Let  f(x)=7tan8⁡x+7tan6⁡x−3tan4⁡x−3tan2⁡x for all x∈−π2,π2. Then the correct expression is  

Let  f(x)=7tan8x+7tan6x3tan4x3tan2x for all xπ2,π2. Then the correct expression is  

  1. A

    0π/4xf(x)dx=112 and 0π/4f(x)dx=0

  2. B

    0π/4f(x)dx=2

  3. C

    0π/4xf(x)dx=16

  4. D

     0π/4f(x)dx=1

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    Solution:

    We have,

    f(x)=7tan6x1+tan2x3tan2x1+tan2x=7tan6x3tan2xsec2x

     0π/4f(x)dx=0π/47tan6x3tan2xsec2xdx

     0π/4f(x)dx=017t63t2dt, where t=tanx.

    0π/4xf(x)dx=xtan7xtan3x0π/40π/4tan7xtan3xdx=0π/4tan3xtan4x1dx

    =0π/4tan3xtan2x1tan2x+1dx=0π/4tan3xtan2x1sec2xdx

    =01t3t21dt, where t=tanx

    =t66t4401=112

     

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