∫0π2200sinx + 100cosxsinx + cosxdx&nbs

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 Multiple Choice QuestionsMultiple Choice Questions

921.

If fx = 1x23x2t - 3f'tdt, then f't, then f'(3) is equal to

  • - 1 2

  • - 13

  • 12

  • 13


922.

dxx + 100x + 99 = fx + c  fx

  • 2(x + 100)1/2

  • 3(x + 100)1/2

  • 2tan-1x + 99

  • 2tan-1x + 100


923.

3 - x21 - 2x + x2exdx = exfx + c  fx

  • 1 + x1 - x

  • 1 - x1 + x

  • 1 - xx - 1

  • x - 11 + x


924.

cotxsinxcosxdx = - fx + c  fx

  • 2tanx

  • - 2tanx

  • - 2cotx

  • 2cotx


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925.

- π2π2log2 - sinθ2 + sinθ is equal to

  • 0

  • 1

  • 2

  • - 1


926.

022x - 22x - x2dx is equal to

  • 0

  • 2

  • 3

  • 4


927.

If sinxcosx1 + cosxdx = f(x) + c, then f(x) is equal to

  • log1 + cosxcosx

  • logcosx1 + cosx

  • logsinx1 + sinx

  • log1 + sinxsinx


928.

x49tan-1x501 + x100dx = ktan-1x502 +c, then k is equal to

  • 150

  • - 150

  • 1100

  • - 1100


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929.

0π2200sinx + 100cosxsinx + cosxdx is equal to

  • 50π

  • 25π

  • 75π

  • 150π


C.

75π

Let   I = 0π2200sinx + 100cosxsinx + cosxdx          = 1000π2sinx + cosx + sinxsinx + cosxdx          = 1000π21dx + 0π2sinxsinx + cosxdxLet I1 = 0π2sinxsinx + cosxdx       ...i         = 0π2sinπ2 - xsinπ2 - x + cosπ2 - x I1 = 0π2cosxsinx + cosxdx       ...iiOn adding Eqs. (i) and (ii), we get    2I1 = 0π21dx = π2 I1 = π4   I = 100π2 +π4 = 100 × 3π4         = 75π


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930.

0πθsinθ1 + cos2θ is equal to

  • π22

  • π23

  • π2

  • π24


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