Jul 19, 2018 · y = arcsec ( tan x ) in [ 0. pi ] tan x = sec y notin ( -1, 1 ), and so, x notin ( -pi/4, pi/4 ), See illustrative extended graph using inversion for piecewise wholesome y = (sec)^(-1)(tan x ): graph{cos y sin x-cos x=0} Graph marked for range y in [ 0, pi ] and tangent at (pi/4, 0): Slide the graph to see how the tangent braces the extended ...

Jan 22, 2017 · C. y= sinx We need to look for asymptotes here. Whenever there are asymptotes, the domain will have restrictions. A: y= cotx can be written as y = cosx/sinx by the quotient identity. There are vertical asymptotes whenever the denominator equals 0, so if: sinx = 0 Then x = 0, pi These will be the asymptotes in 0 ≤ x < 2pi. Therefore, y =cotx is not defined in all the real numbers. B: y = secx ...

Jul 9, 2015 · First, let's find the derivative of #arcsec(x)# by applying the Chain Rule to the equation #sec(arcsec(x))=x# (and using the fact that #d/dx(sec(x))=sec(x)tan(x)#)

Jul 25, 2016 · 1/[4|x+1|sqrt(x(x+2)arcsec(x+1))]. Let, y=g(x)=sqrtu, where, u=arcsect, &, t=x+1. Thus, y is a function of u, u of t, and t of x.

Jul 12, 2017 · Find the value of #tan(arcsec(3x))#? Trigonometry. 1 Answer

Mar 13, 2017 · How do I evaluate the indefinite integral #int(tan^2(x)+tan^4(x))^2dx# ? How do I evaluate the indefinite integral #intx*sin(x)*tan(x)dx# ? See all questions in Integrals of Trigonometric Functions

Mar 20, 2015 · int_(1/5)^oo arc sec (5x) dx diverges. int_(1/5)^oo arc sec (5x) dx = lim_(brarroo)int_(1/5)^b arc sec (5x) dx As xrarroo, the integrand arc sec (5x) rarr pi/2. That is: the line y= pi/2 is a horizontal asymptote. As b rarr oo the area under the arc sec curve from b to b+1 approaches the area of the rectangle with base [b, b+1] and height pi/2. So as b rarr oo the additional area from b to b+1 ...

May 20, 2018 · After using #y=x^2# and #2x*dx=dy# transforms, this integral became #1/2int arcsecy*dy# After using #z=arcsecy# , #y=secz# and #dy=secz*tanz*dz# transforms, it became

Sep 11, 2016 · "arcsec"(x/3)+C We have: int3/(xsqrt(x^2-9))dx We will use the substitution x=3sectheta. Thus dx=3secthetatanthetad theta. =int3/(3secthetasqrt(9sec^2theta-9 ...

Jun 28, 2016 · # = 2*ln(1/2*(x + sqrt(x^2 - 4))) + C# And there we have it, let me know if you need any help with the last bit but it's fairly easy to figure out if you look at a few triangles. Answer link