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Question: \[\int_{}^{}{\mathbf{e}^{\mathbf{x}}\left( \frac{\mathbf{1}\mathbf{-}\mathbf{\sin}\mathbf{x}}{\mathb...

∫ex(1−sin⁡x1−cos⁡x)dx\int_{}^{}{\mathbf{e}^{\mathbf{x}}\left( \frac{\mathbf{1}\mathbf{-}\mathbf{\sin}\mathbf{x}}{\mathbf{1}\mathbf{-}\mathbf{\cos}\mathbf{x}} \right)\mathbf{dx}}

A

−extan⁡x2- e^{x}\tan\frac{x}{2}+ c

B

−excot⁡x2- e^{x}\cot\frac{x}{2}+ c

C

−12extan⁡x2- \frac{1}{2}e^{x}\tan\frac{x}{2}+ c

D

12excot⁡x2\frac{1}{2}e^{x}\cot\frac{x}{2}+ c

Answer

−excot⁡x2- e^{x}\cot\frac{x}{2}+ c

Explanation

Solution

∵\because 1−sin⁡x1−cos⁡x=12sin⁡2x2−2sin⁡x2cos⁡x22sin⁡2x2=cos⁡⥂ec2x2−cot⁡x2\frac{1 - \sin x}{1 - \cos x} = \frac{1}{2\sin^{2}\frac{x}{2}} - \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^{2}\frac{x}{2}} = \cos ⥂ ec^{2}\frac{x}{2} - \cot\frac{x}{2}

∴∫ex(1−sin⁡x1−cos⁡x)dx=∫ex(12cosec2x2−cot⁡x2)dx=−excot⁡x2+c\therefore\int_{}^{}{e^{x}\left( \frac{1 - \sin x}{1 - \cos x} \right)}dx = \int_{}^{}{e^{x}\left( \frac{1}{2}c\text{ose}\text{c}^{2}\frac{x}{2} - \cot\frac{x}{2} \right)}dx = - e^{x}\cot\frac{x}{2} + c