Question

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

Answer 1

The answer is that this integral is DIVERGENT.

First of all I remember that: `cscx=1/sinx`, so the integral becoms:

`int_(pi/2)^picscxdx=int_(pi/2)^pi1/sinxdx`.

Before integrate this function, it is useful to remember the parametric formula of sinus, that says:

`sinx=(2t)/(1+t^2)`, where `t=tan(x/2)`.

Now the integral will be done with the method of substitution:

`tan(x/2)=trArrx/2=arctantrArrx=2arctantrArrdx=2(1/(1+t^2))dt`.

And it is important to change also the two limits of integrations:

if `x=pi/2rArrt=tan((pi/2)/2)=tan(pi/4)=1`, and
if `xrarrpirArrtrarrtan((pi)/2)rarr+oo`.

It is clear that the integral becoms an improper integral. We can do the integral without the limits of integration and than we will do the final count.

Our integral becoms:

`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`.

Now, using the fundamental theorem of Calculus, the final count:

`lim_(Mrarr+oo)[ln|tan(x/2)|]_1^M=lntan(+oo)-ln(tan1)=+oo`

So the integral is divergent.

Did we avoid ALL this counts? The answer is YES!

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.