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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

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

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

$Given, f (x) = \int_{- 1}^{x} |t| dt = \int_{- 1}^{0} |t| dt + \int_{0}^{x} |t| dt = \int_{- 1}^{0} - tdt + \int_{0}^{x} tdt = - {[\frac{t^{2}}{2}]}_{- 1}^{0} + {[\frac{t^{2}}{2}]}_{0}^{x} = - [0 - \frac{1}{2}] + [\frac{x^{2}}{2} - 0] = \frac{1}{2} (1 + x^{2})$