## University Calculus: Early Transcendentals (3rd Edition)

$$\int^{\pi/2}_0\sin x\cos xdx=\frac{1}{2}$$
$$A=\int^{\pi/2}_0\sin x\cos xdx$$ Apply the identity of the product of sine and cosine functions: $$\sin mx\cos nx=\frac{1}{2}[\sin(m-n)x+\sin(m+n)x]$$ we have $$A=\frac{1}{2}\int^{\pi/2}_0(\sin0+\sin2x)dx$$ $$A=\frac{1}{2}\int^{\pi/2}_0\sin2xdx$$ $$A=-\frac{1}{4}\cos2x\Big]^{\pi/2}_0$$ $$A=-\frac{1}{4}(\cos\pi-\cos0)$$ $$A=-\frac{1}{4}(-1-1)=\frac{1}{2}$$
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