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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2012 · Mathematics · Definite Integration

∫_ 0 ^ π / 2 2^ x 2^ x +2^ x d x equals

∫0π/22sin⁡x2sin⁡x+2cos⁡xdx\int_{0}^{\pi / 2} \frac{2^{\sin x}}{2^{\sin x}+2^{\cos x}} d x equals

Options

  1. A2
  2. Bπ\pi
  3. Cπ/4\pi / 4
  4. Dπ/2\pi / 2

Correct answer

C. π/4\pi / 4

Step-by-step solution

I=∫0π/22sin⁡x2sin⁡x+2cos⁡xdx\mathrm{I}=\int_{0}^{\pi / 2} \frac{2^{\sin \mathrm{x}}}{2^{\sin x}+2^{\cos \mathrm{x}} \mathrm{dx}}\mathrm{I}=∫0π/22sin⁡(π/2−x)2sin⁡(π/2−x)+2cos⁡(π/2−x)dx=∫2cos⁡x2cos⁡x+2sin⁡xdx⇒21=∫0π/2dx=π2⇒1=π4=\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}
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