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Trig. Int. or Trig. Sub. ?

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    When I look at the equation, I see a trig integral.

    [tex]\int[/tex]sinxcos3x/[tex]\sqrt{1+(sin^2)x}[/tex] dx

    But the [tex]\sqrt{1++(sin^2)x}[/tex] gets me confused.
    I could tranform it into cos2x if it was sin2x - 1

    3. The attempt at a solution

    [tex]\int[/tex]sinxcos3x/[tex]\sqrt{1++(sin^2)x}[/tex] dx
    =[tex]\int[/tex]sinx - sin2x/[tex]\sqrt{1++(sin^2)x}[/tex]
  2. jcsd
  3. Feb 23, 2010 #2


    Staff: Mentor

    This is really hard to read. You shouldn't mix [ sup] tags in between [ tex] tags. Instead, use code like ^2 or ^{2} for exponents within [ tex] tags.
  4. Feb 23, 2010 #3
    try a using the trig identity [tex]\sin (x)= \frac{1-\cos 2x}{2} [/tex] and [tex] \cos x =\frac{1+\cos 2x}{2} [/tex]
    [tex] \sin x \cos x = 1/2 \sin 2x [/tex]

    it'll look like [tex]\int\frac{1}{2} \sin 2x *\frac{(1+\cos 2x)^{2}}{4} [/tex][tex]\frac{1}{\sqrt{1+1/4*(1+\cos 2x)^2}}[/tex]
    theres is a u sub that makes this a whole lot easier
  5. Feb 23, 2010 #4
    hint u=(1+cos2x) , du=-2sin2x
  6. Feb 24, 2010 #5
    Thank you!
  7. Feb 24, 2010 #6

    Gib Z

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    Homework Helper

    Perhaps even quicker is to use [itex]u=1 + \sin^2 x[/itex] from the start.
  8. Feb 24, 2010 #7
    Recall that :


    Use [tex]t=cos(x)[/tex].

    Then do a trigonometric substitution.
  9. Feb 24, 2010 #8
    Thank you!
  10. Feb 24, 2010 #9
    Now that I am trying your hints, I feel even more lost...

    I cannot use trig identities since my powers are odd (this is what is written in my notebook).

    Well, I'll show you what I have done until now and then tell me if I am on the right way or not.



  11. Feb 24, 2010 #10

    Gib Z

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    None of us even told you to do that substitution. If you actually carried out my suggestion you would have arrived at an elementary integral immediately.
    Last edited: Feb 24, 2010
  12. Feb 25, 2010 #11
    Here is my new try:

    [tex]\int[/tex]sinxcos3x/[tex]\sqrt{1+sin^2(x)}[/tex] dx

    du=1/2 cos2x


    [tex]\int[/tex]1/2 sinxcosx/[tex]\sqrt{1+sin^2(x)}[/tex]

    [tex]\int[/tex]1/4 sin2x/[tex]\sqrt{1+sin^2(x)}[/tex]

    [tex]\int[/tex]1/4 2u/[tex]\sqrt{u^1/2}[/tex]

    [tex]\int[/tex]1/2 u1/2

    1/3 u3/2


    I know something isn't right, but I can't figure out why.

    I know I'm not right since when I differentiate it my answer is sqrt(1+sin(t)^2)*sin(t)*cos(t)
  13. Feb 25, 2010 #12
    youre du is wrong.
    derivative of sin^2(x)=2*sinxcosx
  14. Feb 25, 2010 #13
    i'm going to show you my general method for doing odd powered sinxcosx integrals, it might help you in other problems.
    for some odd numbered natural numbers,k,and n

    [tex]\int \sin^{k}(x)\cos^{n}(x) dx [/tex]
    since [tex]\sin x \cos x = \frac{1}{2}\sin (2x) [/tex]
    and [tex]\sin^{2} x = \frac{1-\cos 2x}{2}[/tex]
    and [tex]\cos^{2} x = \frac{1+\cos 2x}{2}[/tex]
    then [tex]\int \sin^{k}(x)\cos^{n}(x) dx =\int \sin^{k-1}(x)\cos^{n-1}(x)*(\sin x \cos x) dx
    =\int \left(\frac{1-\cos 2x}{2})\right)^{\frac{k-1}{2}}*\left(\frac {1+\cos 2x}{2}\right)^{\frac{n-1}{2}} *(\frac{1}{2}\sin (2x))dx [/tex]
    pick [tex] u=\frac{1-\cos 2x}{2} , du=2\sin 2x dx [/tex]
    then [tex] \frac{1}{2^{ \frac{k+n}{2}+1 } } \int u^{\frac{k-1}{2}} * \left(2-u\right)^{\frac{n-1}{2}} du [/tex]
  15. Feb 25, 2010 #14
    Even though I corrected my du, I don't feel like I'm getting closer the answer...
    Thank you for trying to help the slow me!
  16. Feb 25, 2010 #15
    What do you have now that you've corrected the du?
  17. Feb 25, 2010 #16


    User Avatar
    Staff Emeritus
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    It would really help if you provided more detail on what exactly you did and fixed typos and errors in what you write before posting it. We can't read your mind, so we can only go by what appears here. When there are numerous typos, it's difficult to figure out what you actually meant and what was just a careless mistake. Frankly, it's kind of annoying to have to try figure out what you did or what you mean because you're not willing to put the effort into expressing yourself clearly.

    For example, you wrote
    Obviously, there's an error in the LaTeX code you wrote as the radical sign didn't appear correctly. What's less obvious is that you had a factor of 2 in front of the square root that didn't show up because of that error. The only reason I found this out was because the LaTeX appeared when I happened to hover the pointer over that part of the post.
    In this line, you fixed the LaTeX, and the two now appears. From our standpoint, the two just appeared mysteriously, and we wonder what exactly you did to get that two. So we can guess you did a substitution, replacing the dx by du which resulted in the two, but how are we supposed to know that? You didn't say what substitution you used, and you omitted any mention of dx and du in the integral. It could have just as easily been an algebra error, a mistaken trig identity, etc.

    Finally, you need to follow the normal conventions of mathematics, particularly the use of parentheses. Without having seen where your second line came from, some would interpret your second line to mean

    [tex]1-\frac{\sin^2 x}{2}\sqrt{1+\sin^2 x}[/tex]

    whereas you really (I think) meant

    [tex]\frac{1-\sin^2 x}{2\sqrt{1+\sin^2 x}}[/tex]

    Writing (1-sin2 x)/(2 sqrt(1+sin2 x)) would have accurately conveyed your intent.

    I hope you realize I'm not posting here just to criticize you or give you a hard time. In one sense, I just want you to be able to use this forum effectively. In a broader sense, I'm hoping you will understand how essential it is to be able to express your ideas accurately and clearly.
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