∫0π2dx1 + tanx is equal to from Mathematic

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 Multiple Choice QuestionsMultiple Choice Questions

661.

- π2π2cosx1 + exdx

  • 1

  • 0

  • - 1

  • None of these


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662.

0π2dx1 + tanx is equal to

  • π

  • π2

  • π3

  • π4


D.

π4

Given, I = 0π2dx1 + tanx           I = 0π2cosxsinx + cosxdx            ...i           I = 0π2cosπ2 - xsinπ2 - x + cosπ2 - xdx             = 0π2sinxcosx + sinxdx           ...iiOn adding Eqs. (i) and (ii), we get    2I = 0π2sinx + cosxsinx + cosxdx        = 0π2dx = π2 I = π4


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663.

By Simpson rule taking n = 4, the value of the integral 0111 + x2dx is equal to

  • 0.788

  • 0.781

  • 0.785

  • None of the above


664.

The value of 0πlog1 +cosxdx is

  • - π2log2

  • πlog12

  • πlog2

  • π2log2


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665.

The value of 344 - xx - 3dx is

  • π16

  • π8

  • π4

  • π2


666.

The value of dxxxn + 1 is

  • 1nlogxnxn + 1 + C

  • logxn + 1xn + C

  • 1nlogxn + 1xn + C

  • logxnxn + 1 + C


667.

The value of coslogxdx is

  • 12sinlogx + coslogx + C

  • x2sinlogx + coslogx + C

  • x2sinlogx - coslogx + C

  • 12sinlogx - coslogx + C


668.

The value of ex1 + sinx1 + cosxdx is

  • 12exsecx2 + C

  • exsecx2 + C

  • 12extanx2 + C

  • extanx2 + C


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669.

The value of 13sinx - cosx + 3dx is

  • logtanx2 + 12tanx2 + 1 + C

  • 12log2tanx2 + 1tanx2 + 1 + C

  • log2tanx2 + 1tanx2 + 1 + C

  • 2log2tanx2 + 1tanx2 + 1 + C


670.

The value of sin2xsin4x + cos4xdx is

  • tan-1cot2x + C

  • tan-1cos2x + C

  • tan-1sin2x + C

  • tan-1tan2x + C


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