Integrating $\int_{-4}^0\sin(z(1+w))dz = \frac{-1}{1+w}+\frac{1}{1+w}\cos(4(1+w))$

Question

why is the following integral,

$$\int_{-4}^0\sin(z(1+w))dz = \frac{-1}{1+w}+\frac{1}{1+w}\cos(4(1+w))$$

when I worked it out myself this is what I got and I'm not sure where I am going wrong

$$\int_{-4}^0\sin(z(1+w))dz = -\frac{1}{1+w}\cos(z(1+w))\Bigg|_{-4}^0 \\= \left[ -\frac{1}{1+w} \right] - \left[ -\frac{1}{1+w}\cos(-4(1+w)\right] \\= \left[ -\frac{1}{1+w} \right] - \left[ \frac{1}{1+w}\cos(4(1+w)\right] \\= \frac{-1}{1+w}-\frac{1}{1+w}\cos(4(1+w))$$

It seems to be the second term which I am having trouble with, isn't the following true,

$$\left[ -\frac{1}{1+w}\cos(-4(1+w)\right] = \left[ \frac{1}{1+w}\cos(4(1+w)\right] $$

Answer 1

Cosine is even function, so the third line should contain $+$ instead of $-$. $$\cos(-x)=\cos x$$ and definitely not $-\cos x$.