is this impossible to integrate and if not what is the answer,

substitute u=cos(theta)

>>8639253

Just to eyeball the graph, the proposed integration blows up to positive infinity, and such can be demonstrated in a straightforward manner. So depending on how you are you learning calculus, you might say that the thing "equals infinity", "goes to positive infinity", "does not converge", or another equivalent statement.

Now OP, if you wanted to consider two infinite areas, one positive and one negative, or other species of cases where a function (and particularly its graph) behaves like and /looks like/ an odd function, then you might avail yourself of the Cauchy Principal Value. The Cauchy Principal Value is a tool in analysis for assigning values to /certain/ improper integrals which would (and which are) normally "undefined, does not exist, goes to infinity", etc.

Consider the graph of the tangent function between negative pi over two, through pi over two. This thing is /odd/, meaning that the parts on either side of the origin are /congruent/. So there is a definite sense in which /these two particular infinities do legitimately cancel each other out, from moment to moment, as the limits are taken. It is this notion that the Cauchy Principal Value captures.

Try this: Plug the following into Wolfram alpha:

Int(tan x , x, -pi/2, pi/2)

You will notice toward the bottom of the outputs that a CPV, or "PV" is assigned: it is precisely zero, which makes graphical and intuitive sense.

PVs must NOT be confused with the ordinary rules of integration. The above integral, taken by itself, still "does not converge", "does not exist", etc. the CPV is just a way of getting around that by saying "well yeah, but..." in some cases.

>>8639253
Sec = 1/ cos
Tan = sin/cos

Sec tan = sin / cos^2

= sin / 1 - sin^2

Has no discontinuities.

You can use trig identities to do it. But Idk what it is offhand.

Derivative of sec is sec * tan, so there's your answer

>>8639296
This. Why did it take so long for someone to say this?

>>8639314
Because 99% of /sci/ are underaged idiots still in highschool

do you mean its not integrable, because that is discontinuous at pi/2 and 3pi/2

>>8639318
I'm in high school calc currently and I noticed this. People on here just over think stuff.

>>8639253
answer 1.402*10^4

>>8639253
What discontinuities?

>>8639911
the integral of 4sec(theta)tan(theta) = 4sec(theta)

then you evaluate that function from -pi/4 to 3pi/4.

But sec(theta) is discontinuous at pi/2 (cos(pi/2) = 0 => 1/0).

So it's impossible without the cauchy principal value.

>>8639253
It is integrable as it is continuous in the interval given.

[math] \int_{}^{} 4sec(x)tan(x)dx = 4\int_{}^{} sec(x)tan(x)dx [/math]
Then let [math] u = sec(x), du = sec(x)tan(x)dx [/math] so we get

[math] 4\int_{}^{} du = 4u = 4sec(x) [/math]

Then compute

[math] 4sec(\frac{3 \pi }{4}) - 4sec( \frac {- \pi }{4}) = -4 \sqrt{2} - 4 (\sqrt{5} - 1)
[/math]