Asked by Dave
Evaluate the integral from
[sqrt(pi/2), sqrt(pi)] of x^3*cos(x^2) by first making a substitution and then using integration by parts.
I let u = x^2 and du= 2x dx but then it doesn't equal that in the equation?
[sqrt(pi/2), sqrt(pi)] of x^3*cos(x^2) by first making a substitution and then using integration by parts.
I let u = x^2 and du= 2x dx but then it doesn't equal that in the equation?
Answers
Answered by
Steve
∫x^3 cos(x^2) dx
If you let u=x^2, you have du = 2x dx and the integral is
1/2 ∫[π/2,π] u cos(u) du
Now you can tackle that using integration by parts.
If you let u=x^2, you have du = 2x dx and the integral is
1/2 ∫[π/2,π] u cos(u) du
Now you can tackle that using integration by parts.
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