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Homework Help: Integration Questions

  1. Feb 8, 2005 #1
    Hello all

    1. [itex] \int 2\sin x + 3\cos x dx [/itex] Can you even use substitution in this problem. Or can you directly integrate to get [itex] \int 2\sin x + 3\cos x dx = -2\cos x + 3\sin x [/itex]?

    2. [itex] \int cos^{2} x - cos x dx [/itex] So [itex] u = cos x dx [/itex]. When we integrate we get [itex] \int cos^{2} x - cos x dx = \frac{u^{3}}{3} - \frac{u^{2}}{2} = \frac{(cos x)^3}{3} - \frac{(cos x)^{2}}{2} [/itex] Is this right?

    3. [itex] \int sin 2x\ dx [/itex]. So [itex] u = 2x dx [/itex] Would it be [itex] \frac{1}{2} \-cos x [/itex]?

    4. I need help in doing [itex] \int e^{\sec x} \sec x\tan x dx [/itex]

    Thanks a lot :smile:
     
    Last edited: Feb 8, 2005
  2. jcsd
  3. Feb 8, 2005 #2

    dextercioby

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    The first is okay,the second is wrong.Try the double angle formula.

    Fix the TEX on the last.

    Daniel.
     
  4. Feb 8, 2005 #3

    arildno

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    As for 4., note that [tex]\frac{d}{dx}(\frac{1}{\cos{x}})=\frac{1}{\cos(x)}tan(x)[/tex]
     
  5. Feb 8, 2005 #4

    dextercioby

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    As for 3.,it should be [itex] -\frac{1}{2}\cos 2x + C [/itex]

    And please,remember to put not only the "dx",but also the integration constants... :wink:

    Daniel.
     
  6. Feb 8, 2005 #5
    heh sorry for not putting integration constants.

    ok so I know that the double angle formulas are: [itex] \sin 2x = 2\sin x\cos x [/itex] and [itex] \cos 2x = cos^{2} x - sin^{2} x [/itex]. I take it that I should use the latter. How would I use this? Couldn't I just do a direct substitution?

    Thanks :smile:
     
  7. Feb 8, 2005 #6
    if we have [itex] \int x\cos x^{2} dx [/itex] then [itex] u = x^{2} [/itex] and [itex] du = 2xdx [/itex]. So [itex] \int x\cos x^{2} dx = \int \cos x^{2} x dx [/itex] [itex] \frac{1}{2}\int cos x^{2} 2x dx = \frac{1}{2} \ sin x^{2} + C [/itex]


    For [itex] \int cos^{2} x - cos x dx [/itex] I just let [itex] u = cos x [/itex]. Isnt this right?

    Thanks :smile:
     
    Last edited: Feb 8, 2005
  8. Feb 8, 2005 #7

    dextercioby

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    No.U should use this formula:
    [tex] \cos^{2}x=\frac{\cos 2x +1}{2} [/tex]

    Daniel.
     
  9. Feb 8, 2005 #8
    ok i think i got it. is it:

    [itex] \int cos^{2} x - cos x dx = \frac{\sin 2x}{4} + \frac{x}{2} - \sin x + C [/itex]?

    [itex] \cos^2{x} = \frac{1}{2}(\cos 2x + 1) [/itex]
    [itex] \int \cos^{2} x dx = \frac{1}{2}\int (\cos 2x + 1) dx [/itex]
    [itex] \frac{1}{2}\int \cos 2x dx + \frac{1}{2}\int dx [/itex]
    [itex] u = 2x, du = 2dx [/itex]

    Thanks
     
    Last edited: Feb 8, 2005
  10. Feb 8, 2005 #9
    ok for questions like:

    [itex] \int sec^{2} \frac{x}{2} dx [/itex] I put [itex] u = \frac{1}{2} x, du = \frac{1}{2} dx [/itex].. So we have [itex] 2\int sec^{2} u du = 2\tan \frac{1}{2}x dx [/itex]. Is this correct?

    For [itex] \int tan^{3} x sec^{2} x dx [/itex] do we use the identity [itex] \sec^{2} x = 1 + \tan^{2} x [/itex]? Is it:

    [itex] \int tan^{3} x (1 + \tan^{2} x) dx = \int tan^{3} x dx + \int tan^{5} x dx [/itex]

    So [itex] \frac{tan^{4} x}{4} + \frac{tan^{6} x}{6} + C [/itex]
     
    Last edited: Feb 8, 2005
  11. Feb 8, 2005 #10

    dextercioby

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    The last one is not that easy.I can do it using sine and cosine.If u have a better method,then it's only for your good.

    Daniel.
     
  12. Feb 8, 2005 #11

    dextercioby

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    No.What u did was wrong.The integration is wrt "x" and not "tangent of x".

    Daniel.
     
  13. Feb 8, 2005 #12
    well i used the substitution: [itex] sec^{2} x = 1 + tan^{2} x [/itex]
     
  14. Feb 8, 2005 #13

    dextercioby

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    I could see that.I meant the final 2 integrations.They "stink"... :tongue2:

    Daniel.
     
  15. Feb 8, 2005 #14
    so there is no way to do it with [tex] \sec x [/tex]?
     
  16. Feb 9, 2005 #15
    [tex]
    \[
    \int_{}^{} {e^{\sec x} \sec x\tan xdx} = \int_{}^{} {e^{\sec x} d\left( {\sec x} \right)} = \int_{}^{} {e^\theta d\theta }
    \]
    [/tex]
     
  17. Feb 9, 2005 #16

    dextercioby

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    I never said that.Check this out

    [tex] (\sec x)'=\sec x \tan x [/tex] (1)

    [tex](\tan x)'=\sec^{2}x [/tex] (2)

    [tex]I=\int \tan^{3}x \sec^{2}x dx=\int \tan^{2}x \sec x d(\sec x) [/tex](3)

    Using part integration
    [tex]I=\tan^{2}x \frac{\sec^{2}x}{2}-\frac{1}{2}\cdot 2 \int \sec^{4}x \tan x dx [/tex] (4)

    For the last integral,use the definitions of the 2 functions and it will be done in a sec.

    Daniel.
     
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