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Easy looks integral. Help

  1. Feb 21, 2007 #1
    Hello I have this integral and I can't solve it

    [tex]\int\csc^6 x dx[/tex]

    Well i first start with trig identnity [tex]\csc x = \frac{1}{sen x}[/tex] so my first attempt to slove it looks like this

    [tex]\int\frac{dx}{(\sin^2 x)^3}[/tex]

    then i subsituted with [tex]\sin^2 x = \frac{1 - \cos 2x}{2}[/tex]

    and i get

    [tex]8\int\frac{dx}{(1-cos 2x)^3} [/tex] and i saw that i was going to

    nowhere so i try another way

    i took the function and use the trig identity to get this

    [tex]\int\csc^6 x dx[/tex] =[tex]\int \frac{\sec^6 x}{\tan^6 x} dx [/tex]

    [tex]u = \tan x[/tex] [tex]du = \sec^2 x[/tex]


    [tex]\int\frac{du^3}{tan^6 x}[/tex]

    and i get stucked again. any ideas?
    Last edited: Feb 21, 2007
  2. jcsd
  3. Feb 22, 2007 #2
    That looks most promising.

    If [tex]u = \tan{x}[/tex], then [tex]du = \sec^2{x}dx[/tex] (Don't forget the dx)

    Now, rewrite the integral as


    Now, think of an identity that relates [tex]\sec{x}[/tex] and [tex]\tan{x}[/tex].
    Last edited: Feb 22, 2007
  4. Feb 22, 2007 #3


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    Make [itex] \tan x =t [/itex] and then u'll get 3 simple integrals.
  5. Feb 22, 2007 #4


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    Here's the trick, if the power of csc function is even, you can use the u-substitution u = cot(x).
    And if the power of sec function is even, you can use the u-substitution u = tan(x)
    So, in your problem, the power of csc function is even, we use the u-substitution: u = cot(x)
    du = -csc2(x) dx, so your integral will become:
    [tex]\int \csc ^ 6 (x) dx = \int ( \csc ^ 4 (x) \times \csc ^ 2 (x) ) dx = \int {( \csc ^ 2 (x) )} ^ 2 \times \csc ^ 2 (x) dx = \int {\left( 1 + \cot ^ 2 (x) \right)} ^ 2 \times \csc ^ 2 (x) dx[/tex]
    [tex]= -\int {(1 + u ^ 2)} ^ 2 du = ...[/tex]
    Can you go from here? :)
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