Question
$\int_{ - \,\pi /2}^{\,\pi /2} {\,\frac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}dx} $ is equal to
$\because \frac{\sin x}{1+{{\cos }^{2}}x}{{e}^{-{{\cos }^{2}}x}}$ is an odd function,
$\therefore$ $I = 0$.
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Where $\alpha \in R$, then the value of $16 \alpha$ is equal to
$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$
$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$
$2 x+(3 \lambda+1) y+3(\lambda-1) z=0$
has non-zero solutions, is