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Integrals

Multiple Choice Questions

421.

If $\int_{1}^{x} \frac{d t}{|t| \sqrt{t^{2} - 1}} = \frac{π}{6}$ , then x can be equal to

$\frac{2}{\sqrt{3}}$
$\sqrt{3}$
2
None of these

422.

$\int (\frac{2 - \sin (2 x)}{1 - \cos (2 x)}) e^{x} dx$ is equal to

$- e^{x} \cot (x) + c$
$e^{x} \cot (x) + c$
$2 e^{x} \cot (x) + c$
$- 2 e^{x} \cot (x) + c$

423.

$If I_{n} = \int \sin^{n} (x) dx, then {nI}_{n} - (n - 1) I_{n - 2}$ equals

$\sin^{n - 1} (x) xcos (x)$
$\cos^{n - 1} (x) \sin (x)$
$- \sin^{n - 1} (x) \cos (x)$
$- \cos^{n - 1} (x) \sin (x)$

424.

If I = $\int \frac{x^{5}}{\sqrt{1 + x^{3}}} dx$ , then I is equal to

$\frac{2}{9} {(1 + x^{3})}^{\frac{5}{2}} + \frac{2}{3} {(1 + x^{3})}^{\frac{3}{2}} + c$
$\log |\sqrt{x} + \sqrt{1 + x^{3}}| + c$
$\log |\sqrt{x} - \sqrt{1 + x^{3}}| + c$
$\frac{2}{9} {(1 + x^{3})}^{\frac{3}{2}} - \frac{2}{3} {(1 + x^{3})}^{\frac{1}{2}} + c$

425.

The value of $\int_{0}^{a} \sqrt{\frac{a - x}{x}} d x$ is

$\frac{a}{2}$
$\frac{a}{4}$
$\frac{πa}{2}$
$\frac{πa}{4}$

426.

The value of $\int_{1}^{4} |x - 3| dx$ is equal to :

1
$\frac{5}{2}$
$\frac{1}{2}$
$\frac{3}{2}$

427.

The value of $\int_{0}^{\infty} \frac{dx}{(a^{2} + x^{2})}$ is equal to

$\frac{π}{2}$
$\frac{π}{2 a}$
$\frac{π}{a}$
$\frac{1}{2 a}$

428.

The value of the integral $\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} dx$ is

$e^{x} (\frac{1 - x}{1 + x^{2}}) + c$
$e^{x} (\frac{1 + x}{1 + x^{2}})$
$\frac{e^{x}}{1 + x^{2}} + c$
$e^{x} (1 - x) + c$

429.
If g(x) = $\int_{0}^{x} \cos^{4} (t) d t$ , then $g (x + π)$ is equsl to

g(x) + g( $π$ )

g(x) - g( $π$ )

g(x) . g( $π$ )

$\frac{g (x)}{g (π)}$

g(x) + g( $π$ )

$∵ g (x) = \int_{0}^{x} \cos^{4} (t) d t ∴ g (x + π) = \int_{0}^{π + x} \cos^{4} (t) d t = \int_{0}^{π} \cos^{4} (t) d t + \int_{π}^{π + x} \cos^{4} (t) d t = I_{1} + I_{2}$

$\Rightarrow I_{1} = g (π) and I_{2} = \int_{π}^{π + x} \cos^{4} (t) d t \Rightarrow I_{2} = \int_{0}^{x} \cos^{4} (y + π) d y = \int_{0}^{x} {[\cos (π + y)]}^{4} dy = \int_{0}^{x} \cos^{4} (y) dy = g (x) ∴ g (π + x) = g (x) + g (π)$