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Before using the Fundamental Theorem of Calculus part I the integral must be split into two integrals and the limits of integration must be reversed on the 2nd integral.
This gives \(\displaystyle F(x) = \int\limits_{0}^{\sin{\left(x \right)}} \left(- 5 t - 7\right)\, dt-\int\limits_{0}^{\cos{\left(x \right)}} \left(- 5 t - 7\right)\, dt\). Now using the Fundamental Theorem of Calculus part I and the chain rule with \(\displaystyle u=\sin{\left(x \right)}\) and \(\displaystyle F(u)=\int\limits_{0}^{u} \left(- 5 t - 7\right)\, dt\) and \(\displaystyle v=\cos{\left(x \right)}\) and \(\displaystyle F(v)=- \int\limits_{0}^{v} \left(- 5 t - 7\right)\, dt\) gives: \(\displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx}= \left(- 5 u - 7\right)\left(\cos{\left(x \right)}\right)+\left(5 v + 7\right)\left(- \sin{\left(x \right)}\right)=\left(- 5 \sin{\left(x \right)} - 7\right) \cos{\left(x \right)} + \left(- 5 \cos{\left(x \right)} - 7\right) \sin{\left(x \right)}\)

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\begin{question}Find the derivative of $F(x) = \int\limits_{\cos{\left(x \right)}}^{\sin{\left(x \right)}} \left(- 5 t - 7\right)\, dt$. 
    \soln{9cm}{Before using the Fundamental Theorem of Calculus part I the integral must be split into two integrals and the limits of integration must be reversed on the 2nd integral.\newline This gives $F(x) = \int\limits_{0}^{\sin{\left(x \right)}} \left(- 5 t - 7\right)\, dt-\int\limits_{0}^{\cos{\left(x \right)}} \left(- 5 t - 7\right)\, dt$. Now using the Fundamental Theorem of Calculus part I and the chain rule with $u=\sin{\left(x \right)}$  and $F(u)=\int\limits_{0}^{u} \left(- 5 t - 7\right)\, dt$ and $v=\cos{\left(x \right)}$  and $F(v)=- \int\limits_{0}^{v} \left(- 5 t - 7\right)\, dt$ gives: $F'(x)=\frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx}= \left(- 5 u - 7\right)\left(\cos{\left(x \right)}\right)+\left(5 v + 7\right)\left(- \sin{\left(x \right)}\right)=\left(- 5 \sin{\left(x \right)} - 7\right) \cos{\left(x \right)} + \left(- 5 \cos{\left(x \right)} - 7\right) \sin{\left(x \right)}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

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HTML for Canvas

<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle F(x) = \int\limits_{\cos{\left(x \right)}}^{\sin{\left(x \right)}} \left(- 5 t - 7\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint%5Climits_%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%5E%7B%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%5Cleft%28-%205%20t%20-%207%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(x) = \int\limits_{\cos{\left(x \right)}}^{\sin{\left(x \right)}} \left(- 5 t - 7\right)\, dt " data-equation-content=" \displaystyle F(x) = \int\limits_{\cos{\left(x \right)}}^{\sin{\left(x \right)}} \left(- 5 t - 7\right)\, dt " /> . </p> </p>

HTML for Canvas

<p> <p>Before using the Fundamental Theorem of Calculus part I the integral must be split into two integrals and the limits of integration must be reversed on the 2nd integral.<br> This gives  <img class="equation_image" title=" \displaystyle F(x) = \int\limits_{0}^{\sin{\left(x \right)}} \left(- 5 t - 7\right)\, dt-\int\limits_{0}^{\cos{\left(x \right)}} \left(- 5 t - 7\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint%5Climits_%7B0%7D%5E%7B%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%5Cleft%28-%205%20t%20-%207%5Cright%29%5C%2C%20dt-%5Cint%5Climits_%7B0%7D%5E%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%5Cleft%28-%205%20t%20-%207%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(x) = \int\limits_{0}^{\sin{\left(x \right)}} \left(- 5 t - 7\right)\, dt-\int\limits_{0}^{\cos{\left(x \right)}} \left(- 5 t - 7\right)\, dt " data-equation-content=" \displaystyle F(x) = \int\limits_{0}^{\sin{\left(x \right)}} \left(- 5 t - 7\right)\, dt-\int\limits_{0}^{\cos{\left(x \right)}} \left(- 5 t - 7\right)\, dt " /> . Now using the Fundamental Theorem of Calculus part I and the chain rule with  <img class="equation_image" title=" \displaystyle u=\sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20u%3D%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle u=\sin{\left(x \right)} " data-equation-content=" \displaystyle u=\sin{\left(x \right)} " />   and  <img class="equation_image" title=" \displaystyle F(u)=\int\limits_{0}^{u} \left(- 5 t - 7\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28u%29%3D%5Cint%5Climits_%7B0%7D%5E%7Bu%7D%20%5Cleft%28-%205%20t%20-%207%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(u)=\int\limits_{0}^{u} \left(- 5 t - 7\right)\, dt " data-equation-content=" \displaystyle F(u)=\int\limits_{0}^{u} \left(- 5 t - 7\right)\, dt " />  and  <img class="equation_image" title=" \displaystyle v=\cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20v%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle v=\cos{\left(x \right)} " data-equation-content=" \displaystyle v=\cos{\left(x \right)} " />   and  <img class="equation_image" title=" \displaystyle F(v)=- \int\limits_{0}^{v} \left(- 5 t - 7\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28v%29%3D-%20%5Cint%5Climits_%7B0%7D%5E%7Bv%7D%20%5Cleft%28-%205%20t%20-%207%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(v)=- \int\limits_{0}^{v} \left(- 5 t - 7\right)\, dt " data-equation-content=" \displaystyle F(v)=- \int\limits_{0}^{v} \left(- 5 t - 7\right)\, dt " />  gives:  <img class="equation_image" title=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx}= \left(- 5 u - 7\right)\left(\cos{\left(x \right)}\right)+\left(5 v + 7\right)\left(- \sin{\left(x \right)}\right)=\left(- 5 \sin{\left(x \right)} - 7\right) \cos{\left(x \right)} + \left(- 5 \cos{\left(x \right)} - 7\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20F%27%28x%29%3D%5Cfrac%7BdF%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdx%7D%20%2B%20%5Cfrac%7BdF%7D%7Bdv%7D%5Cfrac%7Bdv%7D%7Bdx%7D%3D%20%5Cleft%28-%205%20u%20-%207%5Cright%29%5Cleft%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%5Cright%29%2B%5Cleft%285%20v%20%2B%207%5Cright%29%5Cleft%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%5Cright%29%3D%5Cleft%28-%205%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20-%207%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%28-%205%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20-%207%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx}= \left(- 5 u - 7\right)\left(\cos{\left(x \right)}\right)+\left(5 v + 7\right)\left(- \sin{\left(x \right)}\right)=\left(- 5 \sin{\left(x \right)} - 7\right) \cos{\left(x \right)} + \left(- 5 \cos{\left(x \right)} - 7\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx}= \left(- 5 u - 7\right)\left(\cos{\left(x \right)}\right)+\left(5 v + 7\right)\left(- \sin{\left(x \right)}\right)=\left(- 5 \sin{\left(x \right)} - 7\right) \cos{\left(x \right)} + \left(- 5 \cos{\left(x \right)} - 7\right) \sin{\left(x \right)} " /> </p> </p>