Option 1: Evaluate $\int_{0}^{\frac{\pi}{2}}e^{x }\sin x~dx$

Let $I = \int e^x \sin x \, dx$. We will use integration by parts.

Let $u = \sin x$ and $dv = e^x dx$. Then $du = \cos x \, dx$ and $v = e^x$.

Using integration by parts, $\int u \, dv = uv - \int v \, du$, we have:

$I = e^x \sin x - \int e^x \cos x \, dx$

Now, we need to integrate $\int e^x \cos x \, dx$. Let $u = \cos x$ and $dv = e^x dx$. Then $du = -\sin x \, dx$ and $v = e^x$.

$\int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx = e^x \cos x + I$

Substituting this back into the expression for $I$, we get:

$I = e^x \sin x - (e^x \cos x + I)$

$I = e^x \sin x - e^x \cos x - I$

$2I = e^x (\sin x - \cos x)$

$I = \frac{1}{2} e^x (\sin x - \cos x) + C$

Now, we evaluate the definite integral:

$\int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx = \frac{1}{2} [e^x (\sin x - \cos x)]_{0}^{\frac{\pi}{2}}$

$= \frac{1}{2} [e^{\frac{\pi}{2}} (\sin \frac{\pi}{2} - \cos \frac{\pi}{2}) - e^0 (\sin 0 - \cos 0)]$

$= \frac{1}{2} [e^{\frac{\pi}{2}} (1 - 0) - 1 (0 - 1)]$

$= \frac{1}{2} [e^{\frac{\pi}{2}} + 1]$

Option 2: Find $\int\frac{1}{\cos(x-a)\cos(x-b)}dx$

Multiply and divide by $\sin(a-b)$:

$\int \frac{1}{\cos(x-a)\cos(x-b)} dx = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b)}{\cos(x-a)\cos(x-b)} dx$

$= \frac{1}{\sin(a-b)} \int \frac{\sin((x-b) - (x-a))}{\cos(x-a)\cos(x-b)} dx$

$= \frac{1}{\sin(a-b)} \int \frac{\sin(x-b)\cos(x-a) - \cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)} dx$

$= \frac{1}{\sin(a-b)} \int \left( \frac{\sin(x-b)}{\cos(x-b)} - \frac{\sin(x-a)}{\cos(x-a)} \right) dx$

$= \frac{1}{\sin(a-b)} \int (\tan(x-b) - \tan(x-a)) dx$

$= \frac{1}{\sin(a-b)} [-\ln|\cos(x-b)| + \ln|\cos(x-a)|] + C$

$= \frac{1}{\sin(a-b)} \ln \left| \frac{\cos(x-a)}{\cos(x-b)} \right| + C$