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681.
$If a > 0, then \int_{- π}^{π} \frac{\sin^{2} (x)}{1 + a^{x}} dx = ?$

$\frac{π}{2}$

$π$

$\frac{2 π}{2}$

a $π$

$\frac{π}{2}$

$Let I = \int_{- π}^{π} \frac{\sin^{2} (x)}{1 + a^{x}} dx . . . (i) Put x = - x, we get I = - \int_{π}^{- π} \frac{\sin^{2} (x)}{1 + a^{x}} dx . . . (ii) On adding eqs (i) and (ii), we get 2 I = \int_{π}^{- π} \frac{(1 + a^{x}) \sin^{2} (x)}{(1 + a^{x})} dx = \int_{π}^{- π} \sin^{2} (x) dx = 2 \int_{0}^{π} \sin^{2} (x) dx = \int_{0}^{π} (1 - \cos (2 x)) dx = {[x + \frac{\sin (2 x)}{2}]}_{0}^{π} = (π + \frac{\sin (2 x)}{2} - 0 - 0) \Rightarrow 2 I = π \Rightarrow I = \frac{π}{2}$

682.

$The value of the integral \int_{0}^{4} \frac{dx}{1 + x^{2}} obtained by using trapezoidal rule with h = 1 is$

$\frac{63}{85}$
$\tan^{- 1} (4)$
$\frac{108}{85}$
$\frac{113}{85}$

683.

$Let A = |\begin{matrix} 2 & e^{iπ} \\ - 1 & i^{2012} \end{matrix}|, C = \frac{d}{dx} (\frac{1}{x}) x = 1, D = \int_{e^{2}}^{1} \frac{dx}{x} . If the sum of two roots of the equation {Ax}^{3} + {Bx}^{2} + Cx - D = 0 is equal to zero, then B is equal to$

684.

$\int e^{x} (\frac{2 + \sin (2 x)}{1 + \cos (2 x)}) dx = ?$

$e^{x} \cot (x) + C$
$2 e^{x} \sec^{2} (x) + C$
$e^{x} \cos (2 x) + C$
$e^{x} \tan (x) + C$

685.

$If \int \frac{x - \sin (x)}{1 + \cos (x)} dx = xtan (\frac{x}{2}) + plog |\sec (\frac{x}{2})| + C, then p = ?$

686.

$If \int \frac{dx}{xlog (x - 2) \log (x - 3)} = I + C, then I = ?$

$\frac{1}{x} \log |\frac{\log (x - 3)}{\log (x - 2)}|$
$\log |\frac{\log (x - 3)}{\log (x - 2)}|$
$\log |\frac{\log (x - 2)}{\log (x - 3)}|$
$\log |\log (x - 3) \log (x - 2)|$

687.

$If \int_{0}^{b} \frac{dx}{1 + x^{2}} = \int_{b}^{\infty} \frac{dx}{1 + x^{2}}, then b = ?$

$\tan^{- 1} (\frac{1}{3})$
$\frac{\sqrt{3}}{2}$
$\sqrt{2}$
1

688.

The approximate value of $\int_{1}^{3} \frac{dx}{2 + 3 x}$ using Simpson's rule and dividing the interval [1, 3] into two equal parts is

$\frac{1}{3} \log (\frac{11}{5})$
$\frac{107}{110}$
$\frac{22}{110}$
$\frac{119}{440}$

689.

$If \int \frac{dx}{\sqrt{\sin^{3} (x) \cos (x)}} = g (x) + c, then g (x) = ?$

$- \frac{2}{\sqrt{\cot (x)}}$
$- \frac{2}{\sqrt{\tan (x)}}$
$\frac{2}{\sqrt{\cot (x)}}$
$\frac{2}{\sqrt{\tan (x)}}$

690.

$If \int \frac{dx}{(1 + \sqrt{x}) \sqrt{x - x^{2}}} = \frac{A \sqrt{x}}{\sqrt{1 - x}} + \frac{B}{\sqrt{1 - x}} + C, where C is real constant, then A + B = ?$