Integral Calculus (2017 edition)
- Introduction to trigonometric substitution
- Substitution with x=sin(theta)
- More trig sub practice
- Trig and u substitution together (part 1)
- Trig and u substitution together (part 2)
- Trig substitution with tangent
- More trig substitution with tangent
- Long trig sub problem
- Trigonometric substitution
More trig substitution with tangent
Another practice problem replacing x with tan(theta) in an integral. Created by Sal Khan.
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- Hi at (2:00to about2:48) Sal starts by expressing that he wants to use "1+tan^2(theta) = sec^2(theta)" My question is: What do I look for in the starting integral, in order for me to find the best trig substitution.
In this example Sal chose to use "1+tan^2(theta)" but why couldn't he use "sin^2(theta)+cos^2(theta) = 1"(8 votes)
- Same here. Expect it also contains the useful identity sections as well, such as:
a^2-a^2 〖sin〗^2 θ= a^2 〖cos〗^2 θ)
a^2+a^2 〖tan〗^2 θ=a^2 〖sec〗^2 θ
a^2 〖sec〗^2 θ-a^2=a^2 〖tan〗^2 θ(1 vote)
- isn´t the integral just ln(36+x^2)*(1/2x)? if you differentiate that you´ll get the same integral again!(5 votes)
- You don't get the same integrand again. You forgot to use the product rule. The derivative of ln(36+x^2)*(1/2x) is
That's because what you are doing isn't allowed. You are using the reverse chain rule when the derivative of the inside function is not present. Also you can not modify the equation to include it as you can only multiply the integrand with constants, so the x needed to use the reverse chain rule remains absent. The integrand would have to be of the form
bx/(36+x^2) (where b is a constant) in order to use the reverse chain rule.
Then the answer would be similar to what you described, (b/2)*ln(36+x^2)+C.(7 votes)
- The way u have done is quite complicated and involve a high risk of doing a mistake...we can just substitute "x=6tan(theta)" in the beginning and move on with it..the answer is same and saves time too(4 votes)
- For this, couldn't you just do 1/(36+x^2)=1/36*1/(1+1/36x^2) and then do u=1/6x and then use the inverse tan derivative to get the answer?(3 votes)
- Yes, you could! (And I think it's a lot easier too.) :) Good eye. I think Sal was just using this as another example of trig substitution.(1 vote)
- can someone explain the difference b/w sin^-1=arcsin and 1/sin=csc? doesn't sin^-1=1/sin????(1 vote)
- The first term is sin^-1(x). The second term is (sin(x))^-1(3 votes)
- what happens if you have a value scaling the x^2?
integral of (1/(5x^2+7))
- This is a little bit more complicated than shown in the video, but if we want to scale the x^2-term with integer a, you get the integral ArcTan[(Sqrt[a] x)/Sqrt]/(Sqrt Sqrt[a]). If we want to scale the x^2-term with 5 we get ArcTan[Sqrt[5/7] x]/Sqrt. You can play around with the numbers and get integrals such as following:
a=2: ArcTan[Sqrt[2/7] x]/Sqrt
a=3: ArcTan[Sqrt[3/7] x]/Sqrt
A pattern appear, see if you can continue with a=4 and try to understand the relations.(2 votes)
- why cant this just be evaluated as the integral of 1/36 + 1/x^2?(1 vote)
- That's some incorrect algebra.
When you add fractions they need to have the same denominator and so the same rule applies to splitting them up.(3 votes)
- I understood nothing about this subtitution.. is there
simpler way to do this??(1 vote)
- why is the integral of d(theta)= (theta)(1 vote)
- imagine seeing "integral of (1)d(theta)". You want the integral of 1. When it is taken with respect to theta, the integral of 1 is theta. The notation requires you to include the "d(theta)". So, "integral of d(theta)" is just another way of saying "integral of (1) d(theta)".(3 votes)
- Can't we just substitute u for (36+x^2), and then get ln(u)/2 --> ln(36+x^2)/2(1 vote)
- No, you can't. Because you cannot substitute "dx" with a reasonable form of "du" then; which YOU didn't mention in your solution. Because there is no '2x' as Numerator. SAL just mentioned it at the beginning of the video.(2 votes)
Let's say we have the indefinite integral of 1 over 36 plus x squared d x. Now, as you can imagine, this is not an easy integral to solve without trigonometry. I can't do u substitution, I don't have the derivative of this thing sitting someplace. This would be easy if I had a 2x sitting there. Than I would say, oh the derivative of this is 2x, I could do u substitution and I'd be set. But there is no 2x there, so how do I do it? Well, I resort to our trigonometric identities. Let's see what trig identity we can get here. The first thing I always do, this is just the way my brain works, I always like it-- I can see this is a constant plus something squared, which tells me I should use a trigonometric identity. But I always like it in terms of 1 plus something squared. I'm just going to rewrite my integral as being equal to, let me write the dx in the numerator. This is just times dx. Let me write a nicer integral than that. This is equal to the integral of d x over 36 times 1 plus x squared over 36. 1 plus x squared over 36, that's another way to write my integral. Let's see if any of our trig identities can somehow be substituted in here for that that would somehow simplify the problem. So the one that springs to mind, and if you don't know this already, I'll write it down right here, is 1 plus tangent squared of theta. Let's prove this one. Tangent squared of theta, this is equal to 1 plus just the definition of tangent sine squared of theta over cosine squared of theta. Now 1 is just cosine squared over cosine squared. So I can rewrite this as equal to cosine squared of theta over cosine squared of theta, that's 1, plus sine squared theta over cosine squared of theta, now that we have a common denominator. Now what's cosine squared plus sine squared? Definition of the unit circle. That equals 1 over cosine squared of theta. Or we could say that that equals 1 over cosine squared. One over cosine is secant. So this is equal to the secant squared of theta. If we make the substitution, if we say let's make this thing right here equal to tangent of theta, or tangent squared of theta. Then this expression will be 1 plus tangent squared of theta. Which is equal to secant squared. Maybe that'll help simplify this equation a bit. We're going to say that x squared over 36 is equal to tangent squared of theta. Let's take the square root of both sides of this equation and you get x over 6 is equal to the tangent of theta, or that x is equal to 6 tangent of theta. If we take the derivative of both sides of this with respect to theta we get d x d theta is equal to-- what's the derivative of the tangent of theta? I could show it to you just by going from these basic principles right here. Actually let me do it for you just in case. So the derivative of tangent theta-- never hurts to do it on the side, let me do it right here. It's going to be 6 times the derivative with respect to theta of tangent of theta. Which we need to figure, so let's figure it out. The derivative of tangent of theta, that's the same thing as d d theta of sine of theta over cosine of theta. That's just the derivative of tangent. Or this is just the same thing as the derivative with respect to theta, let me scroll to the right a little bit. Because I never remember the quotient rule, I've told you in the past that it's somewhat lame, of sine of theta times cosine of theta to the minus 1 power. What is this equal to? We could say it's equal to, well the derivative of the first expression or the first function we could say, which is just cosine of theta. This is equal to cosine of theta, that's just the derivative of sine of theta times our second expression. Times cosine of theta to the minus 1. I've put these parentheses, and put the minus 1 out there because I didn't want to put the minus 1 here and make you think that I'm talking about an inverse cosine or an arccosine. So that's the derivative of sine times cosine and now I want to take plus the derivative of cosine. Not just cosine, the derivative if cosine to the minus 1. So that is minus 1 times cosine to the minus 2 power of theta. That's the derivative of the outside times the derivative of the inside. Let me scroll over more. So that's the derivative of the outside. If the cosine theta was just an x, you would say x to the minus 1 derivative is minus 1 x to the minus 2. Now times the derivative of the inside. Of cosine of theta with respect to theta. So that's times minus sine of theta. I'm going to multiply all of that times sine of theta. The derivative of this thing, which is the stuff in green, times the first expression. So what does this equal? These cosine of theta divided by cosine of theta, that is equal to 1. And then I have a minus 1 and I have a minus sine of theta. That's plus plus. What do I have? I have sine squared, sine of theta time sine of theta over cosine squared. So plus sine squares of theta over cosine squared of theta. Which is equal to 1 plus tangent squared of theta. What's 1 plus tangent squared of theta? I just showed you that. That's equal to secant squared of theta. So the derivative of tangent of theta is equal to secant squared of theta. All that work to get us fairly something-- it's nice when it comes out simple. So d x d theta, this is just equal to secant squared of theta. If we want to figure out what d x is equal to, d x is equal to just both sides times d theta. So it's 6 times secant squared theta d theta. That's our d x. Of course, in the future we're going to have to back substitute, so we want to solve for theta. That's fairly straightforward. Just take the arctangent of both sides of this equation. You get that the arctangent of x over 6 is equal to the theta. We'll save this for later. So what is our integral reduced to? Our integral now becomes the integral of d x? What's d x? It is 6 secant squared theta d theta. All of that over this denominator, which is 36 times 1 plus tangent squared of theta. We know that this right there is secant squared of theta. I've shown you that multiple times. So this is secant squared of theta in the denominator. We have a secant squared on the numerator, they cancel out. So those cancel out. So are integral reduces to, lucky for us, 6/36 which is just 1/6 d theta. Which is equal to 1/6 theta plus c. Now we back substitute using this result. Theta is equal to arctangent x over 6. The anti-derivative 1 over 36 plus x squared is equal to 1/6 times theta. Theta's just equal to the arctangent x over 6 plus c. And we're done. So that one wasn't too bad.