Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cumbersome Trig integral

  1. Jun 20, 2011 #1
    So this is not an homework question, and I have solved the integral...

    Just it took alot of time and was very tiresome.

    A quick outline on how I did it. I mainly used two formulas

    [tex]\sin(A)cos(B)=\sin(A+B)+\sin(A-B)[/tex]

    [tex]\sin(A)\sin(B)\sin(C)=\frac{1}{4} \left[ \sin(A+B-C) - \sin(A-B-C) + \sin(A-B+C) - \sin(A+B+C) \right][/tex]

    A short outline, the steps omitted are those that take alot of time...


    [tex] I = \int\limits_{\pi /6}^{\pi /2} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2}dx} [/tex]
    [tex] I = \int\limits_{\pi /6}^{\pi /2} {{{\left[ {\sin \left( {2x} \right)\cos \left( {3x} \right)} \right]}^2}\sin \left( {2x} \right)dx} [/tex]
    [tex] I = \frac{1}{4}\int\limits_{\pi /6}^{\pi /2} {{{\left[ {\sin \left( {2x - 3x} \right) + \sin \left( {2x + 3x} \right)} \right]}^2}\sin \left( {2x} \right)} dx [/tex]
    [tex] I = \frac{1}{4}\int\limits_{\pi /6}^{\pi /2} {\left[ {\sin {{\left( x \right)}^2} - 2\sin \left( x \right)\sin \left( {5x} \right) + \sin {{\left( {5x} \right)}^2}} \right]\sin \left( {2x} \right)} dx [/tex]
    [tex] I = \frac{1}{{16}}\int\limits_{\pi /6}^{\pi /2} {6\sin \left( {2x} \right) - 3\sin \left( {4x} \right) - 2\sin \left( {6x} \right) + 3\sin \left( {8x} \right) - \sin \left( {12x} \right)dx} [/tex]
    [tex] I = \frac{1}{{16}}\left[ { - \frac{6}{2}\cos \left( {2x} \right) + \frac{3}{4}\cos \left( {4x} \right) + \frac{2}{6}\sin \left( {6x} \right) - \frac{3}{8}\cos \left( {8x} \right) + \cos \left( {12x} \right)} \right] + C[/tex]
    [tex] I = \frac{1}{{16}}\left[ {\left( { - 3\cos \left( \pi \right) + \frac{3}{4}\cos \left( {2\pi } \right) + \frac{1}{3}\sin \left( {3\pi } \right) - \frac{3}{8}\cos \left( {4\pi } \right) + \cos \left( {6\pi } \right)} \right) - \left( { - 3\cos \left( {\frac{\pi }{3}} \right) + \frac{3}{4}\cos \left( {\frac{{2\pi }}{3}} \right) + \frac{1}{3}\sin \left( \pi \right) - \frac{3}{8}\cos \left( {\frac{{4\pi }}{3}} \right) + \cos \left( {2\pi } \right)} \right)} \right] [/tex]
    [tex] I = \frac{1}{{16}}\left[ {\left( {3 + \frac{3}{4} + 0 - \frac{3}{8} + 1} \right) - \left( { - \frac{3}{2} - \frac{3}{8} + 0 + \frac{3}{8}\left( { - \frac{1}{2}} \right) + 1} \right)} \right] [/tex]
    [tex] I = \frac{1}{{16}}\left[ {\left( {\frac{{35}}{8}} \right) - \left( { - \frac{{11}}{{16}}} \right)} \right] = \frac{1}{{16}}\left[ {\frac{{70 + 11}}{{16}}} \right] = \frac{{81}}{{256}} = {\left( {\frac{3}{4}} \right)^4} [/tex]
    [tex] I = \int\limits_{\pi /6}^{\pi /2} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2}dx} = {\left( {\frac{3}{4}} \right)^4} [/tex]
    [tex] \sqrt {{3^4} + {4^3} - 1} = \sqrt {81 + 64 - 1} = \sqrt {144} = \fbox{12} [/tex]

    Step 4 to 5 took me forever to figure out. Had to "invent" the formula above, after many failed attempts at simplifying the expression.

    So anyone have a simpler, easier solution?
     
  2. jcsd
  3. Jun 20, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    An alternate method might be to apply the formulas

    [tex]\cos^2(3x)=\frac{1+\cos(6x)}{2}[/tex]

    and

    [tex]\sin^3(2x)=\frac{3\sin(2x)-\sin(6x)}{4}[/tex]

    This has the benifit that you don't need to work with exponents. But it's probably as much calculation...
     
  4. Jun 20, 2011 #3

    I like Serena

    User Avatar
    Homework Helper

  5. Jul 8, 2011 #4
    You've got a few errors in your integration. Integrating -2sin6x should be 1/3 cos6x and -sin12x should be 1/12 cos12x. The final answer seems alright surprisingly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cumbersome Trig integral
  1. Integration and trig (Replies: 3)

  2. Trig Integrals (Replies: 6)

  3. Integrals + Trig (Replies: 9)

Loading...