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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
By using properties of definite integral, evaluate the following integral in Exercise:
$\int^{\pi}_{0}\frac{\text{x}\ \text{dx}}{1+\sin\text{x}}$
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Answer

$\text{Let}\ \text{I}=\int^{\pi}\limits_{0}\frac{\text{x}\ \text{dx}}{1+\sin\text{x}}\text{dx}$ $ \Rightarrow\ \ \text{I}=\int^{\pi}\limits_{0}\frac{\pi-\text{x}}{1+\sin(\pi-\text{x)}}\text{dx}=\int^{\pi}\limits_{0}\frac{\pi-\text{x}}{1+\sin\text{x}}\text{dx}$Adding eq.(i) and (ii),
$21=\int^{\pi}\limits_{0}\bigg(\frac{\text{x}}{1+\sin\text{x}}+\frac{\pi-\text{x}}{1+\sin\text{x}}\bigg)\text{dx}=\int^{\pi}\limits_{0}\bigg(\frac{\text{x}+\pi-\text{x}}{1+\sin\text{x}}\bigg)\text{dx}=\int^{\pi}\limits\limits_{0}\bigg(\frac{\pi}{1+\sin\text{x}}\bigg)\text{dx}=\pi\int^{\pi}\limits_{0}\bigg(\frac{1}{1+\sin\text{x}}\bigg)\text{dx}$
$\Rightarrow\ \ 21=2\pi\int^{\frac{\pi}{2}}\limits_{0}\frac{\text{dx}}{1+\sin\text{x}}\ \ \bigg[\because\int^{2\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}=2\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx},\text{if}\ \text{f}(2\text{a}-\text{x})=\text{f}\text{(x)}\bigg]$
$\Rightarrow\ \ 21=2\pi\int^{\frac{\pi}{2}}\limits_{0}\frac{\text{dx}}{1+\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)} \ \ \ \bigg[\because\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}=\int^{\text{a}}\limits_{0}\text{f}(\text{a}-\text{x})\text{dx}=\bigg]$
$\Rightarrow\ \ 21=2\pi\int^{\frac{\pi}{2}}\limits_{0}\frac{\text{dx}}{1+\cos\text{x}}$
$\Rightarrow\ \ \text{I}=\pi\int^{\frac{\pi}{2}}\limits_{0}\frac{\text{dx}}{2\cos^{2}\frac{\text{x}}{2}}=\frac{\pi}{2}\int^{\frac{\pi}{2}}_\limits{0}\sec^{2}\frac{\pi}{2}\text{dx}=\frac{\pi}{2}\Bigg[\frac{\tan\frac{\text{x}}{2}}{\frac{1}{2}}\Bigg]^{\frac{\pi}{2}}_{0}$
$=\pi\bigg(\tan\frac{\pi}{4}-\tan0^{\text{o}}\bigg)=\pi(1-0)=\pi$

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