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$\int_0^\pi \frac{\mathrm{d} x}{4+3 \cos x}=$
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Verified Answer
The correct answer is:
$\frac{\pi}{\sqrt{7}}$
Let $\mathrm{I}=\int_0^\pi \frac{\mathrm{dx}}{4+3 \cos x}$
Put $\tan \frac{x}{2}=\mathrm{t}$
$\therefore \quad \mathrm{d} x=\frac{2 \mathrm{dt}}{1+\mathrm{t}^2}$ and $\cos x=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}$
$\begin{aligned} \therefore \quad \mathrm{I}=\int_0^\pi \frac{\mathrm{d} x}{4+3 \cos x} & =\int_0^{\infty} \frac{2 \mathrm{dt}}{7+\mathrm{t}^2} \\ & =\left[\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{\mathrm{t}}{\sqrt{7}}\right)\right]_0^{\infty} \\ & =\frac{2}{\sqrt{7}}\left[\tan ^{-1} \infty-0\right] \\ & =\frac{2}{\sqrt{7}} \cdot \frac{\pi}{2}=\frac{\pi}{\sqrt{7}}\end{aligned}$
Put $\tan \frac{x}{2}=\mathrm{t}$
$\therefore \quad \mathrm{d} x=\frac{2 \mathrm{dt}}{1+\mathrm{t}^2}$ and $\cos x=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}$
$\begin{aligned} \therefore \quad \mathrm{I}=\int_0^\pi \frac{\mathrm{d} x}{4+3 \cos x} & =\int_0^{\infty} \frac{2 \mathrm{dt}}{7+\mathrm{t}^2} \\ & =\left[\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{\mathrm{t}}{\sqrt{7}}\right)\right]_0^{\infty} \\ & =\frac{2}{\sqrt{7}}\left[\tan ^{-1} \infty-0\right] \\ & =\frac{2}{\sqrt{7}} \cdot \frac{\pi}{2}=\frac{\pi}{\sqrt{7}}\end{aligned}$
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