trigonometric integrals

J

JMSNMSJTS

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Nov 6, 2008
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The problems is integrating (sin x)^3dx from to pi/4. Here's my work: (sin x)^2sin x = (1-cos^2 x)(sin x), u = cos x, du = -sin x
new upper bound is square root 2/2 and new lower bound is 1, new integral is -(1 - u^2)du
need upper and lower bounds to be changed: changed upper bound is now 1 and changed lower bound is square root 2/2
new integral is (1 - u^2)du
evaluate u - (u^3)/3 at new upper and lower bounds
my answer is 2/3 - (5square root 2)/12
book's answer is 2/3 - (5sqaure root 2)/6

Who's right?
 
\(\displaystyle \int_{0}^{\pi/4}sin^{3}(x)dx \ = \ \frac{2}{3}-\frac{5\sqrt2}{12}, \ Trusty \ TI-89.\)

\(\displaystyle Hence, \ your \ book \ is \ wrong \ as \ the \ TI-89 \ tells \ no \ lies.\)
 
Happens all the time. I once spent a whole evening trying to get a problem to jell with the answer in the back of the book only to find out I was right all along as the answer in the back of the book was a typo.
 
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