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    (Original post by Indeterminate)
    Looks as though problem 2 has now been dealt with

    A hint for problem 1
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    Notice that we can rewrite our integral as

    \displaystyle \int \dfrac{1}{2} \cdot \left(\dfrac{2\sin^2 \left(\frac{x}{2}\right) \cdot 2 \sin \left(\frac{x}{2}\right) \cdot \cos \left(\frac{x}{2}\right)}{2\cos^  2 \left(\frac{x}{2}\right)}\right) \left(\dfrac{1}{\sqrt{\cos^3 x + \cos^2 x + \cos x}}\right) \ dx

    and now there's some simplification that can be done

    Ahh thanks for the hint but I got to that part yesterday.. I simplified the left part to (1-cosx)^2 all over sinx

    I got stuck there :/
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    (Original post by Student403)
    Ahh thanks for the hint but I got to that part yesterday.. I simplified the left part to (1-cosx)^2 all over sinx

    I got stuck there :/
    That's absolutely fine!

    Here's another hint:
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    Weierstrass
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    (Original post by Indeterminate)
    Looks as though problem 2 has now been dealt with

    A hint for problem 1
    Spoiler:
    Show


    Notice that we can rewrite our integral as

    \displaystyle \int \dfrac{1}{2} \cdot \left(\dfrac{2\sin^2 \left(\frac{x}{2}\right) \cdot 2 \sin \left(\frac{x}{2}\right) \cdot \cos \left(\frac{x}{2}\right)}{2\cos^  2 \left(\frac{x}{2}\right)}\right) \left(\dfrac{1}{\sqrt{\cos^3 x + \cos^2 x + \cos x}}\right) \ dx

    and now there's some simplification that can be done

    Looks like it's down to me to wrap things up.

    Solution 1


    Letting I denote our integral, and continuing from the above, we get

    I = \displaystyle \dfrac{1}{2} \int  \dfrac{\sin x(1-\cos x)}{(1+\cos x)\sqrt{cos^3 x + \cos^2 x + \cos x}} \ dx

    u=\cos x gives us

    I = \displaystyle -\dfrac{1}{2} \int \dfrac{1-u}{(1+u)\sqrt{u^3+u^2 + u}} \ du

     = \displaystyle -\dfrac{1}{2} \int \dfrac{1-u^{-2}}{(u + u^{-1} + 2)\sqrt{u+u^{-1} + 1}} \ du

    Now let v^2 = u + u^{-1} + 2 to get

    I = \displaystyle \dfrac{1}{2} \int \left(\dfrac{1}{v}\right) \dfrac{2v}{1+v^2} \ dv

     = \arctan(v) + C

    And so our solution in terms of x is

    I = \arctan \sqrt{\cos x + \sec x + 1} + K

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    (Original post by Indeterminate)
    Looks like it's down to me to wrap things up.

    Solution 1


    Letting I denote our integral, and continuing from the above, we get

    I = \displaystyle \dfrac{1}{2} \int  \dfrac{\sin x(1-\cos x)}{(1+\cos x)\sqrt{cos^3 x + \cos^2 x + \cos x}} \ dx

    u=\cos x gives us

    I = \displaystyle -\dfrac{1}{2} \int \dfrac{1-u}{(1+u)\sqrt{u^3+u^2 + u}} \ du

     = \displaystyle -\dfrac{1}{2} \int \dfrac{1-u^{-2}}{(u + u^{-1} + 2)\sqrt{u+u^{-1} + 1}} \ du

    Now let v^2 = u + u^{-1} + 2 to get

    I = \displaystyle \dfrac{1}{2} \int \left(\dfrac{1}{v}\right) \dfrac{2v}{1+v^2} \ dv

     = \arctan(v) + C

    And so our solution in terms of x is

    I = \arctan \sqrt{\cos x + \sec x + 1} + K

    Oh my god I would never have got that :cry2:
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    (Original post by Student403)
    Oh my god I would never have got that :cry2:
    :console:

    But these integrals were far from easy and, given the lack of guidance, more difficult than any STEP II question on the topic
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    (Original post by Indeterminate)
    :console:

    But these integrals were far from easy and, given the lack of guidance, more difficult than any STEP II question on the topic
    Thanks for the encouragement Don't worry - it was definitely fun to try!

    I think one of my main weaknesses (which many STEP takers seem to be great at) is choosing my own substitutions to make. Ah well - hopefully that will improve with time and practice
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    (Original post by Student403)
    Thanks for the encouragement Don't worry - it was definitely fun to try!

    I think one of my main weaknesses (which many STEP takers seem to be great at) is choosing my own substitutions to make. Ah well - hopefully that will improve with time and practice
    Yes, practice makes perfect!
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    (Original post by Indeterminate)
    Yes, practice makes perfect!
    Thank you for the thread I will be sure to try harder in the next one!
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    (Original post by Student403)
    Thank you for the thread I will be sure to try harder in the next one!
    Glad you liked it Next time, I think I'll post a few problems that concern other areas of Maths (just for the sake of variety), as well as a couple of integrals!
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    (Original post by Indeterminate)
    Glad you liked it Next time, I think I'll post a few problems that concern other areas of Maths (just for the sake of variety), as well as a couple of integrals!
    Sounds good!
 
 
 
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