∫02πsin6xcos5xdx  = ? from Mathematics Int

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 Multiple Choice QuestionsMultiple Choice Questions

651.

0π2 dx1 + tan3x = ?

  • π

  • π2

  • π4

  • 3π2


652.

- 11 coshx1 +e2xdx = ?

  • 0

  • 1

  • e2 - 12e

  • e2 + 22e


653.

If ex - 1ex + 1dx = fx + c, then fx is equal to

  • 2logex - 1

  • 2loge2x - 1

  • 2logex + 1 - x

  • 2loge2x + 1


654.

tan-11 - x1 + xdx = ?

  • 12xcos-1x - 1 - x2 +c

  • 12xcos-1x + 1 - x2 +c

  • 12xsin-1x - 1 - x2 +c

  • 12xsin-1x + 1 - x2 +c


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655.

sinx + 8cosx4sinx + 6cosxdx = ?

  • x + 12log4sinx + 6cosx + c

  • 2x + log2sinx + 3cosx + c

  • x + log2sinx + 3cosx + c

  • 12log4sinx + 6cosx + c


656.

If ft = - tte- x2dx, then limtft = ?

  • 1

  • 12

  • 0

  • - 1


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657.

02πsin6xcos5xdx  = ?

  • 2π

  • π2

  • 0

  • - π


C.

0

Let I = 02πsin6xcos5xdx       I =  20πsin6xcos5xdx      f(2π - x) = fxLet   f(x) = sin6xcos5x              f(π - x) = - fxfπ - x = sin6π - xcos5π - x               = sin6x- cos5x                = - sin6xcos5x                  = - fx             I = 0


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658.

If ex1 - sinx1 - cosxdx = fx + constant, then f(x) is equal to

  • excotx2 + c

  • e-xcotx2 + c

  • - excotx2 + c

  • - e- xcotx2 + c


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659.

If Inxnecxdx for n  1, then cIn + n . In - 1 is equal to

  • xnecx

  • xn

  • ecx

  • xn + ecx


660.

If ex1 + x . sec2xexdx = f(x) + constant, then f(x) is equal to

  • cosxex

  • sinxex

  • 2tan-1x

  • tanxex


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