STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

Putting $x = - t$ in $\int_{ - \pi /2}^0 {\frac{{\cos x}}{{1 + {e^x}}}dx} $, we get

$I = \int_{ - \pi /2}^0 {\frac{{\cos x}}{{1 + {e^x}}}dx = \int_0^{\pi /2} {\frac{{{e^x}\cos x}}{{1 + {e^x}}}dx} } $

$I = \int_0^{\pi /2} {\frac{{{e^x}\cos x}}{{1 + {e^x}}}dx + \int_0^{\pi /2} {\frac{{\cos x}}{{1 + {e^x}}}dx} } $

$ = \int_{\,0}^{\,\pi /2} {\frac{{(1 + {e^x})\cos x\,dx}}{{(1 + {e^x})}}} $

$ = \int_0^{\pi /2} {\cos x\,dx = [\sin x]_0^{\pi /2} = 1} $.