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Integration by Substitution

AP Calculus AB· difficulty 2/5

∫sin⁡3xcos⁡x dx=\displaystyle\int \sin^3 x \cos x\,dx =

  • A

    sin⁡3x3+C\dfrac{\sin^3 x}{3} + C

  • B

    −cos⁡4x4+C-\dfrac{\cos^4 x}{4} + C

  • C

    cos⁡4x4+C\dfrac{\cos^4 x}{4} + C

  • D

    sin⁡4x4+C\dfrac{\sin^4 x}{4} + C

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Explanation

u=sin⁡xu = \sin x. ∫u3 du=u4/4\int u^3\,du = u^4/4.

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