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Course: Integral Calculus>Unit 1

Lesson 16: Trigonometric substitution

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.

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• 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?
• 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 :)
• We need videos for x^2-a^2 using asec(theta) and videos for partial fractions method of solving integrals please please please.
• 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.
• 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?
• 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
• 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.
• 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.
• 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?
• 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?
• You must know trigonometry. You will be getting into some rather involved uses of the trigonometric functions.
• 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.
• Yes, those inverse trigonometric functions do exist.
• At , why Sal makes x equal to 3 tan of Theta? where that 3 comes from?
• 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.
• 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.
• 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