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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??
• 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)
• Did you just assume that the hypotenuse is 2, or is there some way to tell that the hypotenuse is 2.
• 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.

https://en.wikiversity.org/wiki/Trigonometric_Substitutions
• 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