int_0^{pi} (sin^2 frac{x}{2} - cos^2 frac{x}{ | vedclass

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(B) We know that the trigonometric identity is $\cos^2 	heta - \sin^2 	heta = \cos 2	heta$.Therefore, $\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(

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$0$

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(B) We know that the trigonometric identity is $\cos^2 	heta - \sin^2 	heta = \cos 2	heta$.Therefore, $\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos(2 \cdot \frac{x}{2}) = -\cos x$.Now, the integral becomes $I = \int_0^{\pi} -\cos x \, dx$.Integrating $-\cos x$ with respect to $x$, we get $-\sin x$.Applying the limits from $0$ to $\pi$, we have $I = [-\sin x]_0^{\pi}$.$I = -(\sin \pi - \sin 0)$.Since $\sin \pi = 0$ and $\sin 0 = 0$, we get $I = -(0 - 0) = 0$.

$\int_0^{\pi} (\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}) dx = $ . . . . . . .

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