# Calculate integral (0-pie/2)1/f(subscript1)(x) dx if f(subscript n)(x) =1/((cos )^n+(sinx)^n), 0<equal x<equal pie/2?

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You need to evaluate the following definite integral, such that:

`int_0^(pi/2) 1/(f_1(x)) dx`

The problem provides the function `f_n(x) = 1/(cos^n x + sin^n x)` , hence, you need to evaluate `f_1(x)` replacing 1 for n in equation of the function, such that:

`f_1(x) = 1/(cos x + sin x)`

`int_0^(pi/2) 1/(f_1(x)) dx = int_0^(pi/2) 1/(1/(cos x + sin x)) dx`

`int_0^(pi/2) 1/(f_1(x)) dx = int_0^(pi/2) (cos x + sin x) dx`

Using the property of linearity of integral, yields:

`int_0^(pi/2) 1/(f_1(x)) dx = int_0^(pi/2) cos x dx + int_0^(pi/2) sin x dx`

`int_0^(pi/2) 1/(f_1(x)) dx = sin x|_0^(pi/2) - cos x|_0^(pi/2)`

Using the fundamental theorem of calculus, yields:

`int_0^(pi/2) 1/(f_1(x)) dx = sin(pi/2) - sin 0 - cos(pi/2) + cos 0`

`int_0^(pi/2) 1/(f_1(x)) dx = 1 - 0 - 0 + 1`

`int_0^(pi/2) 1/(f_1(x)) dx = 2`

**Hence, evaluating the given definite integral, under the given conditions, yields `int_0^(pi/2) 1/(f_1(x)) dx = 2` .**