∫dx5+4cos⁡x=atan−1⁡btan⁡x2+C, then

I=∫dx5+4cos⁡x=∫dx5sin2⁡x2+cos2⁡x2+4cos2⁡x2−sin2⁡x2=∫dx9cos2⁡x2+sin2⁡x2=∫sec2⁡x29+tan2⁡x2dx Now, I=∫2dt9+t2=23tan−1⁡t3+C∴ I=23tan−1⁡tan⁡x23+C Hence, a=23,b=13

∫dx5+4cos⁡x=atan−1⁡btan⁡x2+C, then

I=∫dx5+4cos⁡x=∫dx5sin2⁡x2+cos2⁡x2+4cos2⁡x2−sin2⁡x2=∫dx9cos2⁡x2+sin2⁡x2=∫sec2⁡x29+tan2⁡x2dx Now, I=∫2dt9+t2=23tan−1⁡t3+C∴ I=23tan−1⁡tan⁡x23+C Hence, a=23,b=13

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$\int \frac{d x}{5 + 4 \cos x} = a \tan^{- 1} (b \tan \frac{x}{2}) + C, then$

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$a = \frac{2}{3}, b = - \frac{1}{3}$

$a = \frac{2}{3}, b = \frac{1}{3}$

$a = - \frac{2}{3}, b = \frac{1}{3}$

$a = - \frac{2}{3}, b = - \frac{1}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int \frac{d x}{5 + 4 \cos x} \\ = \int \frac{d x}{5 (\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2}) + 4 (\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2})} \\ = \int \frac{d x}{9 \cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2}} = \int \frac{\sec^{2} \frac{x}{2}}{9 + \tan^{2} \frac{x}{2}} d x \end{array}$

$\begin{array}{l} Now, I = \int \frac{2 d t}{9 + t^{2}} = \frac{2}{3} \tan^{- 1} (\frac{t}{3}) + C \\ ∴ I = \frac{2}{3} \tan^{- 1} (\frac{\tan (\frac{x}{2})}{3}) + C \end{array}$