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प्रश्न
\[\int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx\]
विकल्प
- \[- e^x \tan\frac{x}{2} + C\]
- \[- e^x \cot\frac{x}{2} + C\]
- \[- \frac{1}{2} e^x \tan\frac{x}{2} + C\]
- \[- \frac{1}{2} e^x \cot\frac{x}{2} + C\]
MCQ
उत्तर
\[- e^x \cot\frac{x}{2} + C\]
\[\text{Let }I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right)dx\]
\[ \Rightarrow \int e^x \left( \frac{1}{1 - \cos x} - \frac{\sin x}{1 - \cos x} \right)dx\]
\[ \Rightarrow \int e^x \left( \frac{1}{2 \sin^2 \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right)dx\]
\[ \Rightarrow \int e^x \left( \frac{1}{2} {cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \right)dx\]
\[\text{As, we know that }\int e^x \left\{ f\left( x \right) + f'\left( x \right) \right\} dx = e^x f\left( x \right) + C\]
\[ \Rightarrow \int e^x \left( \frac{1}{2 \sin^2 \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right)dx\]
\[ \Rightarrow \int e^x \left( \frac{1}{2} {cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \right)dx\]
\[\text{As, we know that }\int e^x \left\{ f\left( x \right) + f'\left( x \right) \right\} dx = e^x f\left( x \right) + C\]
\[ \therefore I = - e^x \cot \left( \frac{x}{2} \right) + C\]
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