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I am trying to compute $$ \int\sec^2 x \tan x\, dx. $$ I substituted $u =\sec^2 x$ to get the integral as $$ \frac{\sec^4x}{2} $$ as my answer, but according to the textbook I am using, I'm wrong. Can anyone help me with the correct answer? According to my textbook, the correct option is between (a) $2\sec^2 x$ (b) $\tan x$ (c) $1/2 \sec x$ or (d) $\csc x \cot x$

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    $\begingroup$ What is the derivative of $\tan x$? $\endgroup$ – Chinny84 Sep 15 '15 at 13:09
  • $\begingroup$ Exactly, why don't you put $u = \tan x$ instead ? $\endgroup$ – Shailesh Sep 15 '15 at 13:09
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    $\begingroup$ "according to the textbook I'm wrong": you didn't try checking your answer yourself by differentiating your answer to see if you got back the original function? You should. $\endgroup$ – Najib Idrissi Sep 15 '15 at 13:35
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When you do a substitution like that in an integral, you want to end up with something that looks like this:

$$\int u \; du.$$

If $u = \sec^2 x$, what is $du$? If it is not $\tan x\;dx$ then you will not be able to change $\sec^2 x \tan x \; dx$ to $u \; du$.

Try a different substitution. There are not many choices, in fact I see only three likely possibilities and $u = \sec^2 x$ is the only one of the three that does not work.

EDIT: John Joy, in another answer, shows that the substitution $u = \sec^2 x$ actually does work if you do it correctly. You do not get anything in the form of $\int u\;du$ that way; instead, you get something easier. I should have acknowledged earlier that while $\int u\;du$ is one form you might hope to achieve from a substitution, really the point is just to get the integral into some form you know how to solve.

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Notice, we have $$\int\sec^2x \tan x dx$$ Let $\tan x=t\implies \sec^2 xdx=dt$ $$=\int tdt$$ $$=\frac{t^2}{2}+C$$ setting the value $t=\tan x$ $$=\frac{(\tan x)^2}{2}+C$$$$=\color{red}{\frac{\tan^2x}{2}+C}$$

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Your substitution $u=\sec^2 x$ is actually a good one, but it looks like you made an error somewhere.

$$\int\sec^2x\tan x dx$$ $$u = \sec^2 x\implies du = 2\sec x\cdot\sec x\tan xdx = 2\sec^2x\tan x dx$$ so the integral becomes $$\int\sec^2x\tan x dx = \int\frac{1}{2}\cdot 2\sec^2 x\tan xdx=\int \frac{1}{2}du=\frac{1}{2}u+C=\frac{1}{2}\sec^2x+C$$

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  • $\begingroup$ Thanks for pointing this out. My answer was too quick to suggest that $\int u\;du$ is "the" form that one wants the integral to take after substitution. $\endgroup$ – David K Sep 15 '15 at 14:18
  • $\begingroup$ It works, but its kind of like cheating because it presupposes that you already know the answer. The best substitution here: $u=\frac{1}{2}\sec^2x$ hehe $\endgroup$ – John Joy Sep 15 '15 at 14:24
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    $\begingroup$ Yes, one way to do this is if you already know the answer. Or you can just guess that the substitution might work, calculate $du$, and see what you can make of it. And oh, look, lucky guess. It seems to me that a lot of integration is like that--you recognize an integral you've seen before, or you try a particular trick to see if it works. At least that was my experience with first-year calculus, and thankfully after that I didn't have to remember most of that stuff except when I was teaching it. $\endgroup$ – David K Sep 15 '15 at 18:18
  • $\begingroup$ The guess $u = \frac12 \sec^2 x$ actually makes a lot of sense even before you know the answer, because you can already see there's probably going to be a factor of $2$ in the derivative of $\sec^2 x$, so by multiplying by $\frac12$ you trade off a more complicated $u$ for a simpler $du$ (which is often a good tradeoff). $\endgroup$ – David K Sep 15 '15 at 18:21
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The last two answers, from Harish and egreg, are the same. Integrating egreg's construction produces $\frac{\sec^2x}{2} + C$, and that from Harish produces $\frac{tan^2x}{2} + C_1$. Choosing $C_1 = 1/2 + C$ for the latter, yields $\frac{tan^2x+1}{2} + C = \frac{sec^2x}{2} + C,$ as it should.

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∫Sec^2x.Tanx dx But we know d(Secx)/dx = Secx.Tanx

By substituting we change the integral to the form,

∫Secx d(Secx)

This is of the form

∫xdx

..

So we have ,

Sec^2x/2 +C

Where C is an arbitrary constant. Feel free to edit

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You don't need to remember complicated formulas: just recall $$ \sec x=\frac{1}{\cos x},\qquad \tan x=\frac{\sin x}{\cos x} $$ so your integral is $$ \int \frac{\sin x}{\cos^3 x}\,dx= \int -\frac{1}{t^3}\,dt $$ with the substitution $t=\cos x$.

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    $\begingroup$ I think it is helpful to remember some of the somewhat more complicated trig integrals and derivatives when you need to come up with a more "creative" method to solve a harder problem, like $\int \sec(x)^m \tan(x)^n dx$ for various positive integers $m,n$. I don't know how I would come up with the procedure for these kinds of problems without thinking things like "substituting $u=\sec(x)$ will absorb a secant and a tangent into the du", "substituting $u=\tan(x)$ will absorb two secants into the du" and "secants and tangents can be interconverted two at a time using $\sec(x)^2=\tan(x)^2+1$". $\endgroup$ – Ian Sep 15 '15 at 13:37
  • $\begingroup$ @Ian I'm not saying that those formulas are useless (but I believe they're sometimes abused); first go with simple methods, I think this is better. $\endgroup$ – egreg Sep 15 '15 at 13:46
  • $\begingroup$ I actually prefer Using $\tan$ and $\sec$ because they seem much simpler to me than converting to rational functions. $(\tan x)' = \sec^2 x$ and $(\sec x)' = \sec x \tan x$ go hand-in-hand $\endgroup$ – Dylan Sep 16 '15 at 22:01

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