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## Calculus 2

### Unit 2: Lesson 4

Trigonometric substitution- 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

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# More trig substitution with tangent

Another practice problem replacing x with tan(theta) in an integral. Created by Sal Khan.

## Want to join the conversation?

- It is just experience that leads to choosing an appropriate identity substitution in the beginning? It seems like picking the wrong identity will lead you down an endless corridor of trig function swapping.(32 votes)
- My calc book has a little table saying: "sqrt[a^2 - x^2] -> x = a sin(theta)" "sqrt[a^2 + x^2] -> x = a tan(theta)" "sqrt[x^2 - a^2] -> x = a sec(theta)" Not sure if that helps, but there's is a method of recognizing which substitution is appropriate, which boils down to determining which trig identity your integrand looks like.(7 votes)

- 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)

- Hi Sal,

Firstly, I have huge admiration for what you've done and what you are doing, and I think that you've begun something that will change the face of education in the world. Thank you for your huge efforts.

Now for a question :-) I'm having a spot of trouble reconciling your answer to this one with a 'standard' integral that gives 1/(a^2+u^2) = arctan(u/a)+c. Since your example is 1 / 6^2+x^2 that would imply that the integral was arctan(x/6). Am I missing something basic?

Regards,

Andrew(10 votes)- Your standard integral solution is wrong. It should be integral of 1/(a^2+u^2) = (1/a)arctan(u/a)+c(10 votes)

- 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

1/(36+x^2)-ln(x^2+36)/(2x^2)

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.(6 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?

i.e

integral of (1/(5x^2+7))

?(2 votes)- 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[7]]/(Sqrt[7] Sqrt[a]). If we want to scale the x^2-term with 5 we get ArcTan[Sqrt[5/7] x]/Sqrt[35]. You can play around with the numbers and get integrals such as following:

a=1: ArcTan[x/Sqrt[7]]/Sqrt[7]

a=2: ArcTan[Sqrt[2/7] x]/Sqrt[14]

a=3: ArcTan[Sqrt[3/7] x]/Sqrt[21]

A pattern appear, see if you can continue with a=4 and try to understand the relations.(2 votes)

- Isn't it easier just to split the function? For example f(x) = 1/(36+x^2) = 1/36 + 1/x^2 = 1/36 + x^-2

Which makes F(x) = 1/36x - x^-1 + c = 1/36x - 1/x + c(0 votes)- Unfortunately, 1/(36+x^2) does not equal 1/36 + 1/x^2. If you plug in an x value you will see this pretty quickly :)(7 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)

## Video transcript

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.