∫52525 - x23x4dx is equal to from Mathematics

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 Multiple Choice QuestionsMultiple Choice Questions

401.

If 0log1 + x21 + x2dx = k01log1 + x1 + x2dx, then k is equal to

  • 4

  • 8

  • π

  • 2π


402.

0xsintdt, where x  2, 2n + 1π, n  N, is equal to

  • 4n - 1 - cos(x)

  • 4n - sin(x)

  • 4n - cos(x)

  • 4n + 1 - cos(x)


403.

Let I101exdx1 + x and I201x2dxex32 - x3 Then, I1I2 is equal to

  • 13e

  • 3e

  • e3

  • 3e


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404.

52525 - x23x4dx is equal to

  • π3

  • 2π3

  • π6

  • 5π6


A.

π3

I = 525 25 - x23x4dxLet x = 5sinθ  dx = 5cosθ I = π6π2 25 - 25sin2θ3x4sin4θ . 5cosθ

        = π6π253cos3θ . 5cosθ54sin4θ        = π6π2cot2θcsc2θ - 1       = π6π2cot2θcsc2θ - π6π2cot2θ       = π6π2cot2θcsc2θ - π6π2csc2θ - 1       = - cot3θ3 + cotθ + θπ6π2      = - 0 + 0 +π2 - - 333 + 3 + π6 = π3


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405.

dxcosx + 3sinx equals

  • 12logtanx2 + π12 + C

  • 13logtanx2 - π12 + C

  • logtanx2 + π6 + C

  • 12logtanx2 - π6 + C


406.

0π2sin2x . logtanxdx

  • 0

  • 2

  • 4

  • 7


407.

If x = 1cosx1 - cosx1 + sinxcosx1 + sinx + cosxsinxsinx1, then 0π4xdx is equal to

  • 14

  • 12

  • 0

  • - 14


408.

02πsinx + sinxdx is equal to

  • 0

  • 4

  • 8

  • 1


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409.

The value of 02x2dx, where [.] is the greatest integer function, is

  • 2 - 2

  • 2 + 2

  • 2 - 1

  • 2 - 2


410.

If l (m,n) = 01tm1 + tndt, then the expression for l (m, n) in terms of l (m + 1, n + 1) is

  • 2nm + 1 - nm + 1 . l m + 1, n - 1

  • nm + 1 . l m + 1, n - 1

  • 2nm + 1 + nm + 1 . l m + 1, n - 1

  • mn + 1 . l m + 1, n - 1


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