Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate: $ \int\limits_0^\pi\frac{\text{x tan x}}{\text{sec x + tan x}}\text{dx}$.
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Answer
$=\pi^{2}-2\pi$ $\text{I } = \int\limits^\pi_{0}\frac{\text{x tan x dx }}{\text{sec x + tan x }}\Rightarrow\text{I }=\int\limits_{0}^{\pi}\frac{(\pi-\text{x) tan x}}{\text{sec x + tan x}}\text{dx}$ $\text{2I } = \int\limits^\pi_{0}\frac{\pi\text{ tan x }}{\text{sec x + tan x }}=\pi\int\limits_{0}^{\pi}\frac{\text{sin x}}{\text{1+sin x}}\text{dx}$ $=\pi\int\limits_0^{\pi}\Bigg(1-\frac{1}{\text{1+sin x}}\Bigg)\text{dx}=\pi\int\limits_0^{\pi}\Bigg(1-\frac{\text{1-sin x}}{\text{cos}^{2}\text{x}}\text{dx}\Bigg)$ $=\pi\int\limits_0^\pi(\text{1-sec}^{2}\text{x}+\text{sec x tan x )}\text{dx}$ $=\pi[\text{x - tan x + sec x }]^{\pi}_{0}=\pi[\pi-1-1]$ $\therefore\text{I}=\frac{{\pi}^{2}}{2}-\pi=\frac{\pi}{2}[\pi-2]$.
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