The value of \( \int_{0}^{\pi/2}\displaystyle\frac{\cos x}{\cos\displaystyle\frac{x}{2}+\sin\displaystyle\frac{x}{2}}\,dx \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2\sqrt{2}-2 \)
By the double angle formula, cos x is a difference of squares:
\( \displaystyle \cos x=\cos^2\frac{x}{2}-\sin^2\frac{x}{2} \)
\( \displaystyle =\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right) \)
So the denominator cancels:
\( \displaystyle \int_{0}^{\pi/2}\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)dx \)
\( \displaystyle =\left[2\sin\frac{x}{2}+2\cos\frac{x}{2}\right]_{0}^{\pi/2} \)
\( \displaystyle =2\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-2(0+1) \)
\( \displaystyle =2\sqrt{2}-2 \)