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I need help with integration, !

  1. Apr 2, 2005 #1
    I need help with integration, urgent!

    Can someone please show me the steps in calculating the indefinite integrals of:

    Sin[x]Cos[x]Cos[x]
    and
    e^(Sqrt[x+1])

    Please show all the steps, and not just the answer, which I can do on my calculator.

    Thank you.
     
    Last edited: Apr 2, 2005
  2. jcsd
  3. Apr 2, 2005 #2
    For the first one, just substitute [itex]u = \cos{x} \Longrightarrow du = -\sin{x} \ dx[/itex].

    For the second, substitute [itex]u = \sqrt{x+1} \Longrightarrow du = \frac{dx}{2\sqrt{x+1}} \Longrightarrow 2u \ du = dx[/itex], and then integrate by parts.
     
    Last edited: Apr 2, 2005
  4. Apr 2, 2005 #3
    thats exactly what I did, but I can't get the right answer. Could you please do it, i'll give you 10 gmail invites, hehe
     
  5. Apr 2, 2005 #4
    Using the substitutions I posted above,

    [tex] \int \sin{x} \cos^2{x} \ dx = -\int u^2 \ du = -\frac{u^3}{3} + C = -\frac{\cos^3{x}}{3} + C[/tex]

    and

    [tex] \int e^{\sqrt{x+1}} \ dx = 2\int ue^u \ du = 2\left[ue^u - \int e^u \ du\right] = 2\left[ue^u - e^u\right] + C= 2e^u(u-1) + C = 2e^{\sqrt{x+1}}\left(\sqrt{x+1}-1\right) + C[/tex]
     
    Last edited: Apr 2, 2005
  6. Apr 2, 2005 #5
    thanks, in the first one, i forgot the sin[x] dissapears, when you integrate, 2nd one still trying myself
     
  7. Apr 2, 2005 #6
    Well, it doesn't dissapear. It's part of the [itex]du[/itex].
     
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