# Introduction to trigonometric substitution (video)

Trigonometric Substitution
Trigonometric Substitution

## Integral Calculus

• 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

Introduction to trigonometric substitution

Introduction to trigonometric substitution.

## Want to join the conversation?

• Was the choice of 2 for the hypotenuse completely arbitrary?(27 votes)
• No. You should choose the hypotenuse so that it matches what is under the radical. The denominator was sqrt(4 – x²), and he chose the hypotenuse to be 2 and one side to be x in order for the third side of the triangle to be sqrt(4 – x²). So for example if the original integral had sqrt(16 – 25x²), then you would choose the hypotenuse to be 4 and one side to be 5x, so that the third side would be sqrt(16 – 25x²).(28 votes)
• Substituting x = 2 cos t I get the answer -arccos x/2 + C
Is that a valid solution?(28 votes)
• That is also a valid solution, yes. Recall the identity`arcsin(u) = π/2 - arccos(u)`. Letting`u = x/2`and observing that the constant`C`may be written as`π/2 + C'`, for some constant`C'`, we get`-arccos(x/2) + C = π/2 - arccos(x/2) + C' = arcsin(x/2) + C',`
which brings us back to the solution Sal arrived at.(23 votes)
• At, Sal says, “If we define this angle as theta…”. Why make this angle theta and not the other acute angle? 2:10(16 votes)
• A similar question was asked that was already answered, but if we do it for the other angle theta, we would get the same answer in a different form of
-arccos(x/2) + C.
From Qeeko:
That is also a valid solution, yes. Recall the identity arcsin(u) = π/2 – arccos(u). Letting u = x/2 and observing that the constant C may be written as π/2 + C’, for some constant C’, we get
-arccos(x/2) + C = π/2 – arccos(x/2) + C’ = arcsin(x/2) + C’,
which brings us back to the solution Sal arrived at.(2 votes)
• At, Sal says that if x=2sin(theta) then dx=2cos(theta)d(theta), but I do not follow. What is the reason? 3:14(6 votes)
• He is differentiating`x`with respect to`θ`.
Before differentiating you have:`x = 2·sin(θ)`. After differentiating with respect to`θ`you have:`dx/dθ = 2·cos(θ)`
Finally, you solve for`dx`:`dx = 2·cos(θ)·dθ`(15 votes)
• I’ve noticed that the question was a lot like the derivative of arcsin(theta) with the exception that instead of 1/sqrt(1-x^2), we had 1/sqrt(4-x^2). The answer, in the end, was just arcsin(x/2). Is there a pattern to it? If it is 1/sqrt(1-x^2), it is arcsin(x/1), and if it is 1/sqrt(9-x^3), the answer should be arcsin(x/3).
Is this actually true? And if so, can you please explain why?
• I hope that you mean 1/√(9 – x²), not ( …x³), the exponent on x needs to be 2 because the Pythagorean Theorem is the key to this technique. Something of the form 1/√(a² – x²) is perfect for trig substitution using x = a · sin θ. That’s the pattern. Sal’s explanation using the right triangle shows why that pattern works, “a” is the hypotenuse, the x-side opposite θ is equal to a · sin θ, and the adjacent side √(a² – x²) is equal to a · cos θ . Using those substitutions, the original integral becomes easy – you just have to remember to restrict the Domain.(0 votes)
• I’m wondering why, when constructing the triangle, you let the opposite side be x and the adjacent be sqrt(4-x^2). If you reverse this, and let x be the adjacent side for example, you get a completely different answer. How are you supposed to know which is the correct orientation??(7 votes)
• Making x the adjacent side and sqrt'(4-x^2)’ the opposite side will be identical with labeling the other angle, ( the one complementary to theta), the new theta. Doing so, and solving for the integral, you will end up with the answer:’-arccos(x/2)+c’, which is another equally possible solution to the problem.
I hope this helped!(1 vote)
• can’t you just factor out a 4 in the denominator and get 2root(1-(x/2)^2). then it’s in the form of the derivative of arcsin theta. that’s what occurred to me after seeing the way in which the integrand was written. what advantage do we get by doing this substitution stuff(5 votes)
• Did you just assume that the hypotenuse is 2, or is there some way to tell that the hypotenuse is 2.(3 votes)
• There is a way to tell.
The Pythagorean Theorem says a² + b² = c², where c is the hypotenuse.
That means one of the sides, lets say a, is equal to sqrt(c² – b²).
Now we have been given an expression that looks just like that: sqrt(4 – x²), which can be rewritten as sqrt(2² – x²).
That means that for this trig sub triangle, one side (the b side) is equal to x, the other side, the a side is equal to sqrt(2² – x²) which means the hypotenuse for this triangle must be 2.
• What if we would’ve chosen the other side as our x in the beginning?(3 votes)
• Hi, tuf62486!
There are two scenarios, and these are as follows:
– Scenario ONE: https://i.imgur.com/atiQ9yT.png
– Scenario TWO: https://i.imgur.com/1AmZgHZ.png
Explanations:
– Scenario ONE: this one is comparatively complex, but it still does make sense. If we choose`x`as the side adjacent to`theta`, then we will end up with`-arccos(x / sqrt(u)) + c`. Although this appears different, if you look at this graph (drag the slider at the top-left at your own leisure) ==> https://www.desmos.com/calculator/3gfueg1fmv <== then you will see that`-arccos(x / sqrt(u)) + pi/2`is actually equal to`arcsin(x / sqrt(u))`. Since after we integrate we are left with a constant of integration (I have used`c`to denote said constant), this “absorbs” the extra`pi / 2`that we would need to add for the graphs to match exactly. Regardless, the curves for`-arccos(x / sqrt(u))`and`arcsin(x / sqrt(u))`are identical except that one is vertical translation of the other, and that is all we are trying to prove with integration, anyway.
– Scenario TWO: This is identical to how Sal solves the problem in the video. We have just swapped`x`and`theta`, but I have solved the problem so that you can see that it will be the same.
I hope that this helps! Cheers,
• (If you mean`√(4 - x²)`and`√(4 - x)`, respectively, you need to include the parentheses.)
In the indefinite integral`∫ √(4 - x) dx`, the substitution`u = 4 - x`will do. The reason we use a trigonometric substitution in`∫ √(4 - x²) dx`, is that the substitution`u = 4 - x²`is not really that helpful. Besides, we know some useful trigonometric identities involving expressions of the form`a² - x²`, which makes a trigonometric substitution sensible.(3 votes)