EDIT #2
The integral of the OP belongs to a broader class of divergent integrals which, first of all, must be given sense by defining how to circumvent the singularities.
Such divergent integrals are surprisingly frequent in integral tables such as Gradshteyn/Ryshik.
In our problem we compare two approaches
1) restrict the parameter m to appropriate "eigenvalues"
2) take the pricipal value
ad 1 "eigenvalues"
We have already shown here that the integral in question is convergent only if m is an integer multiple of 1/2.
In this case we can use the periodicity of the integrand to write the integral as
fye = 1 /2 Integrate[(Sin[y (2 m + 1)]/Sin[y] - 1), {y, -\[Pi]/
2, \[Pi]/2}];
The result is
(* fye = 1/2 (-\[Pi] +
1/2 (HarmonicNumber[-1 - m/2] - HarmonicNumber[-(1/2) - m/2] -
HarmonicNumber[1/2 (-1 + m)] + HarmonicNumber[m/2]) Sin[
m \[Pi]]) *)
For integer m this gives (notice that we have to take the Limit)
Table[Limit[fye, m -> i], {i, -4, 4}]
(* Out[169]= {-\[Pi], -\[Pi], -\[Pi], -\[Pi], 0, 0, 0, 0, 0} *)
ad 2 pricipal value
For general real m we can define the integral by the PrincipalValue.
fyp = 1/2 Integrate[(Sin[y (2 m + 1)]/Sin[ y] - 1), {y, \[Pi]/2,
3 \[Pi]/2}, PrincipalValue -> True]
(* 1/2 (-\[Pi] +
1/(4 m)E^(-3 I m \[Pi]) (-1 + E^(2 I m \[Pi])) (-2 I -
2 I E^(4 I m \[Pi]) + 2 E^(I m \[Pi]) m \[Pi] +
2 E^(3 I m \[Pi]) m \[Pi] - I m \[Pi] Cot[(m \[Pi])/2] +
I m PolyGamma[0, 1/2 - m/2] -
I (1 + E^(4 I m \[Pi])) m PolyGamma[0, m/2] +
I E^(4 I m \[Pi]) m PolyGamma[0, (1 + m)/2])) *)
In the limit of integer m we find
Table[Limit[fyp, m -> i], {i, -4, 4}]
(* Out[166]= {-\[Pi], -\[Pi], -\[Pi], -\[Pi], 0, 0, 0, 0, 0} *)
Comparison
We compare the functions of m obtained by the two approaches in a plot
Plot[{fye, fyp}, {m, -5, 5},
PlotLabel ->
"Comparison of definitions of the divergent integral\n\
\!\(\*FormBox[\(fyp\\\ = \*FractionBox[\(1\), \(2\)]\\\ \
\(\*SubsuperscriptBox[\(\[Integral]\), FractionBox[\(\[Pi]\), \(2\)], \
FractionBox[\(3\\\ \[Pi]\), \(2\)]]\((\*FractionBox[\(sin(\((2\\\ m + \
1)\)\\\ y)\), \(sin(y)\)] - 1)\) \[DifferentialD]y\)\),
TraditionalForm]\)\nPrincipalValue (yellow) and eigenvalue (blue)\n\!\
\(\*FormBox[\(fye\\\ = \\\ \*FractionBox[\(1\), \(2\)]\\\ \
\(\*SubsuperscriptBox[\(\[Integral]\), \(-\*FractionBox[\(\[Pi]\), \
\(2\)]\), FractionBox[\(\[Pi]\), \(2\)]]\((\*FractionBox[\(sin(\((2\\\
\ m + 1)\)\\\ y)\), \(sin(y)\)] - 1)\) \[DifferentialD]y\)\),
TraditionalForm]\)\nResults coincide for integer values of m\n",
Epilog -> {{PointSize[Large], (Point[{#, 0}] &) /@
Range[0, 6]}, {PointSize[
Large], (Point[{#, -\[Pi]}] &) /@ -Range[1, 5]}}]

This shows that the two methods lead to different "interpolating functions" which coincide at integer points but are different at half integer points.
Restriction to domain Integers
Unfortunately, restricting m
to the domain of Integers in the "interpolating functions" leads to wrong results in both cases:
Simplify[fye, {m \[Element] Integers, m > 0}]
(* Out[178]= -(\[Pi]/2) *)
Simplify[fyp, {m \[Element] Integers, m > 0}]
(* Out[179]= 1/4 (-1 + (-1)^m) \[Pi] *)
EDIT #1
I understand that you did not consider my answer as satisfactory because it relied on calculating the integral for some typical explicit integer constants m rather than finding an expression for the whole class of positive integer values of m which then proves to be zero for those m.
Although I don't have the answer of this type the following ideas might be interesting as they point to a difficulty or weakness (or even a bug) in Mathematica.
Changing the integration variable to
x -> y - \[Pi]/2;
the integral becomes
fy := Integrate[Cos[(m + 1) y] Sin[m y]/Sin[y], {y, -\[Pi]/2, 3 \[Pi]/2}]
Now replacing the trig function by complex exponentials we find the identity
FullSimplify[
Cos[(m + 1) y] Sin[m y]/Sin[y] == 1/2 (Sin[y (2 m + 1)]/Sin[y] - 1)]
(* Out[136]= True *)
which leads to a more symmetric form of the integrand.
Furthermore, the integrand is periodic for integer m:
FullSimplify[
Sin[y (2 m + 1)]/
Sin[y] == (Sin[y (2 m + 1)]/Sin[y] /. y -> y + \[Pi]),
m \[Element] Integers]
(* Out[137]= True *)
so that, for integer m, the integral over the range {-\[Pi]/2,\[Pi]/2}
is the same as that over the range {\[Pi]/2,3 \[Pi]/2
}, and hence the integral fy
becomes
fy1 = Integrate[(Sin[y (2 m + 1)]/Sin[y] - 1), {y, -\[Pi]/2, \[Pi]/2}]
(* Out[143]= -\[Pi] + 1/
2 (HarmonicNumber[-1 - m/2] - HarmonicNumber[-(1/2) - m/2] -
HarmonicNumber[1/2 (-1 + m)] + HarmonicNumber[m/2]) Sin[m \[Pi]] *)
in Latex:
$$fy1 = \frac{1}{2} \left(H_{\frac{m}{2}}+H_{-\frac{m}{2}-1}-H_{-\frac{m}{2}-\frac{1}{2}}-H_{\frac{m-1}{2}}\right) \sin (\pi m)-\pi$$
The graph is
Plot[fy1, {m, -5, 10}]

Unfortunately, simplifying fy1
to positive integer m
Mathematica gives a wrong result:
Simplify[fy1, {m \[Element] Integers, m >= 0}]
(* Out[145]= -\[Pi] *)
The correct result is zero.
The reason is that Mathematica does not notice that the zero in Sin[m \[Pi]]
is cancelled by the infinity in HarmonicNumber
for negative integers.
Here is an example to illustrate this weakness (or bug) of Mathematica:
Limit[HarmonicNumber[-k] Sin[\[Pi] k], k -> 1]
(* Out[156]= -\[Pi] *) correct
Simplify[HarmonicNumber[-k] Sin[\[Pi] k], k \[Element] Integers]
(* Out[147]= 0 *) wrong
But in the limit to an integer we find again
Table[Limit[fy1, m -> k], {k, -4, 4}]
(* Out[201]= {-2 \[Pi], -2 \[Pi], -2 \[Pi], -2 \[Pi], 0, 0, 0, 0, 0} *)
A similar example of the weakness, with more familiar functions, is:
Simplify[Gamma[-n] Sin[n \[Pi]], {n \[Element] Integers, n > 0}]
During evaluation of In[203]:= Simplify::infd: Expression Gamma[-n]
simplified to ComplexInfinity. >>
During evaluation of In[203]:= Simplify::infd: Expression Gamma[-n]
Sin[n [Pi]] simplified to Indeterminate. >>
(* Out[203]= Indeterminate *)
Limit[Gamma[-n] Sin[n \[Pi]], n -> #] & /@ {1, 2, 3, 4}
(* Out[204]= {-\[Pi], -(\[Pi]/2), -(\[Pi]/6), -(\[Pi]/24)} *)
Final remark: there should be some much simpler way - using symmetry arguments - to prove the correct result. But, alas, I haven't found it yet.
My original answer
Things become much simpler if you study the integral for explicit integer numbers m, rather than imposing the variable type Integers:
Let
f[m_] := Integrate[
Sec[x] Cos[(m + 1) (x + π/2)] Sin[m (x + π/2)], {x, -π, π}]
Then
Table[{m, f[m]}, {m, -5, 5}]
(* Out[46]= {{-5, -2 π}, {-4, -2 π}, {-3, -2 π}, {-2, -2 π}, {-1, -2 \
π}, {0, 0}, {1, 0}, {2, 0}, {3, 0}, {4, 0}, {5, 0}} *)
And let
f1[m_] := Integrate[
Sec[x] Cos[(m + 1) (x + π/2)] Sin[m (x + π/2)], {x, 0, π}]
Then
Table[{m, f1[m]}, {m, -5, 5}]
(* Out[48]= {{-5, -π}, {-4, -π}, {-3, -π}, {-2, -π}, {-1, -π}, {0,
0}, {1, 0}, {2, 0}, {3, 0}, {4, 0}, {5, 0}} *)
These results verify my comment.
It can easily be seen that due to the factor Sec[x] the integral is convergent only for integer and half integer m.
C.f. your other OP Integral of trigonometric function gives different answer