Integration help please?

How to integrate:


1 Answer

  • ­
    Lv 7
    1 decade ago
    Favorite Answer

    ∫ sin(x)cos(2x)dx

    Simplify product to sum formula:

    ∫ [sin(x + 2x) + sin(x - 2x)]/(2) dx

    ∫ [sin(3x) + sin(-x)]/(2) dx

    Use the opposite angle identity sin(-x )= -sin(x) to simplify:

    ∫ [sin(3x) - sin(x)]/(2)]dx

    ∫ (sin(3x))/(2) - (sin(x))/(2) dx

    The integral of a sum of expressions is equal to the sum of the integrals of each expression:

    ∫ sin(3x)/(2) dx - ∫ sin(x)/(2) dx

    Replace (3x) with u: u = 3x

    Now find an equation relating the value of dx to du: du = 3 dx

    Since du = 3 dx, then dx = du/3

    Replace the value found for dx in terms of u:

    ∫ (sinu/2)*(1/3) du

    ∫ (sinu/6) du

    The integral of (sinu/6) is -(cosu/6)

    Replace u=(3x) in the solved integral:

    -cos(3x)/6 + C

    The integral of -sin(x)/2 is -[-cos(x)/2:

    -cos(3x)/6 -[-cos(x)]/2 + C

    = -cos3x/6 + cosx/2 + C

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