Question

intregation 1/1+cos x dx​

Answer 1

Given ,

The function is

• $\tt f(x) = \frac{1}{1 + cos(x)}$

Integrating wrt to x , we get

$\tt \implies \int{ \frac{1}{1 + cos(x)} } \: \: dx$

Multiplying denominator and numerator by 1 - cos(x) , we get

$\tt \implies \int{ \frac{\{ 1 - cos(x)\}}{ \{1 + cos(x) \} \{ 1 - cos(x)\}} } \: \: dx$

$\tt \implies \int{ \frac{\{ 1 - cos(x)\}}{ {(1)}^{2} - {cos}^{2}(x) } } \: \: dx$

$\tt \implies \int{ \frac{\{ 1 - cos(x)\}}{ {sin}^{2}(x) } } \: \: dx$

$\tt \implies \int{ \frac{1}{ {sin}^{2}(x) } - \frac{cos(x)}{ {sin}^{2}(x) } } \: \: dx$

$\tt \implies \int{ {cosec}^{2} (x) - cot(x)cosec(x)} \: \: dx$

$\tt \implies - cot(x) - \{ - cosec(x) \}$

$\tt \implies - cot(x) + cosec(x)$

Therefore , the anti - derivative is -cot(x) + cosec(x)

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