int_0^{pi} frac{dx}{4+3 cos x} = | vedclass

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(B) Let $I = \int_0^{\pi} \frac{dx}{4+3 \cos x}$.Using the substitution $	an \frac{x}{2} = t$,we have $dx = \frac{2 dt}{1+t^2}$ and $\cos x = \frac{1

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$\frac{\pi}{\sqrt{7}}$

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(B) Let $I = \int_0^{\pi} \frac{dx}{4+3 \cos x}$.Using the substitution $	an \frac{x}{2} = t$,we have $dx = \frac{2 dt}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$.As $x$ goes from $0$ to $\pi$,$t$ goes from $	an(0) = 0$ to $	an(\frac{\pi}{2}) = \infty$.Substituting these into the integral:$I = \int_0^{\infty} \frac{\frac{2 dt}{1+t^2}}{4 + 3(\frac{1-t^2}{1+t^2})} = \int_0^{\infty} \frac{2 dt}{4(1+t^2) + 3(1-t^2)} = \int_0^{\infty} \frac{2 dt}{4 + 4t^2 + 3 - 3t^2} = \int_0^{\infty} \frac{2 dt}{7 + t^2}$.Using the standard integral formula $\int \frac{dx}{a^2+x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$:$I = 2 \left[ \frac{1}{\sqrt{7}} 	an^{-1}(\frac{t}{\sqrt{7}}) ight]_0^{\infty} = \frac{2}{\sqrt{7}} [	an^{-1}(\infty) - 	an^{-1}(0)]$.Since $	an^{-1}(\infty) = \frac{\pi}{2}$ and $	an^{-1}(0) = 0$:$I = \frac{2}{\sqrt{7}} 	imes \frac{\pi}{2} = \frac{\pi}{\sqrt{7}}$.

$\int_0^{\pi} \frac{dx}{4+3 \cos x} = $

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