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Math Help - Integrals Involving Powers of Trig Functions.

  1. #1
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    Integrals Involving Powers of Trig Functions.

    Well new subject in my Calculus 2 class, and I do not understand why I was wrong on my homework. First I'll show you my work and then I'll show the correct answer from wolfram which uses a formula (reduction Formula) I have yet to learn. I only used wolfram to check the answer.

    The Start,

    \int sin^5(x)cos^8(x)dx

    Break them up:
    \int sin^4(x)cos^8(x)sin(x)dx

    Break sin up even further so I can use Trig Identities:
    \int (sin^2(x))^2cos^8(x)sin(x)dx

    Replace sin squared with a trig identity:
    \int (1-cos^2(x))^2cos^8(x)sin(x)dx

    Substitution:
    u=cos(x)

    du=-sin(x)dx

    Replace cos(x) with u:
    -\int (1-u^2)^2u^8(x)du

    FOIL:
    u^8(1-u^2)(1-u^2)

    (1-u^2-u^2+u^4)

    u^8(1-2u^2+u^4)

    (u^8-2u^{10}+u^{12})

    Enter in Newly FOIL'ed equation:
    -\int (u^8-2u^{10}+u^{12})du

    Integrate:
    -(\frac{{u}^{9}}{9}-2\frac{{u}^{11}}{11}+\frac{{u}^{13}}{13})+C

    Revert 'u' back to cos(x):
    (-\frac{1}{9}cos^{9}(x)+\frac{2}{11}cos^{11}(x)-\frac{1}{13}cos^{13}(x))+C

    Now in Short from Wolfram Alpha (the correct answer):

    \frac{-(cos^9(x) (-540 cos(2 x)+99 cos(4 x)+505))}{10296} +constant

    Where did I go wrong? How do I do this integral without referring to reduction formula?
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  2. #2
    MHF Contributor harish21's Avatar
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    You have not done anything wrong. All you need is a little more simplification to get to the desired result.

    (-\frac{1}{9}cos^{9}(x)+\frac{2}{11}cos^{11}(x)-\frac{1}{13}cos^{13}(x))+C

     = \dfrac{-143cos^{9}x+234 cos^{11}x-99cos^{13}x}{1287}+C

    Taking -cos^{9}x as a common term,if you simplify the terms further, you will get the value you have stated at the bottom.
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  3. #3
    Senior Member Educated's Avatar
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    Quote Originally Posted by Zanderist View Post
    (-\frac{1}{9}cos^{9}(x)+\frac{2}{11}cos^{11}(x)-\frac{1}{13}cos^{13}(x))+C

    Now in Short from Wolfram Alpha (the correct answer):

    \frac{-(cos^9(x) (-540 cos(2 x)+99 cos(4 x)+505))}{10296} + \text{Constant}
    You will see that if you type in -\frac{1}{9} \cos^{9}x+\frac{2}{11} \cos^{11}x-\frac{1}{13} \cos^{13}x+C into Wolfram|Alpha and scroll down to the "alternate forms" section, you will see that there will be the equation in the form that came from Wolfram|Alpha's integration.
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