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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

$\frac{2}{abcos (2 x)}$

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

$\frac{2}{ab \sin (2 x)}$

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

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

$\int f (x) \sin (x) . \cos (x) dx = \frac{1}{2 (b^{2} - a^{2})} \log (f (x)) + c LHS = \frac{1}{2} \int f (x) (2 \sin (x) . \cos (x)) dx = \frac{1}{2} \int f (x) . \sin (2 x) dx Here, put f (x) = \frac{2}{(b^{2} - a^{2})} \times \frac{1}{\cos (2 x)} = \frac{1}{2} \int \frac{2}{(b^{2} - a^{2})} . \frac{\sin (2 x)}{\cos (2 x)} dx = \frac{1}{(b^{2} - a^{2})} \int \tan (2 x) dx = \frac{1}{(b^{2} - a^{2})} . \frac{\log (\sec (2 x))}{2} + c_{1} Here, put c_{1} = c + \frac{1}{2 (b^{2} - a^{2})} \log (\frac{2}{(b^{2} - a^{2})}) = \frac{1}{(b^{2} - a^{2})} . \frac{\log (\sec (2 x))}{2} + \frac{1}{2 (b^{2} - a^{2})} \log (\frac{2}{(b^{2} - a^{2})}) + c = \frac{1}{2 (b^{2} - a^{2})} \log [\frac{2}{(b^{2} - a^{2}) \cos (2 x)}] + c Hence, f (x) = \frac{2}{(b^{2} - a^{2}) \cos (2 x)}$