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freckles

  • one year ago

Made a calculus question for fun. \[\int\frac{5 \cos(x)-2 \sin(x)}{ 5 \sin(x)-2 \cos(x)} dx\]

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  1. freckles
    • one year ago
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    By the way I'm not sure if my way is the hard way or not but I like it and I think it is fun.

  2. ganeshie8
    • one year ago
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    I'm starting with \[a\cos x +b\sin x= C\cos(x +\phi)\]

  3. anonymous
    • one year ago
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    @ganeshie8 can you help me with a question please :)

  4. freckles
    • one year ago
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    err that is my way @ganeshie8 :p

  5. freckles
    • one year ago
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    So problem is too easy. :p

  6. anonymous
    • one year ago
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    someone please help :(

  7. anonymous
    • one year ago
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    @ganeshie8 ? @freckles ?

  8. freckles
    • one year ago
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    You could mention your question in chat @geny55 .

  9. anonymous
    • one year ago
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    is moisture supply abiotic or biotic?

  10. freckles
    • one year ago
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    Well this isn't really chat... but okay.

  11. anonymous
    • one year ago
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    lol

  12. anonymous
    • one year ago
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    so ??

  13. freckles
    • one year ago
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    I don't know.

  14. freckles
    • one year ago
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    You might mention your question though in chat by posting a link to your question .

  15. ganeshie8
    • one year ago
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    \[\frac{5 \cos(x)-2 \sin(x)}{ 5 \sin(x)-2 \cos(x)} =\dfrac{\cos(x - \arctan(-2/5))}{ \cos(x-\arctan(-5/2))}\] ?

  16. ganeshie8
    • one year ago
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    idk readily what to next..

  17. anonymous
    • one year ago
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    @ganeshie8 is moisture supply abiotic or biotic?

  18. freckles
    • one year ago
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    \[\frac{\cos(a+x)}{\cos(b+x)} =\frac{\cos(x+b+(a-b))}{\cos(b+x)}\]

  19. freckles
    • one year ago
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    use sum identity on top

  20. ganeshie8
    • one year ago
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    Wow! that simplifies nicely !

  21. ganeshie8
    • one year ago
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    Let \(a = \arctan(-2/5)\) and \(b = \arctan(-5/2)\) \[\begin{align} &\frac{5 \cos(x)-2 \sin(x)}{ 5 \sin(x)-2 \cos(x)}\\~\\ & =\dfrac{\cos(x -a)}{\cos(x-b)}\\~\\ & =\dfrac{\cos((x -b)+(b-a))}{\cos(x-b)}\\~\\ &=\cos(b-a) - \sin(b-a)\tan(x-b) \end{align}\]

  22. freckles
    • one year ago
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    \[\int\limits \frac{ a \cos(x)+b \sin(x)}{c \cos(x)+d \sin(x)} dx \\ \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}\int\limits \frac{\cos(x+A)}{\cos(x+B)} dx \\ \sqrt{\frac{a^2+b^2}{c^2+d^2}} \int\limits \frac{\cos(x+B+(A-B)}{\cos(x+B)} dx \\ \sqrt{\frac{a^2+b^2}{c^2+d^2}} \int\limits \frac{\cos(x+B) \cos(A-B) -\sin(x+B)\sin(A-B)}{\cos(x+B)} dx \\ \sqrt{\frac{a^2+b^2}{c^2+d^2}} ( \int\limits \cos(A-B) dx-\sin(A-B) \int\limits \tan(x+B) dx) \\ \sqrt{\frac{a^2+b^2}{c^2+d^2} } (\cos(A-B)x+\sin(A-B) \ln|\cos(x+B)|) +K\] if I didn't make a mistake we can do a more general form in this way

  23. freckles
    • one year ago
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    \[\sqrt{ \frac{a^2+b^2}{c^2+d^2}} \cos(A-B) x+\sqrt{\frac{a^2+b^2}{c^2+d^2}}\sin(A-B) \ln|\cos(x+B)|+K \\ \text{ where } A=\arctan(\frac{b}{a}) \text{ and } B=\arctan(\frac{d}{c})\]

  24. freckles
    • one year ago
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    there are some restrictions on a,b,c,d

  25. ganeshie8
    • one year ago
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    Nice! that wasn't so hard as it appeared to be..

  26. freckles
    • one year ago
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    There are only ways people presented here but I was very much in love with my way http://math.stackexchange.com/questions/1219016/indefinite-integral-with-sin-and-cos/1219078#1219078

  27. freckles
    • one year ago
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    other*

  28. freckles
    • one year ago
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    and no I didn't steal randomgirl 's thoughts :p

  29. ganeshie8
    • one year ago
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    that is my fav trig identity!

  30. freckles
    • one year ago
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    mine too!! it is super cute.

  31. ganeshie8
    • one year ago
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    wolf's solution isn't bad, it looks pretty long tho

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