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Integrals

Multiple Choice Questions

471.

The function f(x) = $\int_{0}^{x} \log (\frac{1 - x}{1 + x}) dx$ is

an even function
an odd function
a periodic function
None of these

472.

$\int \frac{\sin (2 x)}{1 + \cos^{2} (x)} dx$ =

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

473.

$\int \frac{e^{x} (1 + \sin (x))}{1 + \cos (x)} dx$ =

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

474.

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

$e^{- x} \tan (x) + c$
$e^{- x} \sec (x) + c$
$e^{x} \sec (x) + c$
$e^{x} \tan (x) + c$

475.

$\int_{\frac{π}{4}}^{\frac{π}{2}} \csc^{2} (x) dx$ is equal to

- 1
1
0
$\frac{1}{2}$

476.

$\int_{0}^{\frac{π}{4}} \log (1 + \tan (x)) dx$ is equal to

$\frac{π}{8} \log_{e} (2)$
$\frac{π}{4} \log_{2} (e)$
$\frac{π}{4} \log_{e} (2)$
$\frac{π}{8} \log_{e} (\frac{1}{2})$

477.

$\int \frac{(x^{3} + 3 x^{2} + 3 x + 1)}{{(x + 1)}^{5}} dx$ is equal to

$- \frac{1}{(x + 1)} + c$
$\frac{1}{5} \log (x + 1) + c$
$\log (x + 1) + c$
$\tan^{- 1} (x) + c$

478.

$\int \frac{\csc (x)}{\cos^{2} (1 + \log (\tan (\frac{x}{2})))} dx$ is equal to :

$\sin^{2} (1 + \log (\tan (\frac{x}{2}))) + c$
$\tan (1 + \log (\tan (\frac{x}{2}))) + c$
$\sec^{2} (1 + \log (\tan (\frac{x}{2}))) + c$
$- \tan (1 + \log (\tan (\frac{x}{2}))) + c$

479.
$\int \frac{dx}{x \sqrt{x^{6} - 16}}$ is equa to :

$\frac{1}{3} \sec^{- 1} (\frac{x^{3}}{4}) + c$

$\cosh^{- 1} (\frac{x^{3}}{4}) + c$

$\frac{1}{12} \sec^{- 1} (\frac{x^{3}}{4}) + c$

$\sec^{- 1} (\frac{x^{3}}{4}) + c$

$\frac{1}{12} \sec^{- 1} (\frac{x^{3}}{4}) + c$

$Let I = \int \frac{dx}{x \sqrt{x^{6} - 16}} = \frac{1}{3} \int \frac{3 x^{2}}{x^{3} \sqrt{{(x^{3})}^{2} - 4^{2}}} dx Put x^{3} = t \Rightarrow 3 x^{2} dx = dt ∴ I = \frac{1}{3} \int \frac{dt}{\sqrt{t^{2} - 4^{2}}} = \frac{1}{3 \times 4} \sec^{- 1} (\frac{t}{4}) + c = \frac{1}{12} \sec^{- 1} (\frac{x^{3}}{4}) + c$