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Antiderivatives and Indefinite Integrals

AP Calculus AB· difficulty 4/5

∫0π/2sin⁡2x dx=\displaystyle \int_0^{\pi/2} \sin^2 x \, dx =

  • A

    π4\frac{\pi}{4}

    check_circle
  • B

    π2\frac{\pi}{2}

  • C

    π8\frac{\pi}{8}

  • D

    11

Explanation

sin⁡2x=(1−cos⁡2x)/2\sin^2 x = (1-\cos 2x)/2, so the integral equals [x/2−(sin⁡2x)/4]0π/2=π/4[x/2 - (\sin 2x)/4]_0^{\pi/2} = \pi/4.

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