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Homework Help: Integration help needed!

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data
    I have various integration questions to do, and I do not want answers at all, just pointers on where to start.. I have 6 ones, and have only got anywhere with 2 of them (nt the two below)
    Find integral of with respect to x:
    ((x-1)/(x+1))1/2 * 1/x

    1/(2 - sin2x)(2 + sinx - sin2x)

    2. Relevant equations
    integration by substitution?

    3. The attempt at a solution
    For the first I tried u = x +1 but this doesn't go anywhere- I wouldn't know how to integrate it even without the extra 1/x term so really don't know where to start.

    For the second I tried the substitution u = sin x thinking maybe this would lead to something where I could use partial fractions (it doesn't), and then there's no cos x to even allow the substitution in the first place.

    My integration is REALLY rusty, so although I can manage integration by parts and simpler stuff, I seem to have completely lost the technique of what to substitute where.

    One of the other ones is to find the integral of x * arcsin x dx. For this I get:
    1/2(x2 * arcsinx + x (1-x2)1/2 - arcsin x ) + C

    I can't find a problem in the working (I used integration by parts where u = x, u' = 1, v' = arcsin x, v = x * arcsin x + (1-x2)1/2), but the maths program I use occasionally to check things came up with different coefficients in the answer.
  2. jcsd
  3. Sep 14, 2010 #2


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    Homework Helper
    Gold Member

    Try the substitution [itex]u=\sqrt{x+1}[/itex] instead and show your work.

    It's not clear exactly what the 2nd one is supposed to be....do you mean [tex]\int\frac{2+\sin x-\sin^2x}{2-\sin^2 x}dx[/tex] or [tex]\int\frac{dx}{(2-\sin^2 x)(2+\sin x-\sin^2 x)}[/tex] ?

    In any case, there is a well-known trig identity that relates [itex]\cos x[/itex] to [itex]\sin x[/itex], so there is nothing to stop you from making your substitution.

    Post your workings then, because your final result is incorrect.
    Last edited: Sep 14, 2010
  4. Sep 14, 2010 #3


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    Homework Helper

    I think that for the trigonometric integral, you can use partial fractions.
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