Calculus
posted by J .
How do I evaluate this integral: x^9cos(x^5)dx
Hint: First make a substitution and then use integration by parts.

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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