MCQ
$\int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx}$ is equal to
- A${e^x}\tan \frac{x}{2} + C$
- B${-e^x}\tan \frac{x}{2} + C$
- ✓${-e^x}\cot \frac{x}{2} + C$
- D${e^x}\cot \frac{x}{2} + C$
✓
Answer
Correct option: C.
${-e^x}\cot \frac{x}{2} + C$
c
$I=\int e^{x}\left(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2-\sin ^{2} \frac{x}{2}}\right) d x$
$I=\int e^{x}\left(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2-\sin ^{2} \frac{x}{2}}\right) d x$
$ = \int {{{\rm{e}}^{\rm{x}}}} \left( {\frac{1}{2}\cos e{c^2}\frac{{\rm{x}}}{2} - \cot \frac{{\rm{x}}}{2}} \right){\rm{dx}}$
$ = - \int {{{\rm{e}}^{\rm{x}}}} \left( {\cot \frac{{\rm{x}}}{2} - \frac{1}{2}\cos e{c^2}\frac{{\rm{x}}}{2}} \right){\rm{dx}}$
$=-e^{x} \cdot \cot \frac{x}{2}+c$
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