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Integration by Substitution

AP Calculus AB· difficulty 3/5

∫0πsin⁡xcos⁡x dx=\displaystyle\int_0^{\pi}\sin x \cos x\,dx =

  • A

    π\pi

  • B

    −1-1

  • C

    11

  • D

    00

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Explanation

u=sin⁡xu = \sin x. ∫00u du=0\int_0^0 u\,du = 0. (Or note sin⁡xcos⁡x=12sin⁡(2x)\sin x \cos x = \tfrac{1}{2}\sin(2x); over [0,π][0, \pi] integrates to 00.)

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