Question

${\int }_0^{\frac{\pi }{2}}{e}^{x}sinxdx=$ [Roorkee 1978]

• A. $\frac{1}{2}(e^{\frac{x}{2}}-1)$
• B. $\frac{1}{2}(e^{\frac{x}{2}}+1)$
• C. $\frac{1}{2}(1-e^{\frac{x}{2}})$
• D. $2(e^{\frac{x}{2}}+1)$

Answer 1

The correct option is B $\frac{1}{2}(e^{\frac{x}{2}}+1)$

$Let{\int }_0^{\frac{\pi }{2}}{e}^{x}sinxdx=$
$=-[e^{x}cosx{]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx$
$=-[e^{x}cosx{]}_0^{\frac{\pi }{2}}+[e^{x}sinx{]}_0^{\frac{x}{2}}-{\int }_0^{\frac{\pi }{2}}{e}^{x}sinxdx$
$\therefore 2I=\left[e^x\right(sinx-cosx){]}_0^{\frac{\pi }{2}}=(e^{\frac{\pi }{2}}+1)$
$Hence{\int }_0^{\frac{\pi }{2}}{e}^{x}sinxdx=\frac{1}{2}(e^{\frac{\pi }{2}}+1)$