Integrate $\int\sec ^2 x \tan x\, dx$
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) $\csc x \cot x$
$\endgroup$ 36 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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 $t=\tan x$ $$=\frac{(\tan x)^2}{2}+C$$$$=\color{red}{\frac{\tan^2x}{2}+C}$$
$\endgroup$ $\begingroup$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 $$\int\sec^2x\tan x dx = \int\frac{1}{2}\cdot 2\sec^2 x\tan xdx=\int \frac{1}{2}du=\frac{1}{2}u+C=\frac{1}{2}\sec^2x+C$$
$\endgroup$ 4 $\begingroup$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.
$\endgroup$ $\begingroup$∫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
$\endgroup$ $\begingroup$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 $t=\cos x$.
$\endgroup$ 3
