For 0 < a < 1, the value of the integral ∫ dx/1−2acosx+a^2; x ∈ [0, π] is :

Question

For \(0<\mathrm{a}<1\), the value of the integral \(\int_{0}^{\pi} \frac{d x}{1-2 a \cos x+a^{2}} \) is :

(1) \(\frac{\pi^{2}}{\pi+a^{2}}\)

(2) \(\frac{\pi^{2}}{\pi-a^{2}}\)

(3) \(\frac{\pi}{1-a^{2}}\)

(4) \(\frac{\pi}{1+a^{2}}\)

Answer 1

Correct option is (3) \(\frac{\pi}{1-a^{2}}\)

\(\mathrm{I}=\int_{0}^{\pi} \frac{\mathrm{dx}}{1-2 \mathrm{a} \cos \mathrm{x}+\mathrm{a}^{2}} ; 0<\mathrm{a}<1\)

\(I=\int_{0}^{\pi} \frac{\mathrm{dx}}{1+2 a \cos x+a^{2}}\)

\(2 I=2 \int_{0}^{\pi / 2} \frac{2\left(1+a^{2}\right)}{\left(1+a^{2}\right)^{2}-4 a^{2} \cos ^{2} x} \mathrm{dx}\)

\(\Rightarrow \mathrm{I}=\int_{0}^{\pi / 2} \frac{2\left(1+\mathrm{a}^{2}\right) \cdot \sec ^{2} \mathrm{x}}{\left(1+\mathrm{a}^{2}\right)^{2} \cdot \sec ^{2} \mathrm{x}-4 \mathrm{a}^{2}} \mathrm{dx}\)

\(\Rightarrow \mathrm{I}=\int_{0}^{\pi / 2} \frac{2 \cdot\left(1+\mathrm{a}^{2}\right) \cdot \sec ^{2} \mathrm{x}}{\left(1+\mathrm{a}^{2}\right)^{2} \cdot \tan ^{2} \mathrm{x}+\left(1-\mathrm{a}^{2}\right)^{2}} \mathrm{dx}\)

\(\Rightarrow \mathrm{I}=\int_{0}^{\pi / 2} \frac{\frac{2 \cdot \sec ^{2} \mathrm{x}}{1+\mathrm{a}^{2}} \cdot \mathrm{dx}}{\tan ^{2} \mathrm{x}+\left(\frac{1-\mathrm{a}^{2}}{1+\mathrm{a}^{2}}\right)^{2}}\)

\(\Rightarrow \mathrm{I}=\frac{2}{\left(1-\mathrm{a}^{2}\right)}\left[\frac{\pi}{2}-0\right]\)

\(I=\frac{\pi}{1-a^{2}}\)