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Good Evening. I am having a problem with a step in the calculation of the following integral:

$\int_{0}^x \frac{dt}{\sin^2 t +\cos t} =\int_{0}^x \frac{dt}{-\cos^2 t +\cos t+1} =-\int_{0}^x \frac{dt}{\cos^2 t -\cos t-1} $

Let $X=\cos t$

$X^2-X-1=0 \Leftrightarrow X=\frac{1-\sqrt{5}}{2} \mbox{and } X=\frac{1+\sqrt{5}}{2}$

$\frac{1}{X^2-X-1}=\frac{\frac{\sqrt{5}}{5}}{X-(\frac{1+\sqrt{5}}{2})}-\frac{\frac{\sqrt{5}}{5}}{X-(\frac{1-\sqrt{5}}{2})}$

Therefore: $-\int_{0}^x \frac{dt}{\cos^2 t -\cos t-1}= -\int_{0}^x \frac{\frac{\sqrt{5}}{5}}{\cos t-(\frac{1+\sqrt{5}}{2})}dt + \int_{0}^x \frac{\frac{\sqrt{5}}{5}}{\cos t-(\frac{1-\sqrt{5}}{2})}dt$

I am having trouble calculating the two last integrals. Please help

Thank you in advance.

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    trigonometric substitution2012-11-06

1 Answers 1

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The $\tan(t/2)$ method is one way to go (though a bit cumbersome). Let $y = \tan(t/2)$. We get that $dy = \sec^2(t/2) \dfrac{dt}2 \implies dt = \dfrac{2 dy}{1+y^2}$ Also, recall that $\sin(t) = \dfrac{2y}{1+y^2} \,\,\,\text{ and }\,\,\, \cos(t) = \dfrac{1-y^2}{1+y^2}$

Then we get that $I = \int \dfrac{dt}{\sin^2(t) + \cos(t)} = \int \dfrac{2dy}{\dfrac{4y^2}{1+y^2} + 1-y^2} = \int \dfrac{2(1+y^2) dy}{1+4y^2 - y^4}$ I leave the rest to you since you should be able to finish it by splitting into appropriate partial fractions.

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    I am unable to integrate the final expression. Can somebody help me ?2012-11-07