Q70 / 125 · BITSAT 2012 · Mathematics

$\int_{0}^{\pi / 2} \frac{2^{\sin x}}{2^{\sin x}+2^{\cos x}} d x$ equals

  • A
    2
  • B
    $\pi$
  • Correct answer
    C
    $\pi / 4$
  • D
    $\pi / 2$

Solution

$\mathrm{I}=\int_{0}^{\pi / 2} \frac{2^{\sin \mathrm{x}}}{2^{\sin x}+2^{\cos \mathrm{x}} \mathrm{dx}}$ $\mathrm{I}=\int_{0}^{\pi / 2} \frac{2^{\sin (\pi / 2-\mathrm{x})}}{2^{\sin (\pi / 2-\mathrm{x})}+2^{\cos (\pi / 2-\mathrm{x})} \mathrm{dx}}$ $=\int \frac{2^{\cos \mathrm{x}}}{2^{\cos \mathrm{x}}+2^{\sin \mathrm{x}}} \mathrm{dx} \Rightarrow 21=\int_{0}^{\pi / 2} \mathrm{~d} \mathrm{x}=\frac{\pi}{2} \Rightarrow 1=\frac{\pi}{4}$

Hence, the correct answer is option: 'C'.

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