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Q.∫dxa+bcos⁡x\int \dfrac{dx}{a + b \cos x} equals

a
2a2−b2tan⁡−1(a−ba+btan⁡x2)+C\dfrac{2}{\sqrt{a^2 - b^2}} \tan^{-1} \left( \sqrt{\dfrac{a-b}{a+b}} \tan \dfrac{x}{2} \right) + C
b
1a2−b2tan⁡−1(a−btan⁡(x/2)a+b)+C\dfrac{1}{\sqrt{a^2 - b^2}} \tan^{-1} \left( \dfrac{\sqrt{a-b} \tan(x/2)}{\sqrt{a+b}} \right) + C
c
log⁡∣a+bcos⁡x∣+C\log |a + b \cos x| + C
d
tan⁡(x/2)+C\tan(x/2) + C

Correct Answer: Option A

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

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