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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Evaluate:
$\int\limits^{\pi}_{0} \frac{\text{x} \tan \text{x}}{\sec \text{x} + \tan \text{x}}\text{dx}$
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Answer

$\text{I} = \int\limits^{\pi}_{0} \frac{\text{x} \tan \text{x}}{\sec \text{x} + \tan \text{x}} \text{dx} = \int\limits^{\pi}_{0} \frac{(\pi - \text{x)} \tan \text{x}}{\sec \text{x} + \tan \text{x}} \text{dx}$
$\Rightarrow \text{2I} = \pi \int\limits^{\pi}_{0} \frac{{\tan \text{x}}}{\sec {\text{x} + \tan \text{x}}} \text{dx} = \pi \int\limits^{\pi}_{0} \tan \text{x} (\sec \text{x} - \tan \text{x}) \text{dx}$
$\text{I} = \frac{\pi}{2} \int\limits^{\pi}_{0} {(\sec \text{x} \tan \text{x} - \sec^{2} \text{x + 1) dx}}$
$= \frac{\pi}{2} [ \sec \text{x} - \tan \text{x + x}]^{\pi}_{0}$
$= \frac{\pi(\pi - 2)}{2}$

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