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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS3 Marks
Question
Evaluate the definite integral in Exercise:
$\int\limits_\frac{\pi}{2}^{\pi}\text{e}^{\text{x}}\bigg(\frac{1-\sin\text{x}}{1+\cos\text{x}}\bigg)\text{dx}$
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Answer

$\text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\bigg(\frac{1-\sin\text{x}}{1+\cos\text{x}}\bigg)\text{dx}$ $\text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\Bigg[\frac{1-2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\sin^{2}\frac{\text{x}}{2}}\Bigg]\text{dx}$ $\text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\Bigg[\frac{\text{cosec}^{2}\frac{\text{x}}{2}}{2}-\cot\frac{\text{x}}{2}\Bigg]\text{dx}$ $\text{Let f (x)}=-\cot\frac{\text{x}}{2}$ $\Rightarrow\text{f (x)}=-\bigg(-\frac{1}{2}\text{cosec}^{2}\frac{\text{x}}{2}\bigg)=\frac{1}{2}\text{cosec}^{2}\frac{\text{x}}{2}$ $\therefore\ \text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\Big[\text{f(x)}+\text{f}'\text{(x)}\Big]\text{dx}$$=\Big[\text{e}^{\text{x}}.\text{f (x) dx}\Big]_{\frac{\pi}{2}}^{\pi}$
$=-\bigg[\text{e}^\text{x}.\cot\frac{\text{x}}{2}\bigg]^{\pi}_{\frac{\pi}{2}}$ $=-\bigg[\text{e}^{\text{x}}\times\cot\frac{\pi}{2}-\text{e}^{\frac{\pi}{2}}\times\cot\frac{\pi}{4}\bigg]$ $=-\bigg[\text{e}^{\text{x}}\times0-\text{e}^{\frac{\pi}{2}}\times1\bigg]$$=\text{e}^{\frac{\pi}{2}}$

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