Integrate the function in Exercise: e^x ( 1+ 1+ ) - Vidyadip

Question

Integrate the function in Exercise:
$\text{e}^\text{x}\Bigg(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Bigg)$

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Answer

$\text{e}^\text{x}\Bigg(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Bigg)$
$=\text{e}^\text{x}\Bigg(\frac{\sin^2\frac{\text{x}}{2}+\cos^2\frac{\text{x}}{2}+2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\Bigg)$
$=\frac{\text{e}^\text{x}\Big(\sin\frac{\text{x}}{2}+\cos\frac{\text{x}}{2}\Big)^2}{2\cos^2\frac{\text{x}}{2}}$
$=\frac{1}{2}\text{e}^\text{x}.\Bigg(\frac{\sin\frac{\text{x}}{2}+\cos\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}\Bigg)^2$
$=\frac{1}{2}\text{e}^\text{x}\Bigg[\tan\frac{\text{x}}{2}+1\Bigg]^2$
$=\frac{1}{2}\text{e}^\text{x}\Bigg[1+\tan\frac{\text{x}}{2}\Bigg]^2$
$=\frac{1}{2}\text{e}^\text{x}\Bigg[1+\tan^2\frac{\text{x}}{2}+2\tan\frac{\text{x}}{2}\Bigg]$
$=\frac{1}{2}\text{e}^\text{x}\Bigg[\sec^2\frac{\text{x}}{2}+2\tan\frac{\text{x}}{2}\Bigg]$
$\frac{\text{e}^\text{x}(1+\sin\text{x})\text{dx}}{(1+\cos\text{x})}=\text{e}^\text{x}\Bigg[\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\Bigg]$
Let $\tan\frac{\text{x}}{2}=\text{f(x)}\Rightarrow\ \text{f}'(\text{x})=\frac{1}{2}\sec^2\frac{\text{x}}{2}$
It is known that, $\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}=\text{e}^\text{x}\ \ \text{f}(\text{x})+\text{C}$
From equation (1), we obtain
$\int\frac{\text{e}^\text{x}(1+\sin\text{x})}{(1+\cos\text{x})}\text{dx}=\text{e}^\text{x}\tan\frac{\text{x}}{2}+\text{C}$

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