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Question

Question: \[\int_{}^{}\frac{dx}{5 + 4\cos x} =\]...

∫dx5+4cos⁡x=\int_{}^{}\frac{dx}{5 + 4\cos x} =

A

23tan⁡−1(13tan⁡x)+c\frac{2}{3}\tan^{- 1}\left( \frac{1}{3}\tan x \right) + c

B

13tan⁡−1(13tan⁡x)+c\frac{1}{3}\tan^{- 1}\left( \frac{1}{3}\tan x \right) + c

C

23tan⁡−1(13tan⁡x2)+c\frac{2}{3}\tan^{- 1}\left( \frac{1}{3}\tan\frac{x}{2} \right) + c

D

13tan⁡−1(13tan⁡x2)+c\frac{1}{3}\tan^{- 1}\left( \frac{1}{3}\tan\frac{x}{2} \right) + c

Answer

23tan⁡−1(13tan⁡x2)+c\frac{2}{3}\tan^{- 1}\left( \frac{1}{3}\tan\frac{x}{2} \right) + c

Explanation

Solution

If a>b,a > b, then

∫dxa+bcos⁡x=2a2−b2.tan⁡−1(a−ba+btan⁡x2)+c∴∫dx5+4cos⁡x=23tan⁡−1(13tan⁡x2)+c\int_{}^{}\frac{dx}{a + b\cos x} = \frac{2}{\sqrt{a^{2} - b^{2}}}.\tan^{- 1}\left( \sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2} \right) + c\therefore\int_{}^{}{\frac{dx}{5 + 4\cos x} = \frac{2}{3}\tan^{- 1}\left( \frac{1}{3}\tan\frac{x}{2} \right) + c}.