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Integrals

Multiple Choice Questions

371.

$\int \frac{1}{16 x^{2} + 9} dx$ is equal to

$\frac{1}{3} \tan^{- 1} (\frac{4 x}{3}) + c$
$\frac{1}{4} \tan^{- 1} (\frac{4 x}{3}) + c$
$\frac{1}{12} \tan^{- 1} (\frac{4 x}{3}) + c$
$\frac{1}{12} \tan^{- 1} (\frac{3 x}{4}) + c$

372.

The value of $\int_{4}^{7} \frac{{(11 - x)}^{2}}{x^{2} + {(11 - x)}^{2}} dx$ is

373.

$\int (\sqrt{\tan (x)} + \sqrt{\cot (x)}) dx$

$\sqrt{2} \tan^{- 1} (\frac{\tan (x)}{\sqrt{\tan (x)}}) + C$
$\sqrt{2} \tan^{- 1} (\frac{\tan (x) - 1}{\sqrt{2 \tan (x)}}) + C$
$\frac{\tan (x)}{\sqrt{2}} . \tan^{- 1} (\frac{\cot (x) + 1}{\sqrt{2 \tan (x)}}) + C$
$\frac{\tan (x)}{\sqrt{2}} . \tan^{- 1} (\frac{\cot (x) + 1}{\sqrt{2 \tan (x)}}) + C$

374.

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

$\frac{\sin (x) + \cos (x)}{xsin (x) + \cos (x)} + C$
$\frac{x \sin (x) - \cos (x)}{xsin (x) + \cos (x)} + C$
$\frac{\sin (x) - xcos (x)}{xsin (x) + \cos (x)} + C$
None of these

375.

$\int_{0}^{x} \frac{xdx}{1 + \cos (α) \sin (x)}, (0 < α < π)$ is equal to

$\frac{πα}{\sin (α)}$
$\frac{πα}{\cos (α)}$
$\frac{πα}{1 + \sin (α)}$
$\frac{πα}{1 - \cos (α)}$

376.

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\cos (x)}{1 + e^{x}} d x$

1
0
- 1
None of these

377.

$\int_{0}^{\frac{π}{2}} \frac{dx}{1 + \tan (x)}$ is equal to

$π$
$\frac{π}{2}$
$\frac{π}{3}$
$\frac{π}{4}$

378.

By Simpson rule taking n = 4, the value of the integral $\int_{0}^{1} \frac{1}{1 + x^{2}} dx$ is equal to

0.788
0.781
0.785
None of the above

379.
The value of $\int_{0}^{π} \log (1 + \cos (x)) dx$ is

$- \frac{π}{2} \log (2)$

$πlog (\frac{1}{2})$

$πlog (2)$

$\frac{π}{2} \log (2)$

$πlog (\frac{1}{2})$

$Let I = \int_{0}^{π} \log (1 + \cos (x)) dx . . . (i) I = \int_{0}^{π} \log (1 + \cos (π - x)) dx . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{π} \{\log (1 + \cos (x)) + \log (1 - \cos (x))\} dx I = \frac{1}{2} \int_{0}^{π} \log (1 - \cos^{2} (x)) dx = \frac{1}{2} \int_{0}^{π} \log (\sin^{2} (x)) dx = \int_{0}^{π} \log (\sin (x)) dx = 2 \int_{0}^{\frac{π}{2}} \log (\sin (x)) dx \{∵ \int_{0}^{2 a} f (x) dx = 2 \int_{0}^{a} f (x) dx, if f (2 a - x) = f (x)\} = 2 \{- \frac{π}{2} \log (2)\} [∵ \int_{0}^{\frac{π}{2}} \log (\sin (x)) dx = - \frac{π}{2} \log (2)] = πlog (\frac{1}{2})$