How can we evaluate ∫ 0 π 1 1 + ( tan ⁡ x ) 2 d x

Question

How can we evaluate      ∫    0    π        1          1      +      (      tan      ⁡      x              )                  2                         d  x

Elisa Spears · Accepted Answer

The       2   is pretty much irrelevant. Notice  I  =      ∫    0    π              d      x              1      +      tan      ⁡      (      x              )                              2                                =      ∫    π    0              d      (              π        2            −      u      )              1      +      (      tan      ⁡      (              π        2            −      u      )              )                              2                                =      ∫    0    π              d      u              1      +      cot      ⁡      (      u              )                              2                              and      1          1      +      tan      ⁡      (      x              )                              2                                =            cos      ⁡      (      x              )                              2                                      cos      ⁡      (      x              )                              2                              +      sin      ⁡      (      x              )                              2                                ,        1          1      +      cot      ⁡      (      u              )                              2                                =            sin      ⁡      (      u              )                              2                                      sin      ⁡      (      u              )                              2                              +      cos      ⁡      (      u              )                              2                                .so  2  I  =      ∫    0    π              d      v              1      +      tan      ⁡      (      v              )                              2                                +      ∫    0    π              d      v              1      +      cot      ⁡      (      u              )                              2                                =      ∫    0    π        [                  cos        ⁡        (        v                  )                                    2                                                  cos        ⁡        (        v                  )                                    2                                      +        sin        ⁡        (        v                  )                                    2                                            +                  sin        ⁡        (        v                  )                                    2                                                  sin        ⁡        (        v                  )                                    2                                      +        cos        ⁡        (        v                  )                                    2                                            ]    d  v  =      ∫    0    π              cos      ⁡      (      v              )                              2                              +      sin      ⁡      (      v              )                              2                                      cos      ⁡      (      v              )                              2                              +      sin      ⁡      (      v              )                              2                                d  vwhich is       ∫    0    π    1  d  v  =  π. Ultimately, this is a symmetry argument.

How can we evaluate int_0^pi 1/(1+(tan x)^sqrt 2)dx

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Answer & Explanation

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