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Peter[Deuce]
United States93 Posts
http://img205.imageshack.us/img205/7389/integrate.jpg ∫(1+4cot(x))/(4-cot(x)) dx from pi/4 to pi/2 Some guy on yahoo answers solved it by going to http://www.wolframalpha.com/input/?i=∫(1+4cot(x))/(4-cot(x)) dx but I'm pretty sure you can solve it by hand. Edit: I did go to the wolframalpha site and looked up the "Show Steps" but it seemed far too complicated to actually be the logical way and human way to solve it.   | ||
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Cambium
United States16368 Posts
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micronesia
United States24651 Posts
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Peter[Deuce]
United States93 Posts
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hype[NZ]
Japan412 Posts
I did do this problem just now, and it's not too much work. One thing to try when you are stuck on these sorts of problems is to decompose the cotangents etc into their basic cosine and sine components, rearrange a bit and then you might see something useful  Good luck! | ||
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Peter[Deuce]
United States93 Posts
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Hiphopapotamus
United States121 Posts
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blue_arrow
1971 Posts
the point is to force everyone to learn the answer and learn about the simplicity and beauty of it all besides, any good teacher doesn't really give a damn about the marks | ||
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coltrane
Chile988 Posts
(u*v)` = u`*v + u*v` ∫(u*v)`dx = ∫(u`*v)dx + ∫(u*v`)dx u*v = ∫(u`*v)dx + ∫(u*v`)dx ∫(u`*v)dx = u*v - ∫(u*v`)dx | ||
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caldo149
United States469 Posts
though not necessarily in that order. Then check for pythagorean identity...? if you don't know: conjugate of 1+4cotx is 1-4cotx, conjugate of 4-cotx is 4+cotx I'm too lazy to actually try this right now... but i hope this helped. hfgl on the final.=) | ||
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Ichigo1234551
United States649 Posts
damn it i took 2 months of calculus and i forgot everything  | ||
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igotmyown
United States4291 Posts
1/sin(x)^2 dx = dy y^2+1=1/sin(x)^2 dx=dy*sin(x)^2 dx=dy*1/(1+y^2) ∫(1+4cot(x))/(4-cot(x)) dx from pi/4 to pi/2 ∫(1+4y)/(4-y)*dy/(1+y^2) from 1 to 0 (it's backwards) This can be solved via logarithms with 4-y, 1+yi, 1-yi, I forget what the method's called Or you can use clever trig substitutions | ||
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igotmyown
United States4291 Posts
∫(1+4cot(x))/(4-cot(x)) dx from pi/4 to pi/2 1+4*cos/(1+sin)/(4-cos/(1+sin)) (1+sin)+4*cos/(4+4sin-cos) cot(a)=1/4 ∫(cot(a)+cot(x))/(1-cot(a)*cot(x)) dx from pi/4 to pi/2 ∫(cot(a+x) dx from pi/4 to pi/2, where a=cot^(-1)(1/4) | ||
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mcgriddle
United States253 Posts
 helps me. | ||
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Hiphopapotamus
United States121 Posts
Multiply top and bottom by sinx and let u = 4sinx - cosx du = sinx + 4cosx dx -> ∫ du/u -> plug into log for answer | ||
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Mah Buckit!
Finland474 Posts
On August 13 2009 17:03 mcgriddle wrote: for tans / cotans i always put in terms of cos and sin helps me. My math teachers always said that cotans are bloody useless and so we never used them  | ||
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