# IntegralWhat is the definite integral of f(x)=1/sin^2x ? x=0,x=pi/4

sciencesolve | Certified Educator

You may integrate directly, using the formula of integration, such that:

`int_0^(pi/4) 1/(sin^2 x)dx = -cot x|_0^(pi/4)`

You need to use fundamental theorem of calculus, such that:

`int_0^(pi/4) 1/(sin^2 x)dx = -cot(pi/4) - (-cot 0)`

`int_0^(pi/4) 1/(sin^2 x)dx = -1 + 1/0`

Since `cot 0` is not a valid value, hence, you cannot evaluate the definite integral `int_0^(pi/4) 1/(sin^2 x)dx` at having the limit of integration `x = 0` .

giorgiana1976 | Student

The definite integral will be evaluated using the Leibniz-Newton formula.

Int f(x)dx = F(b) - F(a), where x = a to x = b

We'll put  f(x) = 1/(sin x)^2

We'll compute the indefinite integral, first:

Int dx/(sin x)^2 = -cot x + C

We'll note the result F(x) = - cot x + C

We'll determine F(a), for a = 0:

F(0) = -cot 0

F(0) = -pi/2

We'll determine F(b), for b = pi/4:

F(pi/4) = -cot pi/4

F(pi/4) = -1

We'll evaluate the definite integral:

Int dx/(sin x)^2 = F(pi/4) - F(0)

Int dx/(sin x)^2 = -1+ pi/2

Int dx/(sin x)^2 = -1+ pi/2, from x = 0 to x = pi/4.