2 Integrals (1 Viewer)

rolpsy · Aug 5, 2012

here's my attempt at number one...

$I = \int \frac{1 - \cot^{n-2}(x)}{\tan(x) + \cot^{n-1}(x)} \, \mathrm{d}x &= \int \frac{\cot(x) - \cot^{n-1}(x)}{1 + \cot^{n}(x)} \, \mathrm{d}x$

$\begin{align*}\textup{Let } u = \cot(x) \textup{, so } \mathrm{d}u &= -(1 + \cot^2(x)) \, \mathrm{d}x\\\mathrm{d}x &= -\frac{1}{1+u^2} \, \mathrm{d}u\end{align*}$

$\begin{align*}I &= \int \frac{u - u^{n-1}}{1 + u^n} \times -\frac{1}{1+u^2} \, \mathrm{d}u\\&= \int \frac{u^{n-1} - u}{(1+u^n)(1+u^2)} \, \mathrm{d}u\\&= \int \frac{u^{-1}}{1+u^2} - \frac{u^{-1}}{1+u^n} \, \mathrm{d}u\\&= \int \frac{1}{u(1+u^2)} - \frac{1}{u(1+u^n)} \, \mathrm{d}u\\&= \int \frac{1}{u} - \frac{u}{1+u^2} - \left (\frac{1}{u} - \frac{u^{n-1}}{1+u^n} \right ) \, \mathrm{d}u\\&= \int \frac{u^{n-1}}{1+u^n} - \frac{u}{1+u^2} \, \mathrm{d}u\\&= \frac{1}{n}\ln(1+u^n) - \frac{1}{2}\ln(1+u^2) + C\\&=\frac{1}{n}\ln(1 + \cot^n(x)) - \frac{1}{2}\ln(1 + \cot^2(x)) + C\end{align*}$

with hindsight I probably should've multiplied by cot²(x)/cot²(x) instead of cot(x)/cot(x) in the first line.. oh well.

asianese · Aug 5, 2012

2) Needs a giant partial fraction and very ugly coefficients - so not bothered haha

juantheron · Aug 12, 2012

Thanks rolpsy and asianese

(2)

asianese · Aug 12, 2012

Thanks for the solution, juantheron.

kayven · Aug 13, 2012

wow, what an ugly integral. BTW what did you use to type that working out?

2 Integrals (1 Viewer)

juantheron

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rolpsy

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asianese

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juantheron

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asianese

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kayven

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