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Integrals

Multiple Choice Questions

801.

The value of the integral $\int_{- \frac{π}{4}}^{\frac{π}{4}} \log (\sec (θ) - \tan (θ)) dθ$ is

0
$\frac{π}{4}$
$π$
$\frac{π}{2}$

802.

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

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

803.

$\int \frac{1}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}} dx$ is equal to

$\frac{- {(1 + x^{4})}^{\frac{1}{4}}}{2 x} + C$
$\frac{- {(1 + x^{4})}^{\frac{1}{4}}}{x} + C$
$\frac{- {(1 + x^{4})}^{\frac{3}{4}}}{x} + C$
$\frac{- {(1 + x^{4})}^{\frac{1}{4}}}{x^{2}} + C$

804.

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

$\int_{0}^{\frac{π}{4}} \log (\frac{\sin (x) + \cos (x)}{\cos (x)}) dx$
$\frac{π}{4} \log (2)$
$\log (2)$
$\frac{π}{2} \log (2)$

805.

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

$\sin (x) + C$
$x + \sin (x) + C$
$\cos (x) + C$
$x - \sin (x) + C$

806.

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

e^x+ C
$e^{x} \tan (\frac{x}{2}) + C$
$e^{x} \sin (x) + C$
$\tan (\frac{x}{2}) + C$

807.

$\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{dx}{1 + \cos (2 x)}$ is equal to

808.

The value of $\int \frac{e^{x} (1 + x)}{\cos^{2} (e^{x} . x)} dx$ is equal to

$- \cot ({ex}^{x})$ + C
$\tan (e^{x} . x) + C$
$\tan (e^{x}) + C$
$\cot (e^{x}) + C$

809.
The vate of $\int \frac{e^{x} (x^{2} \tan^{- 1} (x) + \tan^{- 1} (x) + 1)}{x^{2} + 1} dx$ is equal to

$e^{x} \tan^{- 1} (x) + C$

$\tan^{- 1} (e^{x}) + C$

$\tan^{- 1} (x^{e}) + C$

$e^{\tan^{- 1} (x)} + C$

$e^{x} \tan^{- 1} (x) + C$

$Let I = \int \frac{e^{x} (x^{2} \tan^{- 1} (x) + \tan^{- 1} (x) + 1)}{x^{2} + 1} dx \Rightarrow I = \int e^{x} (\tan^{- 1} (x) + \frac{1}{x^{2} + 1}) dx If f (x) = \tan^{- 1} (x), then f' (x) = \frac{1}{x^{2} + 1} ∴ I = \int e^{x} (\tan^{- 1} (x) + \frac{1}{x^{2} + 1}) dx = e^{x} \tan^{- 1} (x) + C [∵ e^{x} [f (x) + f' (x)] dx = e^{x} f (x) + C]$