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Inverse trigonometric function integration

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I'm struggling to solve the following integral
∫ x/(√27-6x-x2)




my attempt is as follows:
∫x/(√36 - (x+3)2)
= ∫1/ √(36 - (x+3)2) + ∫x+1/ √(36 - (x+3)2)
= arcsin (x + 3)/6 + this is where I got stuck.



 
1,005
64
I'm struggling to solve the following integral
∫ x/(√27-6x-x2)




my attempt is as follows:
∫x/(√36 - (x+3)2)
= -∫1/ √(36 - (x+3)2) + ∫x+1/ √(36 - (x+3)2)


You seem to have missed a sign, which I inserted in red. In the second integral, remember the derivative of [itex]\sqrt{u}[/itex] requires du in the numerator, or -(x+3) in this case (the factor of 2 is canceled by the square root's power rule factor of 1/2). Can you decompose the fractions in an alternate way so that the numerator of the second fraction is x + 3 ? Then you can simply insert the factor of -1 in the usual manner.

You may also consider integration by parts. You know the derivative of x, and you know the integral of [itex]\frac{dx}{\sqrt{36 - (x+3)^2}}[/itex]. Use these two factors in the integration by parts formula.
 
Last edited:
32,773
4,479
I'm struggling to solve the following integral
∫ x/(√27-6x-x2)




my attempt is as follows:
∫x/(√36 - (x+3)2)
= ∫1/ √(36 - (x+3)2) + ∫x+1/ √(36 - (x+3)2)
= arcsin (x + 3)/6 + this is where I got stuck.


@ande, please stop deleted the three parts of the homework template. Its use is required here, and also, deleting it is what's causing your posts to display in bold.
 

epenguin

Homework Helper
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I don't know if you have noticed that your denominator quadratic factorises. You don't need to do this from scratch, you are nearly there by having expressed it as difference of two squares. Maybe you can use something like the method you are trying more effectively on that.

Edit: in fact I'm pretty sure so.
 
Last edited:

hunt_mat

Homework Helper
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Integration by parts perhaps?

You know [tex]\int\frac{dx}{\sqrt{a^{2}-x^{2}}}[/tex]

So go from there.

[edit] Thought of a better answer, integration by substitution, let [tex]x+3=36\sin\theta[/tex]
 
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