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I'm having trouble finding out how to solve this integral.
http://hosted2.webwork.rochester.edu...256128eae1.png
I know the correct answer is: 0.235619, but I want to know if my work is correct. So far, whenever I work it out, I don't get the right answer.
From the integral,
((1/2)(1-cos10x))^2
= (1/4) integral (1-cos10x)^2
=(1/4) integral (1-2cos10x * (cos10x)^2
=(1/4) integral 1-2cos10x * ((1/2)1+cos20x)
From there, I plugged in the numbers, but I'm not getting the correct answer. Help?
$\displaystyle \sin ^4 5x = \frac{(1-\cos 10x)^2}{4} = \frac{1-2\cos 10x +\cos ^2 10x}{4} = \frac{1-2\cos 10x + \frac{\cos 20x +1}{2} }{4} $Quote:
Originally Posted by Kimmy2
$\displaystyle \sin ^4 5x = \frac{2 - 4\cos 10x + \cos 20x +1 }{8} $
$\displaystyle \sin ^4 5x = \frac{3 -4\cos 10x + \cos 20x }{8} $
when you integrate it you will get
$\displaystyle \frac{3\pi}{5\cdot 8}= \frac{3\pi}{40}$
Could you please explain how you arrived to your integral?
Thanks!
$\displaystyle \int _{0}^{\frac{\pi}{5}} \sin ^4 5x \cdot dx $Quote:
Originally Posted by Kimmy2
$\displaystyle \int_{0}^{\frac{\pi}{5}} \frac{3-4\cos 10x + \cos 20x }{8} \cdot dx $
$\displaystyle \left(\frac{1}{8}\right) \left( 3x - \frac{4\sin 10x}{10} + \frac{\sin 20x}{20} \right)\mid^{\frac{\pi}{5}}_{0} $
$\displaystyle \left(\frac{1}{8}\right) \left( \frac{3\pi}{5} - \frac{4\sin 2\pi}{10} + \frac{\sin 4\pi}{20} \right)-\left(\frac{1}{8}\right) \left( 3(0) - \frac{4\sin (0)}{10} + \frac{\sin (0)}{20} \right) $
$\displaystyle \left(\frac{1}{8}\right) \left( \frac{3\pi}{5}\right)$
