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# Introduction to trigonometric substitution

Introduction to trigonometric substitution.

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• Substituting x = 2 cos t I get the answer -arccos x/2 + C
Is that a valid solution? • 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.
• Was the choice of 2 for the hypotenuse completely arbitrary? •  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).
• At , Sal says, "If we define this angle as theta...". Why make this angle theta and not the other acute angle? • 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.
• At , Sal says that if x=2sin(theta) then dx=2cos(theta)d(theta), but I do not follow. What is the reason? • 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``
• 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?
Thank you :) • 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.
• 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?? • Did you just assume that the hypotenuse is 2, or is there some way to tell that the hypotenuse is 2. • What if we would've chosen the other side as our x in the beginning? • 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,
~ Novum Sensum
• So for the integral of √4-x^2, you would use trigonometric substitution but why wouldn't you use trigonometric substitution for something like √4-x ? • (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. 