int {{e^x}left( {frac{{1 - sin x}}{{1 - cos x | vedclass

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(B) Let $I = \int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx} $. Using trigonometric identities $1 - \cos x = 2\sin^2(x/2)$ and $\sin

Vedclass · Accepted Answer

$- {e^x}\cot \left( {x/2} \right)$

Vedclass · Answer

(B) Let $I = \int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx} $. Using trigonometric identities $1 - \cos x = 2\sin^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$,we get: $I = \int {{e^x}\left( {\frac{{1 - 2\sin(x/2)\cos(x/2)}}{{2\sin^2(x/2)}}} \right)dx} $. $I = \int {{e^x}\left( {\frac{1}{{2\sin^2(x/2)}} - \frac{{2\sin(x/2)\cos(x/2)}}{{2\sin^2(x/2)}}} \right)dx} $. $I = \int {{e^x}\left( {\frac{1}{2}\csc^2(x/2) - \cot(x/2)} \right)dx} $. We know that $\int {{e^x}(f(x) + f'(x))dx} = {e^x}f(x) + C$. Here,let $f(x) = -\cot(x/2)$,then $f'(x) = -(-\csc^2(x/2) \cdot 1/2) = \frac{1}{2}\csc^2(x/2)$. Thus,$I = {e^x}(-\cot(x/2)) + C = -{e^x}\cot(x/2) + C$.

$\int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)\,dx} $ is equal to

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