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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2024 · Paper 1 · Mathematics · Definite Integration

∫_0^ (π)/(2) ( ((π)/(4)+x)+ ((3 π)/(4)+x))/( x+ x) d x=

∫0π2sin⁡(π4+x)+sin⁡(3π4+x)cos⁡x+sin⁡xdx=\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x=

Options

  1. Aπ2\frac{\pi}{\sqrt{2}}
  2. Bπ22\frac{\pi}{2 \sqrt{2}}
  3. Cπ32\frac{\pi}{3 \sqrt{2}}
  4. Dπ42\frac{\pi}{4 \sqrt{2}}

Correct answer

B. π22\frac{\pi}{2 \sqrt{2}}

Step-by-step solution

Let I=∫0π2sin⁡(π4+x)+sin⁡(3π4+x)cos⁡x+sin⁡xdx⇒I=∫0π/2sin⁡(3π4−x)+sin⁡(5π4−x)sin⁡x+cos⁡xdx⇒I=∫0π/2sin⁡(π4+x)−sin⁡(3π4+x)cos⁡x+sin⁡xdxI=\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x\Rightarrow I=\int_0^{\pi / 2} \frac{\sin \left(\frac{3 \pi}{4}-x\right)+\sin \left(\frac{5 \pi}{4}-x\right)}{\sin x+\cos x} d x\Rightarrow I=\int_0^{\pi / 2} \frac{\sin \left(\frac{\pi}{4}+x\right)-\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x Now I+I=∫0π/22sin⁡(π4+x)cos⁡x+sin⁡xdx⇒I=∫0π/212cos⁡x+12sin⁡xcos⁡x−sin⁡xdx=π22I+I=\int_0^{\pi / 2} \frac{2 \sin \left(\frac{\pi}{4}+x\right)}{\cos x+\sin x} d x\Rightarrow I=\int_0^{\pi / 2} \frac{\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x}{\cos x-\sin x} d x=\frac{\pi}{2 \sqrt{2}}
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