Posted by **J** on Sunday, February 22, 2009 at 9:37pm.

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

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

- Calculus -
**drwls**, Monday, February 23, 2009 at 2:31am
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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