Evaluate the following integrals: _ 6 ^ 3 ( + )^2dx - Vidyadip

Question

Evaluate the following integrals:$\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}(\tan\text{x}+\cot\text{x})^2\text{dx}$

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Answer

$\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}(\tan\text{x}+\cot\text{x})^2\text{dx}$$=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\big(\tan^2\text{x}+\cot^2\text{x}+2\tan\text{x }\cot\text{x})\text{dx}$
$=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\big(\sec^2\text{x}-1+\text{cosec}^2\text{x}-1+2\big)\text{dx}$
$=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\sec^2\text{x dx}+\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\text{cosec}^2\text{x dx}$
$=\big[\tan\text{x}+(-\cot\text{x})\big]^{\frac{\pi}{3}}_\frac{\pi}{6}$
$=\Big(\tan\frac{\pi}{3}-\tan\frac{\pi}{6}\Big)-\Big(\cot\frac{\pi}{3}-\cot\frac{\pi}{6}\Big)$
$=\Big(\sqrt{3}-\frac{1}{\sqrt{3}}\Big)-\Big(\frac{1}{\sqrt{3}}-\sqrt{3}\Big)$
$=2\sqrt{3}-\frac{2}{3}$
$=\frac{4}{\sqrt{3}}$

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