The integral ∫sec2⁡x(sec⁡x+tan⁡x)9/2dx equals (for some arbitrary constant K)

The integral sec2x(secx+tanx)9/2dx equals (for some arbitrary constant K)

  1. A

    1(secx+tanx)11/211117(secx+tanx)2+K

  2. B

    1(secx+tanx)11/211117(secx+tanx)2+K

  3. C

    1(secx+tanx)11/2111+17(secx+tanx)2+K

  4. D

    1(secx+tanx)11/2111+17(secx+tanx)2+K

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    Solution:

    We have,

    I=sec2x(secx+tanx)9/2dx

    Let secx+tanx=t.

    . Then, secxtanx=1/t and,

    secx(secx+tanx)dx=dt

     sccxdx=1tdt and, secx=12t+1t

     I=121tt+1tt9/2dt=121t9/2+1t13/2dt

     I=17t7/2111t11/2+k I=1t11/2t27+111+k I=-1(secx+tanx)11/2111+17(secx+tanx)2+k

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