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Q.∫02πdxa+bcos⁡x\displaystyle \int_{0}^{2\pi} \dfrac{dx}{a + b \cos x} (a>∣b∣>0)(a > |b| > 0) equals

a
2πa2−b2\dfrac{2\pi}{\sqrt{a^2 - b^2}}
b
πa2−b2\dfrac{\pi}{\sqrt{a^2 - b^2}}
c
2πa−b\dfrac{2\pi}{a - b}
d
πa−b\dfrac{\pi}{a - b}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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