Expert answers to "Integrating ∫1cos⁡(θ)dθ"

gaitaprepeted05u

Answered question

2022-04-21

Integrating

\int \frac{1}{\cos (θ)} d θ

Cynthia Herrera

Beginner2022-04-22Added 16 answers

I think it should be

\int \frac{1}{\cos {x}} d x = \int \frac{\cos {x}}{\cos^{2} x} d x = \int \frac{1}{1 - \sin^{2} x} d (\sin {x}) =

= \frac{1}{2} \int (\frac{1}{1 - \sin {x}} + \frac{1}{1 + \sin {x}}) d (\sin {x}) =

= \frac{1}{2} \ln \frac{1 + \sin {x}}{1 - \sin {x}} + C = \ln \sqrt{\frac{1 + \sin {x}}{1 - \sin {x}}} + C =

= \ln \sqrt{\frac{1 - \cos (\frac{π}{2} + x)}{1 + \cos (\frac{π}{2} + x)}} + C = \ln \sqrt{\tan^{2} (\frac{π}{4} + \frac{x}{2})} + C = \ln | \tan (\frac{π}{4} + \frac{x}{2}) | + C

1 + \tan^{2} θ = \frac{1}{\cos^{2} θ}

haarplukxjf

Beginner2022-04-23Added 16 answers

\int \frac{1}{\cos θ} d θ = \int \frac{\cos θ}{\cos^{2} θ} d θ

= \int \frac{d (\sin θ)}{1 - \sin^{2} θ}

= \frac{1}{2} \ln (\frac{1 + \sin x}{1 - \sin x}) + C

= \ln \sqrt{\frac{1 + \sin x}{1 - \sin x}} + C

= \ln \sqrt{\frac{1 + \sin x}{1 - \sin x} \cdot \frac{1 + \sin x}{1 + \sin x}} + C

= \ln \sqrt{{(\frac{1 + \sin x}{\cos x})}^{2}} + C

= \ln | \frac{1 + \sin x}{\cos x} | + C

= \ln | \tan x + \sec x | + C

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