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Integrals

Multiple Choice Questions

961.

$If I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} (dθ) for n = 1, 2, 3 . . ., then I_{n + 1} + I_{n + 1} = ?$

0
1
$\frac{1}{n}$
$\frac{1}{n + 1}$

962.

$Let f (0) = 1, f (0.5) = \frac{5}{4}, f (1) = 2, f (1.5) = \frac{13}{4}, and f (2) = 5 . Using Simson' s rule, \int_{0}^{2} f (x) dx = ?$

$\frac{14}{3}$
$\frac{7}{6}$
$\frac{14}{9}$
$\frac{7}{9}$

963.

$\int \frac{dx}{x^{2} \sqrt{4 + x^{2}}} =$

$\frac{1}{4} \sqrt{4 + x^{2}} + C$
$- \frac{1}{4} \sqrt{4 + x^{2}} + C$
$- \frac{1}{4 x} \sqrt{4 + x^{2}} + C$
$\frac{9}{4 x} \sqrt{4 + x^{2}} + C$

964.

$\int \sec^{2} (x) \csc^{4} (x) dx = ?$

965.

$\int \frac{dx}{\sqrt{x - x^{2}}} = ?$

$2 \sin^{- 1} (\sqrt{x}) + C$
$2 \sin^{- 1} (x) + C$
$2 x \sin^{- 1} (x) + C$
$\sin^{- 1} (\sqrt{x}) + C$

966.
$If a > 0, then \int_{- π}^{π} \frac{\sin^{2} (x)}{1 + a^{x}} dx = ?$

$\frac{π}{2}$

$π$

$\frac{2 π}{2}$

a $π$

$\frac{π}{2}$

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

967.

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

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

968.

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