Question

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

• A. –e^x tan $\left( \frac{x}{2} \right)$ + c
• B. –e^x cot $\left( \frac{x}{2} \right)$ + c
• C. –$\frac{1}{2}$ e^x tan $\left( \frac{x}{2} \right)$ + c
• D. $\frac{1}{2}$ e^x cot $\left( \frac{x}{2} \right)$ + c

Answer 1

Answer: (B)

–e^x cot $\left( \frac{x}{2} \right)$ + c

Answer 2

LetI = $\left( \frac{1 - \sin x}{1 - \cos x} \right)$ dx =$\int_{}^{}{e^{x}\left( \frac{1 - \sin x}{2\sin^{2}x/2} \right)}$dx

ŽI =$\left( \frac{1}{2}\cos ec^{2}\frac{x}{2} - \cot\frac{x}{2} \right)$dx

= $\frac { 1 } { 2 }$ $\int_{}^{}e^{x}$cosec²$\frac{x}{2}$dx –$\int_{}^{}e^{x}$cot $\frac{x}{2}$dx

= $\frac { 1 } { 2 }$ $\int_{}^{}e^{x}$cosec²$\frac{x}{2}$ dx – e^x cot $\frac{x}{2}$ – $\frac { 1 } { 2 }$ $\int_{}^{}e^{x}$cosec² $\frac{x}{2}$dx

\ I = e^x $\left( - \cot\frac{x}{2} \right)$+ c

= – e^x cot $\frac{x}{2} + c$