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Calculate the definite integral

$$ I=\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\;\mathrm dx $$

given that $a>b>0$.

My Attempt:

If we replace $x$ by $C$, then

$$ I = \int_{0}^{\pi}\frac{\sin^2 C}{a^2+b^2-2ab\cos C}\;\mathrm dC $$

Now we can use the Cosine Formula ($A+B+C=\pi$). Applying the formula gives

$$ \begin{align} \cos C &= \frac{a^2+b^2-c^2}{2ab}\\ a^2+b^2-2ab\cos C &= c^2 \end{align} $$

From here we can use the formula $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ to transform the integral to

$$ \begin{align} I &= \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\;\mathrm dC\\ &= \int_{0}^{\pi}\frac{\sin^2A}{a^2}\;\mathrm dC\\ &= \int_{0}^{\pi}\frac{\sin^2 B}{b^2}\;\mathrm dC \end{align} $$

Is my process right? If not, how can I calculate the above integral?

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  • $\begingroup$ The cosine formula is $ \cos C = \frac{a^2+b^2-c^2}{2ab} $ where a,b,c are sides of a triangle and the uppercase C is the angle contained in between the sides a and b. I think you must have confused the 2 C's here. You can't use that formula here. $\endgroup$ – skankhunt42 Nov 25 '14 at 15:11
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We have $$\eqalign{I_m(a,b)= \int_0^\pi\frac{\cos(mx)}{a^2-2ab\cos x+b^2}\,\mathrm dx &=\left\{\matrix{\frac{\pi}{a^2-b^2}&\hbox{if}&m=0\\\cr \frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^m&\hbox{if}&m\ne0 }\right.}$$ Proof can be seen here. Hence \begin{align} \int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\,\mathrm dx&=\frac{1}{2}\int_0^{\pi} \frac{1-\cos2 x}{a^2+b^2-2ab \cos x}\,\mathrm dx\\ &=\frac{1}{2}\left[\frac{\pi}{a^2-b^2}-\frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^2\right]\\ &=\frac{\pi}{2 a^2} \end{align}

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  • $\begingroup$ Thanks Prof. @OmranKouba. This formula is really helpful. I owe you big. (ɔ◔︣‿◔︣)ɔ ❤ $\endgroup$ – Anastasiya-Romanova 秀 Nov 25 '14 at 17:08
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Such integral is related with the topological degree of a closed curve. We have:

$$ I = \frac{1}{2}\int_{0}^{2\pi}\frac{\sin^2 x}{\|a-be^{ix}\|^2}\,\mathrm dx = -\frac{i}{2}\oint_{\|z\|=1}\frac{\left(\frac{z-z^{-1}}{2}\right)^2}{(a-bz)(az-b)}\,\mathrm dz$$ hence we just have to consider the residues of the last integrand function in $z=0$ and $z=\dfrac{b}{a}$ (since $a>b>0$). That just leads to: $$ I = \frac{\pi}{2a^2}.$$

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Using the formula $$a^2+b^2-ab\cos C = c^2\;,$$ Now If $C\leftrightarrow x\;,$ Then $a^2+b^2-2ab\cos x= c^2$

So Integral Convert into $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\mathrm dC\;,$$ Now Using Sin formula $\displaystyle \frac{\sin C}{c} = \frac{\sin A}{a}.$

So $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 A}{a^2}\mathrm dC\;,$$ Now $A+B+C = \pi\;,$ Then $\mathrm dC = 0-\mathrm dA-\mathrm dB$

Now when $C\rightarrow 0,$ Then $A\rightarrow \pi$ and $B\rightarrow 0$

When $C\rightarrow 0,$ Then $A\rightarrow 0$ and $B\rightarrow 0$

So $$\displaystyle I = \int_{\pi}^{0}\frac{\sin^2 A}{a^2}(-\mathrm dA)+\int_{0}^{0}\frac{\sin^2 A}{a^2}(-\mathrm dB)$$

So $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 A}{a^2}\mathrm dA = \frac{2}{2a^2}\int_{0}^{\frac{\pi}{2}}\left[1-\cos 2A\right]\mathrm dA = \frac{\pi}{2a^2}$$ which agrees with the other solutions.

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Let $p=a^2+b^2,q=2ab$ so that $p^2-q^2=(a^2-b^2)^2$ and the integrand can be rewritten as $$\frac{\sin^2x}{p-q\cos x} = \frac{1}{q}\left(\cos x+\frac{p}{q}+\frac{1}{q}\cdot\frac{q^2-p^2}{p-q\cos x} \right) $$ and thus the desired integral is equal to $$0+\frac{p\pi}{q^2}+\frac{q^2-p^2}{q^2}\int_{0}^{\pi}\frac{dx}{p-q\cos x} $$ The integral above is famously equal to $\pi/\sqrt{p^2-q^2}$ and hence the value of the integral in question is equal to $$\frac{\pi} {q^2}\cdot(p-\sqrt{p^2-q^2})=\frac{\pi}{2a^2}$$

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