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anonymous

  • 5 years ago

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

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  1. watchmath
    • 5 years ago
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    Is that \(\sin(2x)\) or \(\sin^2(x)\)?

  2. anonymous
    • 5 years ago
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    former

  3. anonymous
    • 5 years ago
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    you workin on it?

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

  5. anonymous
    • 5 years ago
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    i cant get the answer..:-(

  6. watchmath
    • 5 years ago
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    \(\frac{1}{(5-u)(5+u)}=\frac{1}{10}(\frac{1}{5-u}+\frac{1}{5+u})\)

  7. anonymous
    • 5 years ago
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    how about \[du/5^{2}-u ^{2}=1/10 \log(u+5)/(u-5)\]

  8. watchmath
    • 5 years ago
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    yes, that is correct!

  9. anonymous
    • 5 years ago
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    i cant work it through... asked some other guys but had no luck

  10. watchmath
    • 5 years ago
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    well then your answer is \(\frac{1}{40}\ln|\frac{4\sin x-4\cos x+5}{4\sin x-4\cos x-5}|\) and then you plug in \(x=\pi/2\) and \(x=0\).

  11. anonymous
    • 5 years ago
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    my book says the answer is 1/10 log3

  12. watchmath
    • 5 years ago
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    \(\frac{1}{40}(\ln(9)-\ln(1/9))=\frac{1}{40}(4\ln(3))=\frac{1}{10}\ln 3\)

  13. anonymous
    • 5 years ago
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    just a little simplification would hav done it.. thanx a lot mate

  14. anonymous
    • 5 years ago
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    Seriously, 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

  15. anonymous
    • 5 years ago
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    he has got a pretty superior brain i guess...lol

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