∫0xsintdt, where x ∈ 2nπ, 2n

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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


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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)


D.

4n + 1 - cos(x)

I = 0xsintdt

  = 02sintdt + 2xsintdt= 2n0πsintdt + 2xsintdt       x lies in either first or second quadrant= 2n- cost0π + - cost2x= 4n - cosx + 1


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403.

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

  • 13e

  • 3e

  • e3

  • 3e


404.

52525 - x23x4dx is equal to

  • π3

  • 2π3

  • π6

  • 5π6


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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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