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evaluate the integral by multiplying by a form of 1 and using a substitution if necessary to reduce to standard form

integral cos x / 1 - sin x dx

I have tried to work this and come up with this answer not sure if I did it correctly
-ln | sec x + tan x | + ln | sin x | +C

2007-03-15 21:17:49 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

well whenever I see 1-sinx or 1 - cos x, I expect to be able to use the 1-sin^2 x = cos^2 x identity. I would jsut multiply numberator and denominator by 1+ sin x, which gives (cos x + sinx cosx)/ 1-sin^2 x, then use that identity and get the very simple sec x+ sinx, which means u just have to find the integral of secant (which is ln(secx + tanx)). so it seems like the answer is just ln (secx + tanx) + cos x...looks different than yr answer a little. Maybe i made mistake dunno 15 yrs since i did integrals. But as general advice, before trying to work an integral especially when its some fraction, first see if you can simplify what you are integrating especially to get seperate terms, because seperate terms are always easier to integrate. im a little confused however because i dont see how substitution could be any simpler, and that was hinted at in the question.

2007-03-15 22:23:33 · answer #1 · answered by Anonymous · 0 0

I = ∫ cos x / (1 - sin x) dx
Let u = 1 - sin x
du = - cos x dx
I = - ∫ (1/ u ) du
I = - log u + c
I = - log (1 - sinx ) + c

2007-03-15 21:46:52 · answer #2 · answered by Como 7 · 1 0

f(x) = -x³ + 13x² - 36x = -x * ( x² - 13x + 36) = -x * ( x-4) * (x-9) enable f1(x) = x² - 13x + 36 . . . . . . . (-a million) -------- (0) -------- (4) ------- (7) -x =>........| + + + + + | - - - - - -- | - - - - - - | f1(x) =>....| + + + + + | + + + + +| - - - - - - | it proceed unfavorable until 9 -x * f1(x)...| + + + + + | - - - - - - -| + + + + | then ? | f(x) | dx = ?+f(x) (-a million,0) + ? -f(x) (0,4) + ? +f(x) (4,7) you wreck (-a million,7) into (-a million,0) ; (0,4) be conscious the (-) f(x) and (4,7)

2016-12-18 15:03:42 · answer #3 · answered by rocca 4 · 0 0

- log|sin(x) - 1 | + C is what i got.

2007-03-15 21:25:01 · answer #4 · answered by gjmb1960 7 · 1 0

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