It seems like you meant to write:
$$\int \frac{T}{2} \sin(Zx) \cos(?x) dx$$
Now, let's use the product-to-sum formula for sine and cosine:
$$\sin(A) \cos(B) = \frac{1}{2}[\sin(A-B) + \sin(A+B)]$$
In our case, $A = Zx$ and $B = ?x$. So, we have:
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