∫ 0 2 π 1 sin 4 ⁡ x + cos 4 ⁡ x 1) I've tried multiplying by sec 4 ⁡ x and doing u substitution, but It used crazy amounts of upper level maths2) I've tried double/half angle identities to simplify down to 1 1 − 1 2 sin 2 ⁡ 2 x 3) I've mixed strategies and came up with 4 cos ⁡ 2 x 4 cos ⁡ 2 x − cos ⁡ 3 x cos ⁡ x

Question

∫    0          2      π            1                  sin        4            ⁡      x      +              cos        4            ⁡      x      1) I&#039;ve tried multiplying by       sec    4    ⁡  x and doing u substitution, but It used crazy amounts of upper level maths2) I&#039;ve tried double/half angle identities to simplify down to             1              1        −                  1          2                          sin          2                ⁡        2        x            3) I&#039;ve mixed strategies and came up with                     4        cos        ⁡        2        x                    4        cos        ⁡        2        x        −        cos        ⁡        3        x        cos        ⁡        x

Saige Barton · Accepted Answer

Step 1The integral works out quite nicely by converting it to a contour integral over the unit circle in the complex plane and then using residues.Let   z  =      e          i      x       ,   d  x  =  d  z      /    (  i  z  ) and get by using the binomial theorem, that the integral is equal to the following contour integral:  −  i  8      ∮                  |            z              |            =      1                  d      z        z        1                  z        4            +      6      +              z                  −          4                      =  −  i  8      ∮                  |            z              |            =      1        d  z            z      3                      z        8            +      6              z        4            +      1      The poles of the integrand satisfy       z    4    =  −      (    3    ±    2          2        )   . Thus, there are four poles outside the unit circle (+) and four inside (-). We need only consider those inside. Note, however, that the expression for evaluating the residues at these poles is            z      3              8              z        7            +      24              z        3              =      1    8        1                  z        4            +      3      Note also that, for each of the four poles inside the unit circle,       z    4    +  3  =  2      2   . Therefore, by the residue theorem, the value of the integral is  i  2  π  (  −  i  8  )      1    8        4          2              2              =  2      2    π

Lorenzo Aguilar · Accepted Answer

Step 1Alternative solution: factor out a cos4x from the denominator and substitute       cos    4    ⁡      x  . Also, use symmetry to express the integral as being 4 times the integral over   [  0  ,  π      /    2  ] . You end up having  4      ∫    0          π              /            2                  d      x                      cos        4            ⁡              x            +              sin        4            ⁡              x              =  4      ∫    0          ∞        d  y            1      +              y        2                    1      +              y        4            Conveniently, we have            1      +              y        2                    1      +              y        4              =      1          2              y        2            +      2              2            y      +      2        +      1          2              y        2            −      2              2            y      +      2        =      1          1      +      (      1      +              2            y              )        2              +      1          1      +      (      1      −              2            y              )        2            Therefore the integral is  4      1          2                  [      arctan      ⁡              (        1        +                  2                y        )            −      arctan      ⁡              (        1        −                  2                y        )            ]        0          ∞        =  2      2        [          (              π        2            −                        −          π                2            )        −          (              π        4            −              π        4            )        ]    =  2      2    π

int_0^(2pi)1/(sin^4x+cos^4x); 1) I've tried multiplying by sec^4x and doing u substitution, but It used crazy amounts of upper level maths; 2) I've tried double/half angle identities to simplify down to 1/(1-1/2sin^2 2x); 3) I've mixed strategies and came up with (4cos2x)/(4 cos 2x-cos 3x cos x)

Answered question

Answer & Explanation

New Questions in Upper Level Math