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Integrals

Multiple Choice Questions

531.

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

$\log |\tan (x)| + C$
$\log |\sin (2 x)| + C$
$\log |\sec (x)| + C$
$\log |\cos (x)| + C$

532.

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

$x + \log (\tan (x)) + C$
$xlog |\tan (x)| + C$
$xtan (x) + C$
$x + \tan (x) + C$

533.

$\int \frac{1}{8 \sin^{2} (x) + 1} dx$ is equal to

$\sin^{- 1} (\tan (x)) + C$
$\frac{1}{3} \sin^{- 1} (\tan (x)) + C$
$\frac{1}{3} \tan^{- 1} (3 \tan (x)) + C$
$\tan^{- 1} (3 \tan (x)) + C$

534.
$\int_{0}^{\frac{π}{2}} \log (\frac{\cos (x)}{\sin (x)}) dx$ is equal to

$\frac{π}{2}$

$\frac{π}{4}$

$π$

0

$Let I = \int_{0}^{\frac{π}{2}} \log (\frac{\cos (x)}{\sin (x)}) d x = \int_{0}^{\frac{π}{2}} \log (\cot (x)) d x . . . (i) Then, I = \int_{0}^{\frac{π}{2}} \log (\cot (\frac{π}{2} - x)) d x [∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{0}^{\frac{π}{2}} \log (\tan (x)) d x . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{\frac{π}{2}} (\log (\cot (x)) + \log (\tan (x))) d x 2 I = \int_{0}^{\frac{π}{2}} 0 d x = 0 \Rightarrow I = 0$