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klimenkov Group Title

\[\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx\]

  • one month ago
  • one month ago

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  1. ganeshie8 Group Title
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    even/odd stuff ha ?

    • one month ago
  2. BSwan Group Title
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    2014 is order of derivative ?

    • one month ago
  3. hartnn Group Title
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    \(\Large \int_a^b f(x) dx= \int_a^bf(a+b-x)dx \) use this!

    • one month ago
  4. klimenkov Group Title
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    @BSwan 2014 is the power.

    • one month ago
  5. hartnn Group Title
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    after you have proved that your function is even

    • one month ago
  6. BSwan Group Title
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    aha cool :)

    • one month ago
  7. hartnn Group Title
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    \(\Large \int_{-a}^a f(x) dx = 2\int_0^a f(x)dx\) if f(x) is even function

    • one month ago
  8. hartnn Group Title
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    did u get what to do ?

    • one month ago
  9. klimenkov Group Title
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    \[2\int\limits_0^{\frac{\pi}2} \frac{\sin^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx\]What's next?

    • one month ago
  10. hartnn Group Title
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    \(\Large \int_a^b f(x) dx= \int_a^bf(a+b-x)dx\) replace x by pi/2 - x

    • one month ago
  11. hartnn Group Title
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    \(I = 2\int\limits_0^{\frac{\pi}2} \frac{\sin^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx ... ... (A)\) just giving a label, to be used later

    • one month ago
  12. klimenkov Group Title
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    \[2 \int\limits_0^{\frac{\pi}2} \frac{\cos^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx\]

    • one month ago
  13. klimenkov Group Title
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    Oops...

    • one month ago
  14. hartnn Group Title
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    \(I = 2 \int\limits_0^{\frac{\pi}2} \frac{\cos^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx ... ... (B)\) Add (A) and (B)

    • one month ago
  15. klimenkov Group Title
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    \[I = -2 \int\limits_0^{\frac{\pi}2} \frac{\cos^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx\]

    • one month ago
  16. hartnn Group Title
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    what ? why negative ?

    • one month ago
  17. hartnn Group Title
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    dx remains as dx

    • one month ago
  18. hartnn Group Title
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    i am NOT doing any substitution

    • one month ago
  19. klimenkov Group Title
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    Yeah, everything is okay. My fault. \[I + I = 2\int_0^{\frac\pi2}dx,\]\[I=\frac\pi2.\]Very nice, thank you.

    • one month ago
  20. hartnn Group Title
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    welcome ^_^

    • one month ago
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