int_0^pi frac{1}{4+3 cos x} d x= | vedclass

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

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

Vedclass · Answer

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

$\int_0^\pi \frac{1}{4+3 \cos x} d x=$

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