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MCQMediumJEE Main 2024 · 27 January, Shift 2Definite Integrals

Mathematics Question from JEE Main 2024 · 27 January, Shift 2

For 0<a<10 < a < 1, find the value of the integral ∫0πdx1−2acos⁡(x)+a2\int_{0}^{\pi} \frac{dx}{1 - 2a \cos(x) + a^2}:

  • A

    π1−a2\frac{\pi}{1 - a^2}

  • B

    π1+a2\frac{\pi}{1 + a^2}

  • C

    π2π+a2\frac{\pi^2}{\pi + a^2}

  • D

    π2π−a2\frac{\pi^2}{\pi - a^2}

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