Why is it, if you take it the integral the sec^2(x) indigenous 0 come pi, my calculator returns "infinity" as the answer, yet according to the second an essential theorem of calculus, I gained 0 v my own work.

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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 $\infty$.

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