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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

$\frac{5}{3}$

$\frac{5}{4}$

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{1}{6}$

$We have, I = \int_{0}^{\frac{π}{8}} \cos^{3} (4 θ) dθ = \int_{0}^{\frac{π}{8}} \cos^{2} (4 θ) \cos (4 θ) dθ = \int_{0}^{\frac{π}{8}} (\frac{1 + \cos (8 θ)}{2}) dθ = \frac{1}{2} \int_{0}^{\frac{π}{8}} \cos (4 θ) dθ + \frac{1}{2} \int_{0}^{\frac{π}{8}} \cos (8 θ) \cos (4 θ) dθ = \frac{1}{2} {[\frac{\sin (4 θ)}{4}]}_{0}^{\frac{π}{8}} + I_{1} . . . (i) Now, I_{1} = \frac{1}{2} \int_{0}^{\frac{π}{8}} \cos (8 θ) \cos (4 θ) dθ = \frac{1}{2} {[\cos (8 θ) \frac{\sin (4 θ)}{4}]}_{0}^{\frac{π}{8}} - \frac{1}{2} \int_{0}^{\frac{π}{8}} (- 8 \cos (8 θ)) \frac{\sin (4 θ)}{4} dθ = - \frac{1}{8} + \int_{0}^{\frac{π}{8}} \sin (8 θ) \sin (4 θ) dθ = - \frac{1}{8} + {[\sin (8 θ) (- \frac{\cos (4 θ)}{4})]}_{0}^{\frac{π}{8}} - \frac{1}{2} \int_{0}^{\frac{π}{8}} 8 \cos (8 θ) (- \frac{\cos (4 θ)}{4}) dθ = - \frac{1}{8} + 0 + 2 \int_{0}^{\frac{π}{8}} \cos (8 θ) \cos (4 θ) dθ I_{1} = - \frac{1}{8} + 4 I_{1} Now, I_{1} = 1 / 24 . . . (ii) From Eqs . (i) and (ii), we get I = \frac{1}{8} + \frac{1}{24} = \frac{3 + 1}{24} = \frac{4}{24} = \frac{1}{6}$