Easy Semi-Trigonometric Integral

We appeal to the identity $\int e^x \cdot \left ( f(x) + f'(x) \right ) dx = e^x \cdot f(x) + C$

Consider $f(x) = \frac { \sin x}{1+\cos x}$ , then $f'(x) = \frac { 1}{1 + \cos x}$ , which satisfies the integral

Evaluate: $\large e^x \cdot \frac { \sin x}{1+\cos x} ]_0^{ {\pi /2} } = \sqrt{ e^{ \pi }}$

$\left \lceil \sqrt{ e^{ \pi }} \right \rceil = \boxed {5}$

We know that:

$\displaystyle \sin x = \frac{2\tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}$

$\displaystyle \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1+ \tan^2 \frac{x}{2}}$ ,

Replace these values to get,

$\displaystyle \int e^x \bigg(\tan \frac{x}{2} + \frac{1}{2} \sec^2 \frac{x}{2} \bigg) \text{dx} = e^x \tan \frac{x}{2}$

Putting limits, we get $\displaystyle e^{\frac{\pi}{2}}$ .

Note- I have used:

$\displaystyle \int e^x (f(x) + f'(x)) \text{dx} = e^x f(x) - \int e^x f'(x) \text{dx} + \int e^x f'(x) \text{dx} = e^{x} f(x)$ , using by parts. Here check that $f(x) = \tan \frac{x}{2}$