∫0xxdx1 + cosαsinx, 0 < &al

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 Multiple Choice QuestionsMultiple Choice Questions

371.

116x2 + 9dx is equal to

  • 13tan-14x3 + c

  • 14tan-14x3 + c

  • 112tan-14x3 + c

  • 112tan-13x4 + c


372.

The value of 4711 - x2x2 + 11 - x2dx is

  • 1

  • 1/2

  • 3/2

  • 0


373.

tanx + cotxdx

  • 2tan-1tanxtanx + C

  • 2tan-1tanx - 12tanx + C

  • tanx2 . tan-1cotx + 12tanx + C

  • tanx2 . tan-1cotx + 12tanx + C


374.

x2xsinx + cosx2dx is equal to

  • sinx + cosxxsinx + cosx + C

  • xsinx - cosxxsinx + cosx + C

  • sinx - xcosxxsinx + cosx + C

  • None of these


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375.

0xxdx1 + cosαsinx, 0 < α < π is equal to

  • παsinα

  • παcosα

  • πα1 +sinα

  • πα1 - cosα


A.

παsinα

Let I = 0πxdx1 + cosαsinx                  ...i I = 0ππ - x1 + cosα . sinπ - x      ...iiOn adding Eqs. (i) and (ii), we get 2I = π0πdx1 + cosα . sinα        = π0πdx1 + cosα2tanx/21 + tan2x/2        = π0πsec2x/21 + tan2x/2 + cosα2tanx/2Put tanx/2 = t 1/2sec2xdx = dt 2I = π02dt1 + t2 + 2tcosα       I = π0dt1 + t2 + 2tcosα        = π0dt1 + cosα2 + sin2α        = πsinαtan-1t + cosαsinα        = παsinα


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376.

- π2π2cosx1 + exdx

  • 1

  • 0

  • - 1

  • None of these


377.

0π2dx1 + tanx is equal to

  • π

  • π2

  • π3

  • π4


378.

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


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379.

The value of 0πlog1 +cosxdx is

  • - π2log2

  • πlog12

  • πlog2

  • π2log2


380.

The value of 344 - xx - 3dx is

  • π16

  • π8

  • π4

  • π2


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