8
$\begingroup$

$$\sec ^2(x)=\tan ^2(x)+1$$ $$\csc ^2(x)=\cot ^2(x)+1$$

We can evaluate integrals of the form:

$$\int \sec ^m(x) \tan ^n(x) \, dx$$

$$\int \csc ^m(x) \cot ^n(x) \, dx$$

with substitution unless $m$ is odd and $n$ is even. What I am interested to know is why am I not able to solve this with substitution if $m$ is odd and $n$ is even. I am aware that I can solve it by integration by parts. But I do not see the underlying reason for why it is not possible when using substitution?

$\endgroup$
1
$\begingroup$

What if $m$ is odd, $n$ is even. For example, $m=1, n=0$.

$$ \int \sec x\;dx $$ Proceed as the method says. Keep $\sec^2$ and use a pythagorean identity to convert all the rest to $\tan$ ... even if you don't see two secants, you can still do it: $$ \int \sec x\;dx = \int \frac{\sec^2 x}{\sec x}\;dx = \int\frac{\sec^2 x}{\sqrt{1+\tan^2 x}} \;dx $$ then substitute $u = \tan x$ to get $$ \int \frac{du}{\sqrt{1+u^2}} $$ Then do this one "somehow" and substitute back $$ \int \frac{du}{\sqrt{1+u^2}} = \log\big(u+\sqrt{1+u^2}\;\big) + C =\log\big(\tan x + \sqrt{\sec^2 x}\;\big)+ C = \log(\tan x \pm \sec x) + C $$

The difference in these cases ($n$ even and $m$ odd) is that you do not end up with a rational function of $u$ to integrate. You have a square root.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

GEdgar's good answer shows that you can evaluate the integral $$\int \sec^m x \tan^n x \,dx. \qquad m, n \in \Bbb Z_{\geq 0} $$ with substitution, even when $m$ odd and $n$ even.

The spirit of the question is thus closer to

Why do neither of the substitutions $$u = \sec x, \quad du = \sec x \tan x \,dx \qquad \textrm{and} \qquad v = \tan x, \quad \,dv = \sec^2 x \,dx$$ produce integrals of polynomial (or rational) functions when $m$ is odd and $n$ is even?

Notice that substituting $\sec^2 x = \tan^2 x + 1$ may change replace the integrand with one with multiple terms, but the powers of $\sec x$ and $\tan x$ of each resulting term are respectively odd and even.

Making the first substitution leaves an odd number $p$ of powers of $\tan x$, which cannot be written as a polynomial in $u$; indeed, $\tan^p x = (\sec^2 x - 1)^{p / 2} = (u^2 - 1)^{p / 2}$. The argument for the other substitution is similar.

Other substitutions will produce rational integrals, however: The tangent half-angle substitution $x = 2 \arctan t$, $dx = \frac{2 \,dt}{1 + t^2}$ transforms the integral into $$2^{n + 1} \int \frac{(1 + t^2)^{m - 1} t^n dt}{(1 - t^2)^{m + n}} .$$ This integral can in turn be evaluated for particular $m, n$ using the Method of Partial Fractions.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Because when $m$ is even"regardless of $n$", we take the benefit of our knowledge that the $\sec^2(x)$ the anti derivative of $\tan(x)$ so we isolate $\sec^2(x)$ and replace the left with $1+\tan^2(x)$ ....... and when $m$ and $n$ are odds we take the benefit of our knowledge that $\sec(x) \tan(x)$ the anti derivative of $\sec(x)$ so we isolate $\sec(x)\tan(x)$ and replace the $\tan^2(x)=\sec^2(x)-1$ ......... now when $m$ is odd and $n$ is even what is the identity or relation that we is able to be used isolating $\sec^2(x)$ has no meaning if we can't replace the rest of the expressions to tan and isolating $\sec(x)\tan(x)$ has no meaning if we can't replace the rest of the expressions to $\sec(x)$ so what is done as you declared you already know we replace $\tan^2(x)=\sec^2(x)-1$ and continue by parts with the $\sec(x)$ ......... I hope I made it clear, Please feed me back if anything doesn't make sense

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ This answer needs to be edited to include proper MathJaX formatting, and could also be split into paragraphs to improve readability, but it is nice overall. Welcome to the site. $\endgroup$ – The Count Feb 18 '17 at 1:33

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.