freckles
  • freckles
Made a calculus question for fun. \[\int\frac{5 \cos(x)-2 \sin(x)}{ 5 \sin(x)-2 \cos(x)} dx\]
Mathematics
jamiebookeater
  • jamiebookeater
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freckles
  • freckles
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.
ganeshie8
  • ganeshie8
I'm starting with \[a\cos x +b\sin x= C\cos(x +\phi)\]
anonymous
  • anonymous
@ganeshie8 can you help me with a question please :)

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freckles
  • freckles
err that is my way @ganeshie8 :p
freckles
  • freckles
So problem is too easy. :p
anonymous
  • anonymous
someone please help :(
anonymous
  • anonymous
freckles
  • freckles
You could mention your question in chat @geny55 .
anonymous
  • anonymous
is moisture supply abiotic or biotic?
freckles
  • freckles
Well this isn't really chat... but okay.
anonymous
  • anonymous
lol
anonymous
  • anonymous
so ??
freckles
  • freckles
I don't know.
freckles
  • freckles
You might mention your question though in chat by posting a link to your question .
ganeshie8
  • ganeshie8
\[\frac{5 \cos(x)-2 \sin(x)}{ 5 \sin(x)-2 \cos(x)} =\dfrac{\cos(x - \arctan(-2/5))}{ \cos(x-\arctan(-5/2))}\] ?
ganeshie8
  • ganeshie8
idk readily what to next..
anonymous
  • anonymous
@ganeshie8 is moisture supply abiotic or biotic?
freckles
  • freckles
\[\frac{\cos(a+x)}{\cos(b+x)} =\frac{\cos(x+b+(a-b))}{\cos(b+x)}\]
freckles
  • freckles
use sum identity on top
ganeshie8
  • ganeshie8
Wow! that simplifies nicely !
ganeshie8
  • ganeshie8
Let \(a = \arctan(-2/5)\) and \(b = \arctan(-5/2)\) \[\begin{align} &\frac{5 \cos(x)-2 \sin(x)}{ 5 \sin(x)-2 \cos(x)}\\~\\ & =\dfrac{\cos(x -a)}{\cos(x-b)}\\~\\ & =\dfrac{\cos((x -b)+(b-a))}{\cos(x-b)}\\~\\ &=\cos(b-a) - \sin(b-a)\tan(x-b) \end{align}\]
freckles
  • freckles
\[\int\limits \frac{ a \cos(x)+b \sin(x)}{c \cos(x)+d \sin(x)} dx \\ \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}\int\limits \frac{\cos(x+A)}{\cos(x+B)} dx \\ \sqrt{\frac{a^2+b^2}{c^2+d^2}} \int\limits \frac{\cos(x+B+(A-B)}{\cos(x+B)} dx \\ \sqrt{\frac{a^2+b^2}{c^2+d^2}} \int\limits \frac{\cos(x+B) \cos(A-B) -\sin(x+B)\sin(A-B)}{\cos(x+B)} dx \\ \sqrt{\frac{a^2+b^2}{c^2+d^2}} ( \int\limits \cos(A-B) dx-\sin(A-B) \int\limits \tan(x+B) dx) \\ \sqrt{\frac{a^2+b^2}{c^2+d^2} } (\cos(A-B)x+\sin(A-B) \ln|\cos(x+B)|) +K\] if I didn't make a mistake we can do a more general form in this way
freckles
  • freckles
\[\sqrt{ \frac{a^2+b^2}{c^2+d^2}} \cos(A-B) x+\sqrt{\frac{a^2+b^2}{c^2+d^2}}\sin(A-B) \ln|\cos(x+B)|+K \\ \text{ where } A=\arctan(\frac{b}{a}) \text{ and } B=\arctan(\frac{d}{c})\]
freckles
  • freckles
there are some restrictions on a,b,c,d
ganeshie8
  • ganeshie8
Nice! that wasn't so hard as it appeared to be..
freckles
  • freckles
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
freckles
  • freckles
other*
freckles
  • freckles
and no I didn't steal randomgirl 's thoughts :p
ganeshie8
  • ganeshie8
that is my fav trig identity!
freckles
  • freckles
mine too!! it is super cute.
ganeshie8
  • ganeshie8
wolf's solution isn't bad, it looks pretty long tho
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