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Integrals

Multiple Choice Questions

411.

$\int \frac{x + 2}{(x^{2} + 3 x + 3) \sqrt{x + 1}} dx$ is equa to

$\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x}{x + 1}) + C$
$\frac{2}{\sqrt{3}} \tan^{- 1} [\frac{x}{\sqrt{3 (x + 1)}}] + C$
$\frac{2}{\sqrt{3}} \tan^{- 1} [\frac{x}{{(x + 1)}^{2}}] + C$
None of these

412.

$\int \sin^{- 1} (\sqrt{\frac{x}{a + x}}) dx$ is equal to

$[\tan^{- 1} (\sqrt{\frac{x}{a}} + \sqrt{\frac{x}{a}})] + C$
$a [\tan^{- 1} (\sqrt{\frac{x}{a}} - \sqrt{\frac{x}{a}})] + C$
$a [\tan^{- 1} (\sqrt{\frac{x}{a}} . \frac{(a + x)}{a})] + C$
$a [\tan^{- 1} (\sqrt{\frac{x}{a}} . \frac{(a + x)}{a} - \sqrt{\frac{x}{a}})] + C$

413.

$\lim_{n \to \infty} \frac{1}{n} (\frac{1}{n + 1} + \frac{2}{n + 2} + . . . + \frac{3 n}{4 n})$ is equal to

log(4)
- log(4)
1 - log(4)
None of these

414.

The value of the integral $\int_{0}^{\frac{π}{2}} \frac{\sin^{2} (x)}{\sin (x) + \cos (x)} dx$ is equal to

$\sqrt{2} (\log (\sqrt{2}))$
$\sqrt{2} (\sqrt{2} + 1)$
$\log (\sqrt{2} + 1)$
None of the above

415.

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

$\frac{1}{3} \tan^{- 1} (\frac{3 \tan (x)}{5}) + c$
$\frac{1}{5} \tan^{- 1} (\frac{\tan (x)}{15}) + c$
$\frac{1}{15} \tan^{- 1} (\frac{\tan (x)}{5}) + c$
$\frac{1}{15} \tan^{- 1} (\frac{5 \tan (x)}{3}) + c$

416.

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

417.
If f(x) = $Asin (\frac{πx}{2}) + B, f' (\frac{1}{2}) = \sqrt{2}$ and $\int_{0}^{1} f (x) dx = \frac{2 A}{π}$ , then A and B are

$\frac{π}{2}, \frac{π}{2}$

$\frac{2}{π}, \frac{3}{π}$

$0, - 4 π$

$\frac{4}{π}, 0$

$\frac{4}{π}, 0$

$f (x) = Asin (\frac{πx}{2}) + B, f' (\frac{1}{2}) = \sqrt{2} \Rightarrow f' (x) = \frac{Aπ}{2} \cos (\frac{πx}{2}) \Rightarrow f' (\frac{1}{2}) = \frac{Aπ}{2} \cos (\frac{π}{4}) = \frac{Aπ}{2 \sqrt{2}} \Rightarrow \frac{Aπ}{2 \sqrt{2}} = \sqrt{2} \{∵ f' (\frac{1}{2}) = \sqrt{2}\} \Rightarrow A = \frac{4}{π} Now, \int_{0}^{1} f (x) dx = \frac{2 A}{π} \Rightarrow \int_{0}^{1} Asin (\frac{πx}{2}) + B . dx = \frac{2 A}{π} \Rightarrow {[\frac{- 2 A}{π} \cos (\frac{πx}{2}) + B]}_{0}^{1} = \frac{2 A}{π} \Rightarrow B + \frac{2 A}{π} = \frac{2 A}{π} \Rightarrow B = 0 Hence, option (d) \frac{4}{π}, 0 is correct .$

418.

Let g(x) = $\int_{0}^{x} f (t) dt$ , where f is such that $\frac{1}{2} \leq f (x) \leq 1$ for t $\in$ [0, 1] and $0 \leq f (t) \leq \frac{1}{2}$ for t $\in$ [1, 2]. Then, g(2) satisfies the inequality