0
$\begingroup$

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) $\csc x \cot x$

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ "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
4
$\begingroup$

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 $t=\tan x$ $$=\frac{(\tan x)^2}{2}+C$$$$=\color{red}{\frac{\tan^2x}{2}+C}$$

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

∫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

$\endgroup$
0
$\begingroup$

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 $t=\cos x$.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.