If ∫sin2tan-11 - x1 + xdx =&nbs

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 Multiple Choice QuestionsMultiple Choice Questions

591.

The value of 01xdx by Trapezoidal rule taking x = 4 is

  • 0.34375

  • 0.5

  • 0.38387

  • 0.353367


592.

dxx4 - 1 is equal to

  • 14logx - 1x + 1 - 12tan-1x + C

  • logx - 1x + 1 + C

  • 14logx - 1x + 1 + 12tan-1x + C

  • logx - 1x + 1 - 12tan-1x + C


593.

If e= 1, e1 = 2.72, e2 = 7.39, e3 = 20.09, e4 = 54.60, then the value of 04exdx using Simpson's rule, will be

  • 5.387

  • 53.87

  • 52.78

  • 53.17


594.

- 12x3 - xdx is equal to

  • 11

  • 4

  • 114

  • 411


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595.

According to Simpson's rule, the value of 17dxx is

  • 1.358

  • 1.958

  • 1.625

  • 1.458


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596.

If sin2tan-11 - x1 + xdx = Asin-1x + Bx1 - x2 + C, then A + B is equal to

  • 10

  • 12

  • 1

  • - 12


C.

1

We have, sin2tan-11 - x1 + xdx = Asin-1x + Bx1 - x2 + CLet  I = sin2tan-11 - x1 + xdxPut x = cos2θ  dx = 2sinθ   I = - sin2tan-11 - cos2θ1 + cos2θ2sin2θ        = - sin2tan-12sin2θ2cos2θ2sin2θ        = - sin2tan-1tan2θ2sin2θ

        = - sin2θ2sin2θ        = - 2sin22θ        = - 1 - cos4θ        = - θ - sin4θ4 + C       = - 12cos-1x + 14 × 2sin2θcos2θ + C       = - 12π2 - sin-1x + 121 - x2 × x + C       = 12sin-1x + x21 - x2 + C - π4But I = Asin-1x + Bx1 - x2 + C A = 12 and B = 12Hence, A + B = 12 + 12 = 1


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597.

1e1logxdx is equal to

  • 1e

  • e

  • 21 - 1e

  • None of the above


598.

limx00x2sintdtx3 is equal to

  • 23

  • 13

  • 0


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599.

By trapezoidal rule, the approximate value of the integral 06dx1 + x2 is

  • 1.3128

  • 1.4108

  • 1.4218

  • None of these


600.

The value of the integral I = tanx + cotxdx, where x  0, π2, is

  • 2sin-1cosx - sinx + C

  • 2sin-1sinx - cosx + C

  • 2sin-1cosx + sinx + C

  • - 2sin-1sinx + cosx + C


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