals two. So we get natural log x, the whole cubed, over three, plus C, okay? Very good! Third form: the derivative over the square root of the function gives two root f of x, plus C, and that is just the second form with n equal to minus one half.

2:52Trick: x²±1 over x⁴+kx²+1

Got it? Next, a pattern that comes up often in JEE: x squared plus or minus one, over x to the power four plus k x squared plus one. The sign with plus over minus means plus or minus. First, divide top and bottom by x squared. Okay? Then put t equals x minus or plus one over x; the sign with minus over plus means minus or plus, the opposite order. So plus one on top goes with x minus one over x. Why the opposite sign? Differentiate and see; this d over d x means the derivative with respect to x. x minus one over x gives one plus one over x squared, exactly the top after dividing by x squared. And x plus one over x gives one minus one over x squared. Clear?

3:42Trap: (x-a)ᵐ(x-b)ⁿ with m+n=2

Now the trap form: d x over x minus a to the power m, times x minus b to the power n, with m plus n equal to two. Which substitution would you try? Use both brackets together! t equals x minus a, over x minus b, or its reciprocal. For this form, it always works. Okay? Why does it always work? d t is a minus b, over x minus b the whole squared, times d x, and the integrand is t to the power minus m, over that same square. So only a power of t remains. Right? In JEE, these powers can be fractions, like three quarters and five quarters, or eleven over thirteen and fifteen over thirteen. Each pair adds up to two. Got it?

4:30Example 1: x cubed over 1 + x8, and 1 over e^x + e^-x

Now, Example one. First: x cubed over one plus x to the power eight, d x. Second: d x over e to the power x plus e to the power minus x. Pause the video and try them first. Done? Okay, let's check. There is x to the power eight at the bottom, so t equals x to the power eight, right? No, no, careful here! Its derivative would need x to the power seven on top. Instead, take t equals x to the power four, so x to the power eight is t squared, and d t equals four x cubed d x: the top is one quarter of d t. Okay? So we get one quarter of the integral of d t over one plus t squared, which is tan inverse t. The small minus one means inverse, not a power. Answer: one quarter times tan inverse of x to the power four, plus C, right? Now the second one. No derivative is visible here. So what should we multiply the top and the bottom by? By e to the power x, yes, exactly! The bottom becomes e to the power two x, plus one. Put t equals e to the power x; then d t equals e to the power x, d x, exactly the top. With me? And e to the power two x is t squared. So again it is d t over one plus t squared: tan inverse of e to the power x, plus C, okay? The same tan inverse again. Very good!

6:10Example 2: x squared + 1 over x4 + 1 using t = x - 1/x

Now a harder one. The integral of x squared plus one, over x to the power four plus one, d x. Pause the video and try it. Hint: use our trick. Done? Okay, let's check. This is our trick with k equal to zero. Divide top and bottom by x squared: one plus one over x squared, over x squared plus one over x squared. Okay? Now, which t? x plus one over x, or x minus one over x? x minus one over x, correct! Its derivative is our top, so d t equals one plus one over x squared, d x. And t squared is x squared, minus two, plus one over x squared, so the bottom is t squared plus two. Clear? Now it is just d t over t squared plus two. Two is root two squared, so the standard tan inverse form gives one over root two, times tan inverse of t over root two, plus C, with me? But the question is in x, so we go back: t is x squared minus one, over x. So the answer is one over root two, times tan inverse of x squared minus one, over root two times x, plus C, okay? That's it!

7:23Quick revision: choosing t, three forms, two patterns

So, quick revision. Spot a function whose derivative, or a multiple of it, is another factor: that is your t. Know the three forms, each with plus C, and n not minus one in the power form, okay? For the x to the power four pattern, divide by x squared, and put t equals x minus or plus one over x. For two brackets with powers adding to two, use their ratio. And always go back to x. Now try the practice problems on this page. All the best!