Question

\(\displaystyle \int_{0}^{\pi/2}\frac{\cos 2x}{\sin x+\cos x}\,dx=\)

• A. \(-1\)
• B. \(0\)
• C. \(1\)
• D. \(\frac{\pi}{2}\)

Hint:

Use \(\cos 2x=\cos^2x-\sin^2x=(\cos x-\sin x)(\cos x+\sin x)\). This helps in cancelling the denominator easily.

Answer 1

The Correct Option is B

We need to evaluate: \[ \int_{0}^{\pi/2}\frac{\cos 2x}{\sin x+\cos x}\,dx. \] First, use the trigonometric identity: \[ \cos 2x=\cos^2x-\sin^2x. \] Now factorize: \[ \cos^2x-\sin^2x=(\cos x-\sin x)(\cos x+\sin x). \] Therefore, \[ \cos 2x=(\cos x-\sin x)(\cos x+\sin x). \] Substitute this in the integral: \[ \int_{0}^{\pi/2}\frac{(\cos x-\sin x)(\cos x+\sin x)}{\sin x+\cos x}\,dx. \] Since \[ \cos x+\sin x=\sin x+\cos x, \] we cancel the common factor: \[ \int_{0}^{\pi/2}(\cos x-\sin x)\,dx. \] Now integrate term by term: \[ \int(\cos x-\sin x)\,dx=\sin x+\cos x. \] So, \[ \int_{0}^{\pi/2}(\cos x-\sin x)\,dx = [\sin x+\cos x]_{0}^{\pi/2}. \] Substitute the upper limit: \[ \sin\frac{\pi}{2}+\cos\frac{\pi}{2}=1+0=1. \] Substitute the lower limit: \[ \sin0+\cos0=0+1=1. \] Therefore, \[ [\sin x+\cos x]_{0}^{\pi/2}=1-1=0. \] Hence, \[ \int_{0}^{\pi/2}\frac{\cos 2x}{\sin x+\cos x}\,dx=0. \]