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\begin{align} \int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2t} dt \end{align} I am trying to evaluate above integral. The results is $2\pi$ according to Mathematica. I want to obtain this result by integrating properly

Can this integral be evaluated using simple trigometric identities?

Do I have to use complex analysis i.e., $\cos(\theta) = \frac{z+\frac{1}{z}}{2}$ and do residue calculus?

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  • $\begingroup$ I would try the Weierstrass substitution $\endgroup$ Oct 31, 2019 at 15:21

3 Answers 3

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Rewrite

\begin{align} I=\int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2t} dt = 2\int_0^{\pi} \frac{2-\cos t}{3\cos^2 t -4\cos t +2} dt \end{align}

and then substitute $\cos t = \frac{1-x^2}{1+x^2}$, along with $dt = \frac{2dx}{1+x^2}$

\begin{align} I= &\ 4\int_0^{\infty} \frac{1+3x^2}{9x^4 -2x^2 +1} dx \\ =&\ 4\int_0^{\infty} \frac{\frac{1}{x^2}+3}{9x^2 +\frac{1}{x^2}-2} dx =4\int_0^{\infty} \frac{d\left(3x-\frac{1}{x}\right)} {\left(3x-\frac{1}{x}\right)^2+4}\\ =&\ 2\tan^{-1}\left[\frac12\left(3x-\frac{1}{x}\right)\right]_0^{\infty}=2\pi \end{align}

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Hint:

Re write the integral as follows:

$-\displaystyle\int\frac{\cos(t)-2}{(\sin^2(t) + (2\cos(t)-1)^2} dt$

and then try to use Weierstrass substitution. In case you don't know it, it is the following:

$$x=\tan\left(\dfrac{t}{2}\right)$$

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Hint: After applying the Weierstrass substitution the indefinite integral will be $$\int \frac{2 \left(3 t^2+1\right)}{(t+1) \left(9 t^3-7 t^2+t+1\right)}dt$$

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  • $\begingroup$ How would you solve this integral? $\endgroup$
    – Martin R
    Nov 1, 2019 at 12:25

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