A) 1
B) 0
C) \[-1\]
D) None of these
Correct Answer: A
Solution :
\[I=\int_{-\pi /2}^{\pi /2}{\frac{\cos x}{1+{{e}^{x}}}dx=\int_{-\pi /2}^{0}{\frac{\cos x}{1+{{e}^{x}}}dx+\int_{0}^{\pi /2}{\frac{\cos x}{1+{{e}^{x}}}dx}}}\] ?..(i) Putting \[x=-t\]in \[\int_{-\pi /2}^{0}{\frac{\cos x}{1+{{e}^{x}}}dx}\], we get \[I=\int_{-\pi /2}^{0}{\frac{\cos x}{1+{{e}^{x}}}dx=\int_{0}^{\pi /2}{\frac{{{e}^{x}}\cos x}{1+{{e}^{x}}}dx}}\] \[I=\int_{0}^{\pi /2}{\frac{{{e}^{x}}\cos x}{1+{{e}^{x}}}dx+\int_{0}^{\pi /2}{\frac{\cos x}{1+{{e}^{x}}}dx}}\] \[=\int_{\,0}^{\,\pi /2}{\frac{(1+{{e}^{x}})\cos x\,dx}{(1+{{e}^{x}})}}\]\[=\int_{0}^{\pi /2}{\cos x\,dx=[\sin x]_{0}^{\pi /2}=1}\].
You need to login to perform this action.
You will be redirected in
3 sec
