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    Hi,
    im struggling to integrate this using the given substitution, Can someone please help.

    Thanks.

    \int_{0}^{2\pi}\frac{1}{7-5sin(x)}

    using u=tan(\frac{x}{2})
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    Ever come across t-substitution?
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    (Original post by belle-belle)
    Hi,
    im struggling to integrate this using the given substitution, Can someone please help.

    Thanks.

    \int_{0}^{2\pi}\frac{1}{7-5sin(x)}

    using u=tan(\frac{x}{2})
    see if you can find ,sinx in terms of u. (sinx=2u/1+u^2.).

    and then sub and integrate
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    i haven't come across t-substitution, no, what does that involve?

    and if i use u with these limits they both come out as 0, is there anything i am doing wrong?
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    You can't "simply" do the t-sub (which is just the tan x/2 sub you're doing) in this case because of the limits.

    However, note that sin(pi-x) = sin(x) so that \displaystyle \int_0^{2\pi}\dfrac{1}{7-5\sin x}\,dx = 2 \int_0^{\pi} \dfrac{1}{7 - 5\sin x}\, dx (and in fact it also equals \displaystyle 4 \int_0^{\pi/2} \dfrac{1}{7 - 5\sin x}\, dx. Both of the latter integrals have "nice" limits after doing the substitution.

    Edit: It has been pointed out this is incorrect. See post #18.
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    (Original post by DFranklin)
    You can't "simply" do the t-sub (which is just the tan x/2 sub you're doing) in this case because of the limits.

    However, note that sin(pi-x) = sin(x) so that \displaystyle \int_0^{2\pi}\dfrac{1}{7-5\sin x}\,dx = 2 \int_0^{\pi} \dfrac{1}{7 - 5\sin x}\, dx (and in fact it also equals \displaystyle 4 \int_0^{\pi/2} \dfrac{1}{7 - 5\sin x}\, dx. Both of the latter integrals have "nice" limits after doing the substitution.
    Thanks very much for your help. However, we are still struggling with the integration. Please could you show us how to do this?
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    (Original post by belle-belle)
    Thanks very much for your help. However, we are still struggling with the integration. Please could you show us how to do this?
    ermm..we?
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    (Original post by rbnphlp)
    ermm..we?
    yeah there is more than one of us, we are doing some work in a group and are sturggling.
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    Post your working.

    Also try googling 't-substitution'

    e.g. http://www.ucl.ac.uk/Mathematics/geo...nt/fisub4.html
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    (Original post by DFranklin)
    Post your working.

    Also try googling 't-substitution'

    e.g. http://www.ucl.ac.uk/Mathematics/geo...nt/fisub4.html
     u=tan\left(\frac{x}{2}\right)


    dx = \frac{2du}{u^2+1}

    eventuallty manipulating so that:


     4\int_{0}^{\frac{\pi}{2}} \frac{10}{7-10u} du

    when evaluating at the limits, gives 49.96.
    which is not even close to what we're looking for.
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    Since u = tan(x/2), the limits will change. (I'm also unconvinced you'd done the manipulation right, but since you haven't posted working I can't check it easily).
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    (Original post by DFranklin)
    Since u = tan(x/2), the limits will change. (I'm also unconvinced you'd done the manipulation right, but since you haven't posted working I can't check it easily).
    what will the limits change to? Can you suggest an appropriate start point to help us manipulate it properly please?
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    http://en.wikipedia.org/wiki/Weierstrass_substitution
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    i think i have figured out how the function integrates.

    i got

     -{\Large \frac{arctan\left(\frac{5-7tan\left(\frac{x}{2}\right)}{2 \sqrt{6}}\right)}{\sqrt{6}}}

    but i dont know what the limits are to find the area.
    Please could you tell me the limits?
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    OK, tell me honestly. Did you do that yourself, or did you use a computer algebra system? Because you don't seem to actually understand what you've just done. You certainly haven't used the method in the link I gave you.
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    I'm not quite sure how to write maths using this program, but i've obtained the limits 0 and 1 after taking 4 out of the original equation, making the original limits 0 and Pi/2.

    my integral is then:

    2/(u^2 - 10u +7) du

    I attempted to factorise using the completing the square method but my answer still came out wrong. Can you please tell me what im doing wrong?
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    Community Assistant
    FP2 Tangent Half Angle Formulae.ppt

    You might find my half tangent Powerpoint helpful.
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    N.B. Amended version of incorrect earlier post. Sorry about that! I think this is now correct.

    You can't "simply" do the t-sub (which is just the tan x/2 sub you're doing) in this case because of the limits.

    However, note that sin(pi-x) = sin(x) so that \displaystyle \int_0^{2\pi}\dfrac{1}{7-5\sin x}\,dx = 2 \int_0^{\pi/2} \dfrac{1}{7 - 5\sin x}\, dx + 2 \int_{3\pi/2}^{2\pi} \dfrac{1}{7 - 5\sin x}\, dx and that this also equals \displaystyle 2 \int_{-\pi/2}^{\pi/2}\dfrac{1}{7 - 5\sin x}\, dx. This final integral has "nice" limits after doing the substitution.
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    (Original post by DFranklin)
    N.B. Amended version of incorrect earlier post. Sorry about that! I think this is now correct.

    You can't "simply" do the t-sub (which is just the tan x/2 sub you're doing) in this case because of the limits.

    However, note that sin(pi-x) = sin(x) so that \displaystyle \int_0^{2\pi}\dfrac{1}{7-5\sin x}\,dx = 2 \int_0^{\pi/2} \dfrac{1}{7 - 5\sin x}\, dx + 2 \int_{3\pi/2}^{2\pi} \dfrac{1}{7 - 5\sin x}\, dx and that this also equals \displaystyle 2 \int_{-\pi/2}^{\pi/2}\dfrac{1}{7 - 5\sin x}\, dx. This final integral has "nice" limits after doing the substitution.
    Thank you!!
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    (Original post by DFranklin)
    You can't "simply" do the t-sub (which is just the tan x/2 sub you're doing) in this case because of the limits.

    However, note that sin(pi-x) = sin(x) so that \displaystyle \int_0^{2\pi}\dfrac{1}{7-5\sin x}\,dx = 2 \int_0^{\pi} \dfrac{1}{7 - 5\sin x}\, dx (and in fact it also equals \displaystyle 4 \int_0^{\pi/2} \dfrac{1}{7 - 5\sin x}\, dx. Both of the latter integrals have "nice" limits after doing the substitution.

    Edit: It has been pointed out this is incorrect. See post #18.
    What's wrong with limits of 2pi and 0 ? Or is it harder to change them when you use the substitution?
 
 
 
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