Question

Evaluate the definite integral: $\int_{\pi}^{\frac{\pi}{2}}e^{x}(\frac{1-sinx}{1-cosx})dx$

Answer 1

$I=\int_{\pi}^{\frac{\pi}{2}}e^{x}(\frac{1-sinx}{1-cosx})dx$

=\int_{\pi}^{\frac{\pi}{2}}$$(\frac{1-2sin\frac{x}{2}cos\frac{x}{2}}{2sin^{2}\frac{x}{2}})dx

=\int_{\pi}^{\frac{\pi}{2}}$$e^{x}(\frac{cosec^{2}\frac{x}{2}}{2}-cot\frac{x}{2})dx

Let$ƒ(x)=-cot\frac{x}{2}$

⇒ƒ'(x)=$$-(-\frac{1}{2}cosec^{2}\frac{x}{2})=\frac{1}{2}cosec^{2}\frac{x}{2}

$∴I=\int_{\pi}^{\frac{\pi}{2}}e^{x}(ƒ(x)+ƒ'(x)]dx$

$=[e^{x}.ƒ(x)dx]_{\pi}^{\frac{\pi}{2}}$

$=-[e^{x}.cot\frac{x}{2}]_{\pi}^{\frac{\pi}{2}}$

$=-[e^{\pi}\times cot\frac{\pi}{2}-e^\frac{π}{2}×cot\frac{π}{4}]$

$=-[e^{\pi}\times 0-e^{\frac{\pi}{2}}×1]$

$=e^{\frac{π}{2}}$