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How do I evaluate this integral: x^9cos(x^5)dx

Hint: First make a substitution and then use integration by parts.

  • Calculus -

    If w = x^5, 5x^4 dx = dw
    The integral you want becomes the integral of
    (1/5) w cos w dw
    Now use integration by parts, with
    u = w dv = cosw dw
    du = dw v = sin w
    The integral becomes
    (1/5)[u v - Integral v du]
    = (1/5)[w sin w - integral of sin w dw]
    = (1/5)[w sin w + cos w]
    = (1/5) [x^5 sin(x^5) + cos(x^5)]

    Check my work. The method seems correct by my algebra is often sloppy.

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