Getting errors when trying to solve an integral equation

Question

I am getting the final integrate value with x range, I need to find the only one real parametera,but I receive an error:

This is the code after José:

Solve[NIntegrate[1-1/(1+a Cos[\[Pi] (x-14.625)/150]^2),{x,0,126}]==0.5,a]

error:

NIntegrate: The integrand 1-1/(1+a Cos[Times[<<3>>]]^2) has evaluated to non-numerical values for all sampling points in the region with boundaries {{0,126}}.

General:Further output of NIntegrate::inumr will be suppressed during this calculation.

Solve: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help.

Out: Solve[NIntegrate[1-1/(1+a Cos[1/150 \[Pi] (x-14.625)]^2),{x,0,126}]==0.5,a]

Or try this: Solve[Integrate[1-1/(1+a Cos[\[Pi] (x-14.625)/150]^2),{x,0,126}]==0.5,a]

error:

Solve: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help.

out: Solve[126+1/Sqrt[(0.338738 -0.940881 I)+(0.338738 -0.940881 I) a] ((-27.4545-39.0638 I) ArcTanh[((1.8303*10^6+2.60425*10^6 I) (-2.-0.18185 a))/Sqrt[(0.338738 -0.940881 I)+(0.338738 -0.940881 I) a]]-(27.4545 +39.0638 I) ArcTanh[((2.69162*10^6+3.82978*10^6 I) (-2.-0.18185 a))/Sqrt[(0.338738 -0.940881 I)+(0.338738 -0.940881 I) a]]+(27.4545 +39.0638 I) ArcTanh[((0.328084 +0.466817 I) (-0.963507+0.416274 a))/Sqrt[(0.338738 -0.940881 I)+(0.338738 -0.940881 I) a]]-(27.4545 +39.0638 I) ArcTanh[((0.165316 +0.23522 I) a)/Sqrt[(0.338738 -0.940881 I)+(0.338738 -0.940881 I) a]])==0.5,a]

Is this possible or other way around to work it out?

Answer 1

f[a_?NumericQ] := 
 NIntegrate[1 - 1/(1 + a Cos[π (x - 14625/1000)/150]^2), {x, 0, 126}]

Using NSolve

soln1 = NSolve[f[a] == 1/2, a][[1]]

(* NSolve::ifun: Inverse functions are being used by NSolve, so some solutions 
    may not be found; use Reduce for complete solution information.

{a -> 0.00868611} *)

Using FindRoot

soln2 = FindRoot[f[a] == 1/2, {a, 1}]

(* {a -> 0.00868611} *)

Verifying

f[a] == 1/2 /. {soln1, soln2}

(* {True, True} *)