∫02πsinx + sinxdx is equal to from Mathe

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 Multiple Choice QuestionsMultiple Choice Questions

451.

0πxdx1 + sinx is equal to :

  • - π

  • π2

  • π

  • None of these


452.

dxxx5 + 1 is equal to :

  • 15logx5x5 + 1 + c

  • 15logx5 + 1x5 + c

  • 15logx5x5 + 1 + c

  • None of these


453.

x + sinx1 +cosxdx is equal to

  • xtanx2 + c

  • xsec2x2 + c

  • logcosx2 + c

  • None of these


454.

0π8cos34θ is equal to

  • 53

  • 54

  • 13

  • 16


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455.

0π2cosx - sinx1 + cosx sinxdx is equal to

  • 0

  • π2

  • π4

  • π6


456.

dxxx7 + 1 is equal to

  • logx7x7 + 1 + c

  • 17logx7x7 + 1 + c

  • logx7 + 1x7 + c

  • 17logx7 + 1x7 + c


457.

- 111 - xdx is equal to

  • - 2

  • 0

  • 2

  • 4


458.

xexdx is equal to

  • 2x - ex - 4xex +c

  • 2x - 4x + 4ex +c

  • 2x + 4x + 4ex +c

  • 1 - 4x ex +c


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459.

dxx2 +2x + 2 is equal to

  • sin-1x +1 +c

  • sinh-1x +1 +c

  • tanh-1x +1 +c

  • tan-1x +1 +c


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460.

02πsinx +sinxdx is equal to

  • 4

  • 0

  • 1

  • 8


A.

4

We have, 02πsinx +sinxdx= 0πsinx +sinxdx + π2πsinx +sinxdx     Using 02afxdx = 0afxdx + 02afxdx = 0πsinx + sinxdx + π2πsinx - sinxdx= 0π2sinxdx +0= 20π2sinxdx = 2- cosx0π= - 2cosπ - cos0 = - 2- 1 - 1 = 4


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