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## Integral Calculus (2017 edition)

### Course: Integral Calculus (2017 edition) > Unit 6

Lesson 6: 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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# Trig substitution with tangent

When you are integrating something which looks like 1+(x^2), try replacing x with tan(theta). Created by Sal Khan.

## Want to join the conversation?

- Now, how on God's green earth did we decide that we can get x = a sin θ out of an integrand of a^2-x^2 and x = a tan θ out of an integrand of a^2 + x^2? Even though Wikipedia has a kinda helpful page all about trig substitution, I'm still not quite sure how these substitutions can be derived or proven or why they make any sense. Can anyone fill me in?(44 votes)
- Watch Sal's video "Introduction to trigonometric substitution" in this same section. I guess it's a new video and probably didn't exist while you were watching this.

I reacted the same way in school as you did just now. but once I watched that video, things started fitting in. Trust me :)(3 votes)

- We need videos for x^2-a^2 using asec(theta) and videos for partial fractions method of solving integrals please please please.(20 votes)
- There are some helpful videos on Youtube by other uploaders for partial fractions, and the sec^2(x) - tan^2(x) = 1 is used in a very similar manner.(4 votes)

- In all these videos, it is said at the beginning that "it might be useful to use [such and such substitution]." in this video it is that you use x = a tan (theta) from seeing 9 + x^2. How do you know to do this?(4 votes)
- Anytime you have to integrate an expression in the form a^2 + x^2, you should think of trig substitution using tan θ. Here's why:

If we have a right triangle with hypotenuse of length y and one side of length a, such that:

x^2 + a^2 = y^2

where x is one side of the right triangle, a is the other side, and y is the hypotenuse.

Drawing our right triangle:

.............../|

............/...|

........./......|

..y. /.........|

..../...........| x

../.............|

/θ..).*____*|

......... a

tan θ = opposite / adjacent

tan θ = x / a

solve for x:

x = a * tan θ(23 votes)

- The way I would do it is to factor out a 9 from the denominator giving
`(1/9) * ∫ dx/(1 + (x/3)^2)`

then u = x/3 and 3 du = dx giving`(3/9) * ∫ du/√(1 + u^2) = (1/3) arctan(u) + c = (1/3) arctan(x/3) + c`

Are there other trig-sub problems that are less straightforward, or have I just gotten used to recognizing the derivative of arctan?(7 votes) - these videos have been unseful b/c I'm in calculus III and we've been doing volumes and hypervolumes with triple integrals, and we keep getting sqrt(a^2 - x^2) all over the place, so I keep getting stuck performing the substitution. I go to the solution's manual and they come up to the point where you need to do trig sub, and skip straight to the final expression as if we could do it all in our heads.(3 votes)
- Since you are now at triple integrals, the assumption is that the trig sub, which you learned in Calc II, is now old hat and you have mastered it. Double and triple integrals, as I am sure you know, are more about finding the limits of integration, re-arranging the order of integration, substitutions/Jacobians and applications like moments and centers of mass etc. The techniques to solve them, in the end (that is, the outside integral), are the same as single integrals. Quite a few of my old texts are similar - the solutions jumping over the stuff already covered. Perhaps check out a couple more textbooks online for more trig sub examples.(7 votes)

- I have an actual broad kind of question: trig substitution is easily the hardest for me. Can I get by in calculus without it or do I have to go back and re-learn trig now?(2 votes)
- You must know trigonometry. You will be getting into some rather involved uses of the trigonometric functions.(7 votes)

- So you get arctan, arccos and arcsin, would one then also get arccot, arcsec and arccsc? I've never come across those terms, just interested in knowing whether they actually exist.(2 votes)
- Yes, those inverse trigonometric functions do exist.(6 votes)

- At1:58, why Sal makes x equal to 3 tan of Theta? where that 3 comes from?(3 votes)
- We choose the substitution that makes things work out as easily as possible. In this case, we want something that will simplify the expression 9 + x². If we choose tan θ, we end up with 9 + tan² θ, which doesn't help much. But when we choose 3 tan θ we get 9 + 9 tan² θ, and that works because we can factor out a 9 and use a trig identity to get 9 sec² θ. The general rule here is that when you have something that looks like a + x², where a is a constant, the substitution you want is √a tan θ.(3 votes)

- I'm not sure if this is the right area, but I'm not aware of any other forums on KA.

I was doing one of the sums given in the exercise for this section. You can see it here : https://imgur.com/3teO4gR

I solved it (correctly) all the way until it was 2*INTEGRAL[cot(theta) dtheta].

Since that's just 2*INTEGRAL[cos(theta)/sin(theta) dtheta], I used u-substitution. u = sin(theta), du = cos(theta) dtheta. The sum was now 2*INTEGRAL[1/u du], so I wrote down the answer as 2*ln(abs(u)) = 2*ln(sin(theta)). This, however, wasn't one of the answers available. Can someone please explain where I went wrong? Is it wrong to mix logarithms and trigonometry? I'd hope not, but once again, I'm at the answerer's mercy at that.

Thanks in advanced :)(2 votes)- You did not substitute correctly and you did not back substitute. Here's how to solve it:

Given: ∫ dx / [x²√(4-x²)]

Let x = 2 sin θ. Thus dx = 2 cos θ dθ

And

½ x = sin θ

θ = arcsin(½ x)

--

∫ dx / [x²√(4-x²)]

= ∫2 cos θ dθ / [2² sin²θ √(2² - 2²sin²θ)]

= ∫2 cos θ dθ / [2²sin²θ √(2²cos²θ)]

= ∫2 cos θ dθ / [2²sin²θ (2cosθ)]

= ∫dθ / [2²sin²θ]

= ∫¼ csc² θ dθ

= ¼ ∫ csc² θ dθ

= -¼ (cot(θ)) + C

= -¼cot(arcsin(½ x)) + C

= -¼(2 [√(1-¼ x²)]/x + C

= -¼([√(4-x²)]/x + C

= [-√(4-x²)]/(4x) + C(3 votes)

- Couldn't you just take the derivative of (9+x^2)^-1? Would you get the same answer?(1 vote)
- We're taking integrals here, so you can't just do the chain rule like you would if you were taking a derivative.(4 votes)

## Video transcript

Let's see if we can evaluate
the indefinite integral 1 over plus 9 plus x squared dx. And we know that if you
have the pattern a squared minus x squared, it
could be a good idea to make the substitution,
x is equal to a sine theta. But we don't see that
pattern over here. Instead, what we see is
a squared plus x squared. And in this context, it tends
to be a good idea-- it's not always going to work, but
it never hurts to try out. This is a little bit of
an art here-- to try out, x is equal to a tangent theta. Now you might say,
Sal, why is that? Well, let's make
that substitution and see how this
thing would simplify. This thing would become a
squared plus a squared tangent squared theta, which
is a squared times 1 plus tangent squared theta. And this right over here,
we could reprove it. Actually, let me just
reprove it for you. This is going to become
a squared times-- this is 1 could be written
as cosine squared theta over cosine squared theta. Tangent squared is
sine squared theta over cosine squared theta. And this is why I picked cosine
squared as the denominator, so that I can add the two. And this is going to become
a squared times cosine squared theta plus
sine squared theta, all of that over
cosine squared theta. This numerator from the
unit circle definition of trig functions becomes 1. So this is 1 over
cosine squared theta. And everything simplifies
to a squared secant squared theta, which
might simplify things. So let's see if we have
what's going on over here. We can rewrite this. So 9 plus x squared
you could rewrite as 3 squared plus x squared. In this case, a
would be equal to 3. So we want to make
the substitution, x is equal to 3
tangent of theta. And if we wanted
to solve for x, you can divide both sides by
3, because we're later going to have to undo
the substitution. x over 3 is equal
to tangent theta, or theta is equal to arctangent
or inverse tangent of x over 3. Now we're also going to have
to figure out what dx is. We're also going to have
to figure out what that is. So let's take the
derivative or we'll write it in differential form. dx is equal to 3 derivative
of tangent theta with respect to theta is secant
squared theta d theta. So now it looks like
we're armed with all of the things necessary to
rewrite this entire integral. It's going to be equal to
the indefinite integral. You're going to
have dx here, which is equal to 3 secant
squared theta d theta. That's the dx. And all of that's going
to be over this business right over here, our a
squared plus our x squared. Now we already know what
that's going to simplify to. Our a squared plus
our x squared is going to-- since we
made this substitution, x is equal to 3
tangent theta is going to simplify to a squared
secant squared theta. So this is going to simplify
to 9 secant squared theta. And you could essentially
go through this logic. You're going to get 9 plus
9 tangent squared theta. Factor out a 9, you get 9
times 1 plus tan squared theta. And so that's going
to be 9 times secant squared theta, exactly
what we have here. Lucky for us, we have
the secant squared. It's canceling out. The secant squared
is canceling out. You have 3 over 9. This whole thing
can be rewritten as 1/3-- that's
just the 3 over 9-- times the indefinite integral
of just d theta, which is equal to 1/3 theta plus c. And now we just have to
put things in terms of x. And we see theta is equal
to arctangent of x over 3. So this is going to be
equal to 1/3 arctangent of x over 3 plus c. And we are done. So now we know how to deal with
cases where we see something like an a squared
minus an x squared and an a squared
plus an x squared. It won't always
work, but it might be a useful-- it'll definitely
allow you to do this thing. It might not always make
the integral solvable, but it's not a bad thing to try. When it looks like
u-substitution isn't working, then you can look
for these patterns and try some trig substitution.