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anonymous
 5 years ago
integrate from 0 to pi/2 (sinx + cosx)/(9+16sin2x) dx
anonymous
 5 years ago
integrate from 0 to pi/2 (sinx + cosx)/(9+16sin2x) dx

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.2Is that \(\sin(2x)\) or \(\sin^2(x)\)?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.2This is very tricky. Notice that \([4(\sin(x)\cos(x))]^2=16(\sin^2(x)+\cos^2(x)\sin(2x))=1616\sin(2x)\). So the denominator of the integreal is equal to \(25(4(\sin x  \cos x))^2=[5(4\sin x 4\cos x )]\cdot [5+(4\sin x4\cos x)]\) Now let \(u=4\sin x4\cos x\). Then \(du=4(\sin x+\cos x)\,dx\). So the integral is equivalent to \[\frac{1}{4}\int \frac{du}{(5u)(5+u)}\] Then you may continue from there :).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i cant get the answer..:(

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.2\(\frac{1}{(5u)(5+u)}=\frac{1}{10}(\frac{1}{5u}+\frac{1}{5+u})\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how about \[du/5^{2}u ^{2}=1/10 \log(u+5)/(u5)\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.2yes, that is correct!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i cant work it through... asked some other guys but had no luck

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.2well then your answer is \(\frac{1}{40}\ln\frac{4\sin x4\cos x+5}{4\sin x4\cos x5}\) and then you plug in \(x=\pi/2\) and \(x=0\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my book says the answer is 1/10 log3

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.2\(\frac{1}{40}(\ln(9)\ln(1/9))=\frac{1}{40}(4\ln(3))=\frac{1}{10}\ln 3\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just a little simplification would hav done it.. thanx a lot mate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Seriously, watchmath, where did you even think to rewrite it like that? I understand how to do the partial fractions/whatever. But I would never think to come up with something like the to rewrite it o.o

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0he has got a pretty superior brain i guess...lol
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