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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following definite integrals:$\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
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Answer

We have,$\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{\sin^2\text{x}+\cot^2\text{x}}{\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{1}{\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
Multiplying numberator and denominator by 2$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{2}{2\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{2}{\sin2\text{x}}\Big)^2\text{dx}$ $\big[\because2\sin\text{x}\cos\text{x}=\sin2\text{x}\big]$
$=4\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\text{cosec}^2\text{x dx}$
$=4\Big[-\frac{\cot2\text{x}}{2}\Big]^{\frac{\pi}{4}}_\frac{\pi}{3}$
$=2\Big[-\cot\frac{\pi}{2}+\cot2\frac{\pi}{3}\Big]$
$=2\Big[\frac{-1}{\sqrt{3}}-0\Big]$
$=\frac{-2}{\sqrt{3}}$
$\therefore\ \int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}=\frac{-2}{\sqrt{3}}$

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