[mage lang="" source="flickr"]stewart calculus 6e answers[/mage]

**Can you help me figure out this difficult integration problem?**

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*

*If f is continuous on [0, pi], use the substitution [u = (pi - x)] to show that*

the integral from 0 to pi of [(x)(f(sin x))dx]

is equal to

[pi/2] times the integral from 0 to pi of [f(sin x)dx].

It’s really difficult to type these integrals onto Yahoo Answers, so it might help if you rewrite the question on paper. I’ve been trying to figure it out for some time now, and any hints would be wonderful.

Thanks.

(James Stewart’s “Calculus” 6E, Section 5.5, Question 86)

After the substitution you are working on the integral from 0 to pi of

(pi – u) f [sin( pi - u)] du. But sin (pi – u) = sin pi * cos u – cos pi * sin u which is just sin u

You eventually find that your original integral is equal to its negative plus pi times the integral of f(sin x) dx.

rearranging gives you the desired results.