anonymous
  • anonymous
An integration question: so the original problem was integral of (sinx)^3/ (cosx)^4, and I simplified this to integral of secx(tanx)^3, which is equal to int. of secx((secx)^2-1). This is where I am stuck. Any help would be much appreciated :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
How about instead of going the secant route...and I'm just tossing out a ball here rewrite to \[\int\limits_{?}^{?}(\sin x)^3 * 1/(\cos x )^4 = (\sin x )^2 * 1/(\cos x)^4 * (\sin x)\] then use the identity sin^2 x + cos^2 x = 1 to replace the first sin^2 x to get \[\int\limits_{?}^{?}1- (\cos x)^2 * 1/ (\cos x)^4 * \sin x\] The cos x's cancel and you get 1/(cos)^2 * sin x and then I'm not sure is that getting simpler? reuse the identity cos^2 = -sin^2 +1 which will give you \[\int\limits_{?}^{?}1/(\sin x)^2 +1 * (\sin x) / 1\]
anonymous
  • anonymous
nope now I'm stuck too...
dumbcow
  • dumbcow
\[\frac{\sin^{3} x}{\cos^{4} x} = \frac{\sin x(1-\cos^{2} x)}{\cos^{4} x}\] Let u = cos x du = -sin x \[\int\limits_{}^{}\frac{-(1-u^{2})}{u^{4}} = \int\limits_{}^{}\frac{u^{2}-1}{u^{4}} = \int\limits_{}^{}\frac{1}{u^{2}}-\frac{1}{u^{4}}\]

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anonymous
  • anonymous
Ahhh I was On the right track...I was look...but then I tried to keep being fancy instead of going for the u sub.
anonymous
  • anonymous
so I kept my first simplification: \[\int\limits_{?}^{?} \sec x \tan ^{3} x\] then used u = sec x du = sec x tan x for \[\int\limits_{?}^{?} (u ^{2}-1)du = 1/3 (u ^{3}) + u +C\] but thanks for the help

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