\( \int _ { 0 } ^ { t } \sin u e ^ { t - u } d u \)

Calculate the integral \( \int_{0}^{t} \sin(u) e^{t-u} du \). Evaluate the integral by following steps: - step0: Evaluate using partial integration formula: \(\int_{0}^{t} \sin\left(u\right)\times e^{t-u} du\) - step1: Evaluate the integral: \(\int \sin\left(u\right)\times e^{t-u} du\) - step2: Prepare for integration by parts: \(\begin{align}&u=\sin\left(u\right)\\&dv=e^{t-u}du\end{align}\) - step3: Calculate the derivative: \(\begin{align}&du=\cos\left(u\right)du\\&dv=e^{t-u}du\end{align}\) - step4: Evaluate the integral: \(\begin{align}&du=\cos\left(u\right)du\\&v=-e^{t-u}\end{align}\) - step5: Substitute the values into formula: \(\sin\left(u\right)\left(-e^{t-u}\right)-\int \cos\left(u\right)\left(-e^{t-u}\right) du\) - step6: Calculate: \(-\sin\left(u\right)\times e^{t-u}-\int -\cos\left(u\right)\times e^{t-u} du\) - step7: Evaluate the integral: \(-\sin\left(u\right)\times e^{t-u}-\cos\left(u\right)\times e^{t-u}-\int \sin\left(u\right)\times e^{t-u} du\) - step8: Write the expression as an equation: \(\int \sin\left(u\right)\times e^{t-u} du=-\sin\left(u\right)\times e^{t-u}-\cos\left(u\right)\times e^{t-u}-\int \sin\left(u\right)\times e^{t-u} du\) - step9: Move the expression to the left side: \(2\times \int \sin\left(u\right)\times e^{t-u} du=-\sin\left(u\right)\times e^{t-u}-\cos\left(u\right)\times e^{t-u}\) - step10: Divide both sides: \(\int \sin\left(u\right)\times e^{t-u} du=-\frac{e^{t-u}\sin\left(u\right)}{2}-\frac{e^{t-u}\cos\left(u\right)}{2}\) - step11: Rewrite the expression: \(-\frac{e^{t-u}\sin\left(u\right)}{2}-\frac{e^{t-u}\cos\left(u\right)}{2}\) - step12: Return the limits: \(\left(-\frac{e^{t-u}\sin\left(u\right)}{2}-\frac{e^{t-u}\cos\left(u\right)}{2}\right)\bigg |_{0}^{t}\) - step13: Calculate the value: \(\frac{e^{t}}{2}-\frac{1}{2}\sin\left(t\right)-\frac{1}{2}\cos\left(t\right)\) The integral of \( \sin(u) e^{t-u} \) from 0 to t is \( \frac{e^{t}}{2} - \frac{1}{2}\sin(t) - \frac{1}{2}\cos(t) \).