Use Wallis's Formulas to evaluate the integral.
\[\int\limits_{0}^{\pi/2} (\sin(x))^5dx\]
according to the book it's the integral = \[\frac{2}{3}\frac{4}{5}\]
I guess that means it's = to 8/15?
I'm not familiar with the formula you're talking about, but I think it's similar to\[\int(1-\cos^4x)\sin x\,dx\]or some such variation, where simple trig substitutions are used.
here I have attached it....
yup 8/15 works.
I'm about to log out. @Ishaan94 Help please. :) I'm indebted to you.
:)
@badreferences here's a tutorial on wallis' formula if you're interested :) http://openstudy.com/updates/4fa2341ae4b029e9dc332b3f
@lgbasallote so apparently I can use this whenever the integral is from 0 to pi/2? The next question is kinda hairy, so I figured this might be of use...
\[\int\limits_{0}^{\pi/2} \frac{\cos(t)}{19+\sin(t)}\]
\[\frac{1}{19}\int\limits_{0}^{\pi/2} \cos(t) \sin(t)^{-1}\]
hmm you cant do that..
no? :(
\[\frac{1}{19} \cos t \sin t ^{-1} = \frac{\cos t}{19\sin t}\]
but good news is you can use u-sub :DDD u = 19 + sin t du = cos t dt
so you use ln thingies
I don't wanna! :p
so I guess it's going to be ln|u| + c ln|19+sin(t)| + c for 0 to pi/2
lol why is it so important to use wa;;os =)) and yup
wallis*
cuz I wanted to be cool :P
\[\ln|19 + \sin(\pi/2)| - \ln|19\sin(0)|\]
\[\ln|20| - \ln|19| = \ln|1| = 0\]
@lgbasallote I guess I did something wrong :(.
ln 20 - ln 19 is not ln 1
oh? haha :P
ln (20 - 19) = ln 1 though
yay that works! :D ty :D
Dang, I've never seen Wallis' formula before. I use reduction and substitution for these sorts of problems. Looking through all my textbooks, there isn't a single mention of Wallis either.
that sucks... I have Larson calculus 9th ed... I could send it to you if you want, it's a pdf.
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