How do I evaluate #int_(pi/2)^picscx dx#?

Question

Scarlett Allison · Accepted Answer

This integral is DIVERGENT, is the response. First of all I remember that: #cscx=1/sinx#, so the integral becoms: #int_(pi/2)^picscxdx=int_(pi/2)^pi1/sinxdx#. Prior to integrating this function, it helps to keep in mind the sinus parametric formula, which states: #sinx=(2t)/(1+t^2)#, where #t=tan(x/2)#. The integral will now be completed using the substitution method: #tan(x/2)=trArrx/2=arctantrArrx=2arctantrArrdx=2(1/(1+t^2))dt#. Additionally, it's critical to alter the two integration limits: if #x=pi/2rArrt=tan((pi/2)/2)=tan(pi/4)=1#, and
if #xrarrpirArrtrarrtan((pi)/2)rarr+oo#. It is evident that the integral becomes an incorrect integral. The final count can be performed after the integral is completed without regard to integration limits. Our essential behaviors: #int1/sinxdx=int1/((2t)/(1+t^2))2(1/(1+t^2))dt=int(1+t^2)/(2t)2(1/(1+t^2))dt=int1/tdt=ln|t|+c=ln|tan(x/2)|+c#. Using the calculus fundamental theorem now, the final count: #lim_(Mrarr+oo)[ln|tan(x/2)|]_1^M=lntan(+oo)-ln(tan1)=+oo# As a result, the integral diverges. Yes, we were able to avoid ALL of these counts. First of all a substitution: #x=pi-trArrdx=-dt# If #x=pi/2rArrpi/2=pi-trArrt=pi/2#, and if #x=pirArrpi=pi-trArrt=0# #int_(pi/2)^pi1/sinxdx=int_(pi/2)^0(1/sin(pi-t))(-dt)=-int_(pi/2)^0(1/sin(pi-t))dt=int_0^(pi/2)1/sintdt# The function #sint# near zero is asymptotic to #t#, and the #int_0^(pi/2)1/tdt# is divergent.

William Abdalla · Answer

undefined To evaluate the integral ∫(π/2)^(π/4) sec(x) dx, we'll first express sec(x) in terms of trigonometric functions. Since sec(x) = 1/cos(x), we can rewrite the integral as ∫(π/2)^(π/4) (1/cos(x)) dx. Next, we'll use a trigonometric identity to simplify the integral. The identity is cos(π/2 - x) = sin(x). Therefore, cos(x) = sin(π/2 - x). Substituting this into the integral, we get ∫(π/2)^(π/4) (1/sin(π/2 - x)) dx. Now, we'll make the substitution u = π/2 - x, which implies du = -dx. As the limits change, when x = π/2, u = 0, and when x = π/4, u = π/4. Therefore, the integral becomes ∫0^(π/4) (1/sin(u)) (-du). This is equivalent to -∫0^(π/4) (1/sin(u)) du. Now, recall that csc(u) = 1/sin(u). So, the integral becomes -∫0^(π/4) csc(u) du. The integral of csc(u) is -ln|csc(u) + cot(u)|. Finally, we evaluate this expression from 0 to π/4 and subtract the result when u = 0 from the result when u = π/4. This yields the final answer.