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Integrals

Multiple Choice Questions

491.

$\int \frac{\sin (x) \cos (x)}{\sqrt{1 - \sin^{4} (x)}} dx$ is equal to

$\frac{1}{2} \sin^{- 1} (\sin^{2} (x)) + C$
$\frac{1}{2} \cos^{- 1} (\sin^{2} (x)) + C$
$\tan^{- 1} (\sin^{2} (x)) + C$
$\tan^{- 1} (2 \sin^{2} (x)) + C$

492.

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

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

493.

$\int \csc (x - a) \csc (x) dx$ is equal to

$\frac{- 1}{\sin (a)} \log |\sin (x) \csc (x - a)| + c$
$\frac{- 1}{\sin (a)} \log [\sin (x - a) \sin (x)] + c$
$\frac{1}{\sin (a)} \log [\sin (x) \csc (x - a)] + c$
$\frac{1}{\sin (a)} \log [\sin (x - a) \sin (x)] + c$

494.

If f(x) = $\int_{- 1}^{x} |t| dt$ , then for any x $\geq$ 0, f(x) is equal to

1 - x²
$\frac{1}{2} (1 + x^{2})$
1 + x²
$\frac{1}{2} (1 - x^{2})$

495.

$\int_{1}^{3} \frac{\sqrt{4 - x}}{\sqrt{x} + \sqrt{4 - x}} dx$ is equal to

496.

If $\int_{0}^{1} f (x) dx = 5, then the value of . . . + \int_{0}^{1} x^{9} f (x^{10}) dx$ is equal to

497.

If $\int f (x) \sin (x) . \cos (x) dx = \frac{1}{2 (b^{2} - a^{2})} \log (f (x)) + c$ , where c is the constant of integration, then f(x) is

$\frac{2}{abcos (2 x)}$
$\frac{2}{(b^{2} - a^{2}) \cos (2 x)}$
$\frac{2}{ab \sin (2 x)}$
$\frac{2}{(b^{2} - a^{2}) \sin (2 x)}$

498.

If $\int \frac{\sqrt{x}}{x (x + 1)} dx = {ktan}^{- 1} (m)$ , then (k, m) is

(2, x)
(1, x)
$(1, \sqrt{x})$
$(2, \sqrt{x})$

499.
$\int_{0}^{\frac{π}{4}} \frac{\sin (x) + \cos (x)}{3 + \sin (2 x)} dx$ is

$\frac{1}{4} \log (3)$

$\log (3)$

$\frac{1}{2 \log (3)}$

$2 \log (3)$

$\frac{1}{4} \log (3)$

$\int_{0}^{\frac{π}{4}} \frac{\sin (x) + \cos (x)}{3 + \sin (2 x)} dx = \int_{0}^{\frac{π}{4}} \frac{\sin (x) + \cos (x)}{4 - (1 - \sin (2 x))} dx = \int_{0}^{\frac{π}{4}} \frac{(\sin (x) + \cos (x))}{4 - {(\sin (x) + \cos (x))}^{2}} dx Put t = (\sin (x) - \cos (x)) dt = (\cos (x) + \sin (x)) dx = \int_{- 1}^{0} \frac{dt}{(4 - t^{2})} = \int_{- 1}^{0} \frac{dt}{(2 + t) (2 - t)} = \frac{1}{4} {[\log (2 + t) - \log (2 - t)]}_{- 1}^{0} = \frac{1}{4} {[\log (\frac{2 + t}{2 - t})]}_{- 1}^{0} = \frac{1}{4} [\log (1) - \log (\frac{1}{3})] = \frac{1}{4} \log (3)$