^ x ( 1- 1- )dx= - Vidyadip

MCQ

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

A
$-\text{e}^{\text{x}}\tan\frac{\text{x}}{2}+\text{C}$
✓
$-\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$
C
$-\frac{1}{2}\text{e}^{\text{x}}\tan\frac{\text{x}}{2}+\text{C}$
D
$-\frac{1}{2}\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$

✓

Answer

Correct option: B.

$-\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$

Let $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$
$\int\text{e}^{\text{x}}\Big(\frac{1}{1-\cos\text{x}}-\frac{\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$
$\int\text{e}^{\text{x}}\bigg(\frac{1}{2\sin^2\frac{\text{x}}{2}}-\frac{2\sin\frac{\pi}{2}\cos\frac{\pi}{2}}{2\sin^2\frac{\pi}{2}}\bigg)\text{dx}$
$\int\text{e}^\text{x}\Big(\frac{1}{2}\text{cosec}^2\frac{\text{x}}{2}-\cot\frac{\text{x}}{2}\Big)\text{dx}$
As, we know that $\int\text{e}^{\text{x}}\{\text{f(x)}+\text{f}'(\text{x})\}\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$
$\therefore\ \text{I}=-\text{e}^{\text{x}}\cot\Big(\frac{\text{x}}{2}\Big)+\text{C}$

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