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Integrals

Multiple Choice Questions

451.

$\int_{0}^{π} \frac{xdx}{1 + \sin (x)}$ is equal to :

$- π$
$\frac{π}{2}$
$π$
None of these

452.

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

$\frac{1}{5} \log (x^{5}) (x^{5} + 1) + c$
$\frac{1}{5} \log (\frac{x^{5} + 1}{x^{5}}) + c$
$\frac{1}{5} \log (\frac{x^{5}}{x^{5} + 1}) + c$
None of these

453.

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

$xtan (\frac{x}{2}) + c$
${xsec}^{2} (\frac{x}{2}) + c$
$\log (\cos (\frac{x}{2})) + c$
None of these

454.

$\int_{0}^{\frac{π}{8}} \cos^{3} (4 θ) dθ$ is equal to

$\frac{5}{3}$
$\frac{5}{4}$
$\frac{1}{3}$
$\frac{1}{6}$

455.
$\int_{0}^{\frac{π}{2}} \frac{\cos (x) - \sin (x)}{1 + \cos (x) \sin (x)} dx$ is equal to

0

$\frac{π}{2}$

$\frac{π}{4}$

$\frac{π}{6}$

$Given that, I = \int_{0}^{\frac{π}{2}} \frac{\cos (x) - \sin (x)}{1 + \cos (x) \sin (x)} dx . . . (i) Putting x = (\frac{π}{2} - x) in Eq . (i), we get I = \int_{0}^{\frac{π}{2}} \frac{\cos (\frac{π}{2} - x) - \sin (\frac{π}{2} - x)}{1 + \cos (\frac{π}{2} - x) \sin (\frac{π}{2} - x)} dx \Rightarrow I = \int_{0}^{\frac{π}{2}} - (\frac{\cos (x) - \sin (x)}{1 + \cos (x) \sin (x)}) dx . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{\frac{π}{2}} (\frac{\cos (x) - \sin (x)}{1 + \cos (x) \sin (x)} - \frac{\cos (x) - \sin (x)}{1 + \cos (x) \sin (x)}) dx = \int_{0}^{\frac{π}{2}} 0 dx = 0 \Rightarrow I = 0$