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Integrals

Multiple Choice Questions

751.

The value of $\int \frac{dx}{\sqrt{2 x - x^{2}}}$ is

$\sin^{- 1} (1 - x) + c$
$\sin^{- 1} (x - 1) + c$
$\sin^{- 1} (1 + x) + c$
$- \sqrt{2 x - x^{2}} + c$

752.

The value of $\int xlog (x^{3}) dx$ is

$\frac{1}{16} (4 x^{4} \log (x) - x^{4} + c)$
$\frac{1}{8} (x^{4} \log (x) - 4 x^{4} + c)$
$\frac{x^{4} \log (x)}{4} + c$
$\frac{1}{16} (4 x^{4} \log (x) + x^{4} + c)$

753.
The value of $\int_{0}^{π} \frac{dx}{5 + 3 \cos (x)}$ is

$\frac{π}{4}$

$\frac{π}{8}$

$\frac{π}{2}$

zero

$\frac{π}{4}$

$\int_{0}^{π} \frac{dx}{5 + 3 \cos (x)} = \int_{0}^{π} \frac{dx}{5 + 3 \{\frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}\}} = \int_{0}^{π} \frac{\sec^{2} (\frac{x}{2})}{5 + 5 \tan^{2} (\frac{x}{2}) + 3 - 3 \tan^{2} (\frac{x}{2})} = \int_{0}^{π} \frac{\sec^{2} (\frac{x}{2})}{8 + 2 \tan^{2} (\frac{x}{2})} = \frac{1}{2} \int_{0}^{π} \frac{\sec^{2} (\frac{x}{2})}{{(2)}^{2} {(\tan (\frac{x}{2}))}^{2}} = \tan (\frac{x}{2}) = t \sec^{2} (\frac{x}{2}) dx = 2 dt = \int_{0}^{\infty} \frac{dt}{{(2)}^{2} + {(t)}^{2}} = {[\frac{1}{2} \tan^{- 1} (\frac{t}{2})]}_{0}^{\infty} = \frac{1}{2} [\frac{π}{2} - 0] = \frac{π}{4}$