How to solve improper integral '1/sin(x)' at region with variables

I'd like to solve

 syms x h
 int(1/sin(x), x, h, 1)   %%when 0<h<1

But matlab can't solve it. Probably because it doesn't consider the region of 'h'. So during the calculation it has infinite value.

Contrastively, the integral works for

 syms x h
 int(sin(x), x, h, 1)
ans =

 cos(h) - cos(1)

Is there any good method to give simply a condition for 'h'?

As a second alternative plan(not really what i prefer,though) , i tried

   syms x h
  h=0.2:0.2:0.8;
  int(1/sin(x), x, h, 1)

But It gave me errors for different reasons. I'll appreciate for any help. Thanks

Answer 1

[Bjorn Gustavsson]

For rational functions of trigonometric functions use the substitution z = tan(x/2) - see for example:

tan(x/2)-substitution

[Chanho Jeong]

You gave me the perfect answer. Thanks But actually I'm trying to solve the following integral

 syms h py 
 int(-3200*(1 - (h - 40)^2/(1600*sin(py)^2))^(1/2), py, -asin(h/40 - 1), 43/25)

I substituted t=tan(py/2) following your answer but it doesn't give me the answer.

 int((-3200*(1 - (h - 40)^2/(1600*(2*t/(1+t^2))^2))^(1/2))*2/(1+t^2), t, tan((-asin(h/40 - 1))/2), tan(43/50))

 ans = 
 int(-(6400*(1 - ((t^2 + 1)^2*(h - 40)^2)/(6400*t^2))^(1/2))/(t^2 + 1), t, -tan(asin(h/40 - 1)/2), 5231182527675565/4503599627370496)

Do you mind helping me a little bit more about this? Thanks

[Bjorn Gustavsson]

If you try pretty on the integrand you'll see that it's a bit complicated with the square root...

pretty((-3200*(1 - (h - 40)^2/(1600*(2*t/(1+t^2))^2))^(1/2))*2/(1+t^2))

That indicates that a symbolic integral might be difficult to find, try some handbooks to see if this is something "well-known"...

HTH