Question

Evaluate: ${\int }_0^{\pi /2}{e}^x(\sin x-\cos x)dx.$

Answer 1

Consider, $I={\int }_0^{\frac{\pi }{2}}{e}^x(sinx-cosx)dx$

$={\int }_0^{\frac{\pi }{2}}{e}^{x}sinxdx-{\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx$

$=[sinx.e^x{]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}cosx.e^{x}dx-{\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx$

$=(sin\frac{\pi }{2}.e^{\frac{\pi }{2}})-2{\int }_0^{\frac{\pi }{2}}{e}^{x}cosdx$

$=e^{\frac{\pi }{2}}-2{\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx$--- (1)

Now we have to find out ${\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx$

${\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx=\left[e^x\right(sinx){]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}{e}^{x}sinxdx$

$=e^{\frac{\pi }{2}}-\left[\right[e^x(-cosx){]}_0^{\frac{\pi }{2}}]-{\int }_0^{\frac{\pi }{2}}{e}^x(-cosx)dx$

$=e^{\frac{\pi }{2}}-1-{\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx$

$2{\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx=e^{\frac{\pi }{2}}-1\Rightarrow {\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx=\frac{1}{2}(e^{\frac{\pi }{2}}-1)$

Putting the value of ${\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx$ in (1), we get

$I=e^{\frac{\pi }{2}}-2\left\{\frac{1}{2}(e^{\frac{\pi }{2}}-1)\right\}=e^{\frac{\pi }{2}}-e^{\frac{\pi }{2}}+1=1$