Using All The Basic Techniques of Integration

Putting $y =\pi - x$ we have $I \; = \; \int_0^\pi \frac{\sin x(\sin x +1)e^{\sin x + \cos x}}{e^{\cos x} + 1}\,dx \; = \; \int_0^\pi \frac{\sin y(\sin y + 1)e^{\sin y - \cos y}}{e^{-\cos y} + 1}\,dy \; = \; \int_0^\pi \frac{\sin x (\sin x + 1) e^{\sin x}}{e^{\cos x} + 1}\,dx$ and hence $\begin{aligned} I & = \; \frac12\int_0^\pi \sin x(\sin x + 1)e^{\sin x}\,dx \; = \; \frac12\int_0^\pi \big(\sin x - \cos^2x + 1\big)e^{\sin x}\,dx \; = \; \frac12\Big[-\cos x e^{\sin x}\Big]_0^\pi + \frac12\int_0^\pi e^{\sin x}\,dx \\ & = \; 1 + \frac12\int_0^\pi e^{\sin x}\,dx \end{aligned}$ making the answer $100\big(1 + \frac14\big) = \boxed{125}$ .

1. Subtracting and adding $1.$

$\begin{aligned} \int_{0}^{\pi} \frac{\sin x(\sin x + 1)e^{\sin x + \cos x}}{e^{\cos x}+1} dx &=\int_{0}^{\pi} \frac{(\sin^2 x \, {\color{#D61F06} -1}+ \sin x \, {\color{#D61F06} +1})e^{\sin x + \cos x}}{e^{\cos x}+1} dx \\ & = \int_{0}^{\pi} \frac{(\sin x - \cos^2 x + 1)e^{\sin x + \cos x}}{e^{\cos x}+1} dx \\ & = \int_{0}^{\pi} \frac{e^{\sin x + \cos x}}{e^{\cos x}+1} dx + \int_{0}^{\pi} \frac{(\sin x - \cos^2 x)e^{\sin x + \cos x}}{e^{\cos x}+1} dx\\ \end{aligned}$

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2. Figuring out the first one.

Let $f(x)=e^{\sin x}$ and $h(x)=\dfrac{e^{\cos x}}{e^{\cos x}+1}.$

Notice that $f(x)=f(\pi-x)$ and $h(x)+h(\pi-x)=\dfrac{e^{\cos x}}{e^{\cos x}+1}+\dfrac{e^{-\cos x}}{e^{-\cos x}+1}=1.$

$\begin{aligned} \int_{0}^{\pi} \frac{e^{\sin x + \cos x}}{e^{\cos x}+1} dx & = \int_{0}^{\pi} f(x)h(x) \, dx \\ & = \int_{0}^{\pi} f(x)(1-h(\pi-x)) \, dx \\ & = \int_{0}^{\pi} f(x) \,dx - \int_{0}^{\pi} f(\pi-x)h(\pi-x) \, dx\\ & = \int_{0}^{\pi}e^{\sin x} dx - \int_{0}^{\pi} f(t)h(t) \, dt && \small \color{#3D99F6} \pi-x=t;~-dx=dt. \\ \end{aligned}$

Therefore, $\displaystyle \int_{0}^{\pi} \frac{e^{\sin x + \cos x}}{e^{\cos x}+1} dx = \frac{1}{2}\int_{0}^{\pi}e^{\sin x} dx.$

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3. Figuring out the second one.

Let $g(x)=(\sin x - \cos^2 x)e^{\sin x}.$

You will notice that $g(x)=g(\pi-x).$

Then using the same method, we obtain $\displaystyle \int_{0}^{\pi} g(x)h(x)\,dx=\frac{1}{2}\int_{0}^{\pi}g(x)\,dx.$

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4. Figuring out the value of $\displaystyle \int_{0}^{\pi}g(x)\,dx.$

We're trying to get $\displaystyle \int_{0}^{\pi} (\sin x - \cos^2 x)e^{\sin x} \, dx,$ but firstly, we'll mess around with

$\begin{aligned} & \int \sin x e^{\sin x} \, dx \\ & = (-\cos x)\cdot (e^{\sin x}) - \int (-\cos x) \cdot (\cos x e^{\sin x}) \, dx + C\\ & = -\cos x e^{\sin x} + \int \cos^2 x e^{\sin x} \, dx + C \end{aligned}$

(!!!) From this, we can see that $\displaystyle \int (\sin x - \cos^2 x)e^{\sin x} \, dx=-\cos x e^{\sin x} + C,$ leading to

$\displaystyle \int_{0}^{\pi}g(x)\,dx=1 - (-1)=2.$

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5. Sum up all the things we figured out.

$\begin{aligned} & \int_{0}^{\pi} \frac{\sin x(\sin x + 1)e^{\sin x + \cos x}}{e^{\cos x}+1} dx \\ & = \int_{0}^{\pi} \frac{e^{\sin x + \cos x}}{e^{\cos x}+1} dx + \int_{0}^{\pi} \frac{(\sin x - \cos^2 x)e^{\sin x + \cos x}}{e^{\cos x}+1} dx \\ & = \int_{0}^{\pi} f(x)h(x) \,dx + \int_{0}^{\pi} g(x)h(x) \, dx \\ & = \frac{1}{2}\int_{0}^{\pi} f(x) \, dx + \dfrac{1}{2}\int_{0}^{\pi} g(x) \, dx \\ & = \frac{1}{2}\int_{0}^{\pi}e^{\sin x} dx + 1. \end{aligned}$

From above, $a=1$ and $b=\dfrac{1}{2}.$

Therefore, $100(a^2+b^2)=\boxed{125}.$