Evaluate the following integrals: ^ x ( 1+ 1+ )dx - Vidyadip

Question

Evaluate the following integrals:

$\int\text{e}^{\text{}x}\Big(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$

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Answer

Let $\text{I}=\int\text{e}^{\text{}x}\Big(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\frac{1}{1+\cos\text{x}}+\frac{\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\bigg(\frac{1}{2\cos\frac{\text{x}}{2}}+\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\text{e}^{\text{}x}\big(\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\big)\text{dx}$
Putting $\text{e}^{\text{x}}\tan\frac{\text{x}}{2}=\text{t}$
Diff. both sides w.r.t.x
$\text{e}^{\text{x}}.\tan\big(\frac{\text{x}}{2}\big)+\text{e}^{\text{x}}\times\frac{1}{2}\sec^{2}\frac{\text{x}}{2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}\big[\tan\frac{\text{x}}{2}+\frac{1}{2}\sec^2\big(\frac{\text{x}}{2}\big)\big]\text{dx}=\text{dt}$
$\therefore\int\text{e}^{\text{x}}\big(\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\big)\text{dx}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\tan\big(\frac{\text{x}}{2}\big)+\text{C}$

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