If int x^2 cos^2 x , dx = frac{1}{6} f(x) + g | vedclass

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(B) We know that $\cos^2 x = \frac{1 + \cos 2x}{2}$.Substituting this into the integral,we get $\int x^2 \left(\frac{1 + \cos 2x}{2}\right) dx = \frac

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$2$

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(B) We know that $\cos^2 x = \frac{1 + \cos 2x}{2}$.Substituting this into the integral,we get $\int x^2 \left(\frac{1 + \cos 2x}{2}\right) dx = \frac{1}{2} \int x^2 dx + \frac{1}{2} \int x^2 \cos 2x dx$.$= \frac{1}{2} \cdot \frac{x^3}{3} + \frac{1}{2} \left[ x^2 \cdot \frac{\sin 2x}{2} - \int 2x \cdot \frac{\sin 2x}{2} dx \right]$$= \frac{x^3}{6} + \frac{x^2 \sin 2x}{4} - \frac{1}{2} \int x \sin 2x dx$.Using integration by parts for $\int x \sin 2x dx = x \left(-\frac{\cos 2x}{2}\right) - \int 1 \cdot \left(-\frac{\cos 2x}{2}\right) dx = -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4}$.Substituting back: $\frac{x^3}{6} + \frac{x^2 \sin 2x}{4} - \frac{1}{2} \left( -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} \right) + c = \frac{x^3}{6} + \left( \frac{x^2}{4} - \frac{1}{8} \right) \sin 2x + \left( \frac{x}{4} \right) \cos 2x + c$.Comparing with $\frac{1}{6} f(x) + g(x) \sin 2x + h(x) \cos 2x + c$,we get $f(x) = x^3$,$g(x) = \frac{x^2}{4} - \frac{1}{8}$,and $h(x) = \frac{x}{4}$.Then $f(1) = 1^3 = 1$,$g(2) = \frac{2^2}{4} - \frac{1}{8} = 1 - \frac{1}{8} = \frac{7}{8}$,and $h(\frac{1}{2}) = \frac{1/2}{4} = \frac{1}{8}$.Finally,$f(1) + g(2) + h(\frac{1}{2}) = 1 + \frac{7}{8} + \frac{1}{8} = 1 + 1 = 2$.

If $\int x^2 \cos^2 x \, dx = \frac{1}{6} f(x) + g(x) \sin 2x + h(x) \cos 2x + c$,then $f(1) + g(2) + h(\frac{1}{2}) = $

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