You are watching: What is the antiderivative of sec^2x

I combined sec^2(x) to get tan(x), then evaluated at a, and b, and took the difference:tan(pi) - tan(0) = 0

I would love come understand just how infinity is an answer that mine "princetoneclub.org tool" got.

You can"t really take the integral top top $<0,\pi/2>$ since $\sec^2x=1/\cos^2 x$ is discontinuous in ~ $\pi/2$. For this reason what we really want is$$\lim_\theta\to\pi/2^-\int_0^\theta\sec^2x\,dx+\lim_\psi\to\pi/2^+\int_\psi^\pi\sec^2x\,dx.$$We have to break-up the integral up around the singularity. In this case, us have$$\lim_\theta\to\pi/2^-\int_0^\theta\sec^2x\,dx=\lim_\theta\to\pi/2^-\tan x\big|^\theta_0=\lim_\theta\to\pi/2^-\tan\theta=\infty\\\lim_\psi\to\pi/2^+\int_\psi^\pi\sec^2x\,dx=\lim_\psi\to\pi/2^+\tan x\big|_\psi^\pi=\infty.$$So the when the integral as you put it doesn"t really work, this need to approximate what your calculator is functioning out.

**Hint**

As you correctly found, the antiderivative of $\sec ^2(x)$ is $\tan (x)$. If the border of integration room $0$ and also $a$, the value of the integral is $\tan (a)$ which way that the outcomes approached infinity as soon as $a$ approached $\pi/2$.

I am sure that you have the right to take from here.

$sec^2(x)$ = $\frac1cos^2(x)$ as $x$ goes from 0 to $\frac\pi2$ what happends? Well, think about this: $cos(\frac\pi2)=0$. Together we method $\frac\pi2$ from one of two people side, we have actually $\frac1cos^2(x) \rightarrow \infty$. Then, if girlfriend think the the integral together measuring the area under the curve, you see why this integral goes come $\infty$.

.

Thanks because that contributing response to princetoneclub.orgematics stack Exchange!

Please be sure to*answer the question*. Provide details and share your research!

But *avoid* …

Use princetoneclub.orgJax to format equations. Princetoneclub.orgJax reference.

See more: Does Dueling Work With A Shield 5E, Fighter: Dueling Style And Shield

To learn more, see our tips on writing an excellent answers.

short article Your price Discard

By clicking “Post her Answer”, girlfriend agree come our terms of service, privacy policy and cookie policy

## Not the prize you're feather for? Browse various other questions tagged integration infinity pi or questioning your very own question.

site architecture / logo © 2021 stack Exchange Inc; user contributions licensed under cc by-sa. Rev2021.10.12.40432

your privacy

By clicking “Accept all cookies”, friend agree ridge Exchange have the right to store cookie on your an equipment and disclose details in accordance with our Cookie Policy.