In = ∫0π4tannxdx, then limn→∞nIn 

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 Multiple Choice QuestionsMultiple Choice Questions

301.

x3sintan-1x41 + x8dx is equal to :

  • 14costan-1x4 + c

  • 14sintan-1x4 + c

  • - 14costan-1x4 + c

  • 14sec-1tan-1x4 + c


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302.

In0π4tannxdx, then limnnIn + In +2 equals :

  • 12

  • 1

  • zero


B.

1

 In = 0π4tannxdxIn + 2 = 0π4tann + 2xdxNow, In + In + 2 = 0π4tannx1 + tan2xdx                          = 0π4sec2xtannxdxLet        tanx = t  sec2xdx = dt In + In + 2 = 01tndt = tn + 1n + 10 1= nn + 1Hence, limnIn + In +2 = limnnn + 1             = limn11 + 1n = 1


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303.

If xfxdx = fx2, then f(x) is equal to :

  • ex

  • e- x

  • log(x)

  • ex22


304.

02x2dx is :

  • 2 - 2

  • 2 + 2

  • 2 - 1

  • - 2 - 3 + 5


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305.

0πcosxdx is equal to :

  • 12

  • - 2

  • 1

  • - 1


306.

sin2xsin3xsin5xdx is equal to :

  • 15logesin5x - 13logesin3x + c

  • 13logesin3x - 15logesin5x

  • 13logesin3x + 15logesin5x

  • - 12cos2x + 13logesin3x


307.

exlogsinx + cotxdx is equal to

  • excot(x) + c

  • exlog(sin(x)) + c

  • exlog(sin(x)) + tan(x) + c

  • ex + sin(x) + c


308.

- 1010loga + xa - xdx is equal to :

  • 0

  • - 2log(a + 10)

  • 2loga + 10a - 10

  • 2log(a + 10)


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309.

Define f(x) = 0xsintdt, x  0, Then :

  • f is increasing only in the interval 0, π2

  • f is decreasing in the interval 0, π

  • f attains maximum at x = π2

  • f attains minimum at x = π


310.

Let f(x) = sin2πx1 + π2. Then, fx + f- xdx is equal to :

  • 0

  • x + c

  • x2 - cosπx2π + c

  • x2 - sin2πx4π + c


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