Loisa M.

asked • 04/13/20

Integrals: Inverse Trigonometric

Integrate: dx/[sqrt(8+6x-9x^2)]

Please show step by step solution. I have 1/3 arcsin (3x-1) + c as an answer, is that right?

1 Expert Answer

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Mark M. answered • 04/13/20

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Recall: ∫[du / √[a2 - u2] = Arcsin(u/a) + C


The given integral is equivalent to ∫dx / √[1 - (-7 - 6x +9x2)]


= ∫dx / √[1 - (9x2 - 6x + 1 - 8)] = ∫dx / √[9 - (3x-1)2]


Let u = 3x-1, then du = 3dx. So, dx = (1/3)du


The integral then can be expressed as (1/3)∫[du / √[32 - u2]


= (1/3)Arcsin(u/3) + C = (1/3)Arcsin[(3x-1)/3] + C


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Loisa M.

Is there really have 1/a beside of arcsin because in my book, there's none. And also i don't know how did you complete the square the quadratic, can you show me how?
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04/13/20

Mark M.

tutor
You are correct, there is no "1/a" in front--my mistake. To complete the square, 9x^2 - 6x = 9(x^2 - 2/3x + 1/9 - 1/9) = 9(x - 1/3)^2 - 1 = (3x - 1)^2 - 1
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04/13/20

Loisa M.

That's my answer too, (3x-1)^2 - 1, but how did it become, 9 - (3x-1)^2 (reffered to your first answer above)?
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04/14/20

Mark M.

tutor
9 = 3^2. So, 9(x-1/3)^2 - 1 = 3^2(x-1/3)^2 -1 = (3x-1)^2 - 1
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04/15/20

Loisa M.

So, if the answer is (3x - 1)^2 - 1, then why is it your a=3, it should be a=1, right? So, therefore, the answer should be 1/3 arcsin 3x-1 + c only, it should not have 3 as denominator.
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04/15/20

Mark M.

tutor
You have to do u-substitution to put the integrand into the form du/(a^2-u^2)^1/2 so that the formula given at the beginning of my solution can be applied.
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04/15/20

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