Evaluate: ^ /2 _ 0 2^ 2^ + 2^ x dx - Vidyadip

Question

Evaluate:
$\int\limits^{\pi/2}_{0} \frac{2^{\sin\text{x}}}{2^{\sin\text{x}} + 2^{\cos \text{x}}}\text{dx}$

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Answer

$\text{I} = \int\limits^{\pi/2}_{0} \frac{2^{\sin\text{x}}}{2^{\sin\text{x}} + 2^{\cos \text{x}}}\text{dx}\dots\dots\dots\dots\dots\text{(i)}$
$ = \int\limits^{\pi/2}_{0}\frac{2^{\sin\bigg(\frac{\pi}{2} - \text{x}\bigg)}}{2^{\sin\bigg(\frac{\pi}{2} - \text{x}\bigg)} + 2^{\cos\bigg(\frac{\pi}{2} - \text{x}\bigg)}} \text{dx}\Bigg[\text{using}\int\limits^{\text{a}}_{0}\text{f (a - x) dx}\Bigg]$
$ = \int\limits^{\pi/2}_{0}\frac{2^{\cos\text{x}}}{2^{\sin \text{x}} + 2^{\cos \text{x}}}\text{dx}\dots\dots\dots\dots\dots\dots\dots\text{(ii)}$
$\text{Adding (i) and (ii),}$
$\text{2 I} \int\limits^{\pi/2}_{0}1 \text{dx} = [\text{x}]^{\pi/2}_{0} = \frac{\pi}{2}$
$\Rightarrow\text{I} = \frac{\pi}{4}$

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