The value of int_{0}^{pi / 2} frac{cos 3 x+1} | vedclass

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(B) Let $I = \int_{0}^{\pi / 2} \frac{\cos 3x + 1}{2 \cos x - 1} dx$. Using the identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we have $\cos 3x + 1 = 4 \c

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(B) Let $I = \int_{0}^{\pi / 2} \frac{\cos 3x + 1}{2 \cos x - 1} dx$. Using the identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we have $\cos 3x + 1 = 4 \cos^3 x - 3 \cos x + 1$. Since $\cos(\pi/3) = 1/2$,we can write $2 \cos x - 1 = 2(\cos x - \cos(\pi/3))$. Also,$4 \cos^3(\pi/3) - 3 \cos(\pi/3) + 1 = 4(1/8) - 3(1/2) + 1 = 1/2 - 3/2 + 1 = 0$. Thus,the numerator is $4 \cos^3 x - 3 \cos x + 1 = 4 \cos^3 x - 3 \cos x - (4 \cos^3(\pi/3) - 3 \cos(\pi/3))$. Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,we simplify the integrand: $\frac{4(\cos^3 x - \cos^3(\pi/3)) - 3(\cos x - \cos(\pi/3))}{2(\cos x - \cos(\pi/3))} = \frac{4(\cos^2 x + \cos x \cos(\pi/3) + \cos^2(\pi/3)) - 3}{2} = \frac{4 \cos^2 x + 2 \cos x + 1 - 3}{2} = 2 \cos^2 x + \cos x - 1$. Using $\cos^2 x = \frac{1 + \cos 2x}{2}$,we get $2(\frac{1 + \cos 2x}{2}) + \cos x - 1 = 1 + \cos 2x + \cos x - 1 = \cos 2x + \cos x$. Now,$I = \int_{0}^{\pi / 2} (\cos 2x + \cos x) dx = [\frac{1}{2} \sin 2x + \sin x]_{0}^{\pi / 2} = (\frac{1}{2} \sin \pi + \sin(\pi/2)) - (0 + 0) = 0 + 1 = 1$.

The value of $\int_{0}^{\pi / 2} \frac{\cos 3 x+1}{2 \cos x-1} d x$ is

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