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$$\text {Let }H(x)=(\int^{\pi}_{\sin(x)}\frac{t}{\cos(t)}dt)^3$$

My steps:

$$H(x) = (-\int^{\sin(x)}_{\pi} \frac{t}{\cos(t)}dt)^3$$ $$H'(x) = 3(-\int^{\sin(x)}_{\pi} \frac{t}{\cos(t)}dt)^2\frac{\sin(x)}{\cos(\sin(x)}\cos(x)$$ $$H'(x) = \frac{3\sin(x)\cos(x)}{\cos(\sin(x))}(-\int^{\sin(x)}_{\pi} \frac{t}{\cos(t)}dt)^2$$

Is this correct.

Also can I pull the negative sign out or not? It's squared, so can I really?

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    Your solution is fine. The minus sign is squared, and so $(-1)^2=1$. – 2017-02-08
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    I think there is a missing minus sign in the second line where you take the derivative of the negative integral, so you should end up with a negative sign out front. – 2017-02-08

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