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The value of ∫0π/2cos⁡xcos⁡x2+sin⁡x2 dx\int_0^{\pi/2} \frac{\cos x}{\cos\frac{x}{2}+\sin\frac{x}{2}}\,dx is

Solution

✅ Correct Option: 2

Write cos⁡x=cos⁡2x2−sin⁡2x2=(cos⁡x2−sin⁡x2)(cos⁡x2+sin⁡x2)\cos x=\cos^2\frac x2-\sin^2\frac x2=\left(\cos\frac x2-\sin\frac x2\right)\left(\cos\frac x2+\sin\frac x2\right). The denominator cancels, leaving ∫0π/2(cos⁡x2−sin⁡x2)dx=[2sin⁡x2+2cos⁡x2]0π/2=22−2\int_0^{\pi/2}\left(\cos\frac x2-\sin\frac x2\right)dx=\left[2\sin\frac x2+2\cos\frac x2\right]_0^{\pi/2}=2\sqrt2-2.

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