If b > a then,∫dxa+bcos⁡x=1b2−a2log⁡b−atan⁡x2+b+ab−atan⁡x2−b+a⇒ I=17log⁡tan⁡x2+7tan⁡x2−7+c

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$\int \frac{d x}{3 + 4 \cos x} = ?$

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$\frac{1}{\sqrt{7}} \log (\frac{\tan (x / 2) - \sqrt{7}}{\tan (x / 2) + \sqrt{7}}) + c$

$\frac{1}{\sqrt{7}} \log (\frac{\tan (x / 2) + \sqrt{7}}{\tan (x / 2) - \sqrt{7}}) + c$

$\frac{1}{\sqrt{7}} \log (\frac{\sqrt{7} + \tan (x / 2)}{\sqrt{7} - \tan (x / 2)}) + c$

$\frac{1}{\sqrt{7}} \log (\frac{\sqrt{7} - \tan (x / 2)}{\sqrt{7} + \tan (x / 2)}) + c$

answer is B.

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Detailed Solution

If b > a then,

$\begin{array}{l} \int \frac{d x}{a + b \cos x} = \frac{1}{\sqrt{b^{2} - a^{2}}} \log [\frac{\sqrt{b - a} \tan \frac{x}{2} + \sqrt{b + a}}{\sqrt{b - a} \tan \frac{x}{2} - \sqrt{b + a}}] \\ \Rightarrow I = \frac{1}{\sqrt{7}} \log (\frac{\tan \frac{x}{2} + \sqrt{7}}{\tan \frac{x}{2} - \sqrt{7}}) + c \end{array}$