$\int \frac{1}{1 + \cos a x} d x =$

$c o t (\frac{a x}{2}) + c$

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$(\frac{1}{a}) \tan (\frac{a x}{2}) + c$

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$\frac{1}{a} (\cos e c a x - c o t a x) + c$

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$\frac{1}{a} (\cos e c a x + c o t a x) + c$

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Solution

The correct option is B
$(\frac{1}{a}) \tan (\frac{a x}{2}) + c$

Explanation for the correct option:
Finding the value of the given integral:
The given equation is
$\int \frac{1}{1 + \cos a x} d x = \int \frac{1}{[2 \cos^{2} \frac{a x}{2}]} d x [\sin c e \cos a x = 2 \cos^{2} \frac{a x}{2} - 1] = \frac{1}{2} \int s e c^{2} \frac{a x}{2} d x = \frac{1}{2} [\frac{\tan \frac{a x}{2}}{\frac{a}{2}}] + c = \frac{1}{a} \tan \frac{a x}{2} + c$
Thus, the correct answer is option (B).

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Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Δ

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