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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
Question
$\int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + {e^x}}}\,dx = } $
✓

Answer

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

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} $.

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