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Evaluate: int(sin^8x-cos^8x)/(1-2sin^2xc...

Evaluate: `int(sin^8x-cos^8x)/(1-2sin^2xcos^2x)\ dx`

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To evaluate the integral \[ I = \int \frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} \, dx, \] we can follow these steps: ...
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Knowledge Check

  • int(sin^8x-cos^8x)/(1-2sin^2xcos^2x)dx is equal to

    A
    `sin2X+C`
    B
    `-1/2sin2x+C`
    C
    `1/2sin2x+C`
    D
    `-sin2x+C`
  • int(sin^(2)x-cos^(2)x)/(sin^(2)xcos^(2)x)dx=

    A
    `tanx+cotx+c`
    B
    `tanx+cosecx+c`
    C
    `-tanx+cotx+c`
    D
    `tanx+secx+c`
  • If int(sin^(8)x-cos^(8)x)/(1-2sin^(2)xcos^(2)x)dx=A sin 2x+B, then A =

    A
    `-(1)/(2)`
    B
    `(1)/(2)`
    C
    `-1`
    D
    1
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