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This question already has an answer here:

$$\int\sec x \,\mathrm dx = \ln\left|\sec{x} + \tan{x}\right|+ C = \ln{\left|\tan\left(\frac{x}{2} + \frac{\pi}{4}\right)\right|} + C$$

My question is how? How are these derived?

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marked as duplicate by Aditya Hase, Milo Brandt, user98602, Hanul Jeon, user147263 Dec 15 '14 at 3:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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It is well-known that

$$ \int \csc x dx = \ln \tan \frac x 2. $$ Shifting the integration variable by $\pi/2$ gives, using the fact that $\sin(x+\pi/2) = \cos x$,

$$ \int \sec x dx= \ln \tan \left({\frac x 2 + \frac \pi 4 }\right). $$

To prove the first integral, write $$ \csc x = \frac{1}{2 \sin(x/2) \cos(x/2) } = \frac 1 2 \left[\cot \frac x 2 + \tan \frac x 2 \right]. $$

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  • $\begingroup$ Thank you :) Its helping.. $\endgroup$ – user196429 Dec 8 '14 at 7:16
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You can rewrite your integrand in the form $$\frac{\cos(x)}{\cos(x)^2}\,.$$ Setting $t=\sin(x)$ gives us $$\int\frac{1}{1-t^2}dt\,.$$

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  • $\begingroup$ Thank you for the answer :) $\endgroup$ – user196429 Dec 8 '14 at 7:16
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For most people, they were derived by someone else, or a textbook, telling them.
To check that they work, differentiate them.
$$\frac d{dx}\ln(\sec x+\tan x)=\frac{\sec x\tan x+\sec^2x}{\sec x+\tan x}$$

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  • $\begingroup$ Thank you for the answer:) $\endgroup$ – user196429 Dec 8 '14 at 7:17

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