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So I was looking at how to obtain the following integral $$\int \sec(x)\,\mathrm dx$$ I saw that, to solve it, they divided and multiplied by $\tan(x)sec(x)/tan(x)sec(x)$. I understand that this is the equivalent to multiplying by $1$ and that they can do this. However, I would like to understand the logic behind why they are multiplying by this instead of some other trig ratio? Also is there any other way to solve this integral? I looked around but the only result I could find was multiplying and dividing by $\tan(x)sec(x)/tan(x)sec(x)$


marked as duplicate by grand_chat, N. F. Taussig, Nosrati, Jyrki Lahtonen, Adrian Keister Sep 21 '18 at 0:08

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    $\begingroup$ the logic behind it is finding it's antiderivative. $\endgroup$ – Nosrati Sep 20 '18 at 15:57
  • $\begingroup$ it is $\sect+\tan t$. $\endgroup$ – Nosrati Sep 20 '18 at 15:58
  • $\begingroup$ Using en.m.wikipedia.org/wiki/… will be more natural way $\endgroup$ – lab bhattacharjee Sep 20 '18 at 16:00
  • $\begingroup$ So they just plugged and chugged until they arrived to the antiderivative, or? @Nosrati $\endgroup$ – M.M Sep 20 '18 at 16:00
  • $\begingroup$ Notice that the integrand is $\dfrac{\sec x \tan x + \sec^{2}x }{\sec x+ \tan x }$ so that you can pull it to a logarithmic form. $\endgroup$ – Narasimham Sep 20 '18 at 22:16

There is some interesting history behind this. A formula for the integral (which is needed for construction of Mercator's projection maps) was discovered in a numerical comparison of two tables of numbers, and published in 1645. To prove it was an important open problem in the mid-17th century. James Gregory found a very complicated proof in 1668, and Barrow a simpler one in 1670. All this was before the Fundamental Theorem of Calculus: once you have that, it's completely elementary to verify by differentiating. See e.g. these notes.


Another way \begin{align} \int\dfrac{1}{\cos x}dx &= \int\dfrac{\cos x}{1-\sin^2x}dx \\ &= \dfrac12\int\dfrac{\cos x}{1+\sin x}dx+\dfrac12\int\dfrac{\cos x}{1-\sin x}dx \\ &= \dfrac12\ln\dfrac{1+\sin x}{1-\sin x}+C \end{align}


Yes, there are ways other than multiplying by $\dfrac{\sec x+\tan x}{\sec x+\tan x}$.

For example, you could use the Weierstrass $t$-substitution $$ \begin{align*} \tan\frac{x}{2}&=t\\ \cos x&=\frac{1-t^2}{1+t^2}\\ \sin x&=\frac{2t}{1+t^2}\\ \mathrm{d}x&=\frac{2\,\mathrm{d}t}{1+t^2} \end{align*} $$ which convert a rational function of trigonometric functions of $x$ to a rational function of $t$, from which there are standard procedure you can follow to find its antiderivative (partial fractions and the worst case involves integrating $\dfrac{a+bu}{(1+u^2)^m}$). So $$ \int\sec x\,\mathrm{d}x=\int\frac{1+t^2}{1-t^2}\,\frac{2\mathrm{d}t}{1+t^2}=\int\left(\frac1{t-1}-\frac1{t+1}\right)\,\mathrm{d}t $$ which you can integrate immediately.


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