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klimenkov

  • 2 years ago

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

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  1. ganeshie8
    • 2 years ago
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    even/odd stuff ha ?

  2. anonymous
    • 2 years ago
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    2014 is order of derivative ?

  3. hartnn
    • 2 years ago
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    \(\Large \int_a^b f(x) dx= \int_a^bf(a+b-x)dx \) use this!

  4. klimenkov
    • 2 years ago
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    @BSwan 2014 is the power.

  5. hartnn
    • 2 years ago
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    after you have proved that your function is even

  6. anonymous
    • 2 years ago
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    aha cool :)

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

  8. hartnn
    • 2 years ago
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    did u get what to do ?

  9. klimenkov
    • 2 years ago
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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?

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

  11. hartnn
    • 2 years ago
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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

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

  13. klimenkov
    • 2 years ago
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    Oops...

  14. hartnn
    • 2 years ago
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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)

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

  16. hartnn
    • 2 years ago
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    what ? why negative ?

  17. hartnn
    • 2 years ago
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    dx remains as dx

  18. hartnn
    • 2 years ago
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    i am NOT doing any substitution

  19. klimenkov
    • 2 years ago
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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.

  20. hartnn
    • 2 years ago
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    welcome ^_^

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