int_0^{frac{pi}{2}} frac{d x}{cos x-sqrt{3} s | vedclass

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Question

(D) Let $I = \int_0^{\frac{\pi}{2}} \frac{dx}{\cos x - \sqrt{3} \sin x}$.Multiply and divide by $2$ to rewrite the denominator:$I = \int_0^{\frac{\pi}

Vedclass · Accepted Answer

$\frac{1}{2} \log (2 \sqrt{3}-3)$

Vedclass · Answer

(D) Let $I = \int_0^{\frac{\pi}{2}} \frac{dx}{\cos x - \sqrt{3} \sin x}$.Multiply and divide by $2$ to rewrite the denominator:$I = \int_0^{\frac{\pi}{2}} \frac{dx}{2(\frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x)} = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{dx}{\sin(\frac{\pi}{6} - x)}$.Using the identity $\sin(	heta) = \cos(\frac{\pi}{2} - 	heta)$, we have $\sin(\frac{\pi}{6} - x) = \cos(\frac{\pi}{2} - (\frac{\pi}{6} - x)) = \cos(x + \frac{\pi}{3})$.Thus, $I = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sec(x + \frac{\pi}{3}) dx$.The integral of $\sec(u)$ is $\log|\sec u + 	an u|$.$I = \frac{1}{2} [\log|\sec(x + \frac{\pi}{3}) + 	an(x + \frac{\pi}{3})|]_0^{\frac{\pi}{2}}$.Evaluating at the limits:At $x = \frac{\pi}{2}$, $\sec(\frac{5\pi}{6}) = -\frac{2}{\sqrt{3}}$ and $	an(\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}}$.At $x = 0$, $\sec(\frac{\pi}{3}) = 2$ and $	an(\frac{\pi}{3}) = \sqrt{3}$.$I = \frac{1}{2} [\log|-\frac{3}{\sqrt{3}}| - \log|2 + \sqrt{3}|] = \frac{1}{2} [\log(\sqrt{3}) - \log(2 + \sqrt{3})]$.This simplifies to $\frac{1}{2} \log(\frac{\sqrt{3}}{2+\sqrt{3}})$.Rationalizing the denominator: $\frac{\sqrt{3}(2-\sqrt{3})}{4-3} = 2\sqrt{3} - 3$.Thus, $I = \frac{1}{2} \log(2\sqrt{3} - 3)$.

$\int_0^{\frac{\pi}{2}} \frac{d x}{\cos x-\sqrt{3} \sin x}=$

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