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Integrals

Multiple Choice Questions

691.

The value of $\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\sin^{- 1} (x)}{{(1 - x^{2})}^{\frac{3}{2}}} dx$ is

$\frac{π}{2} - \log (2)$
$\frac{π}{4} - \frac{1}{2} \log (2)$
$\frac{π}{4} + \frac{1}{2} \log (2)$
$π - \frac{1}{2} \log (2)$

692.

Integral of $\frac{1}{2 + \cos (x)}$

$\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{1}{2} \tan (x)) + C$
$\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}} \tan (\frac{x}{2})) + C$
$- \sin (x) \log (2 + \cos (x)) + C$
$\sin (x) \log (2 + \cos (x)) + C$

693.

$\int \frac{dx}{\sqrt{x^{4} + x^{6}}}$ is equal to

$- \frac{\sqrt{1 + x^{2}}}{x} + C$
$\frac{\sqrt{1 + x^{2}}}{x} + C$
$- \frac{\sqrt{1 - x^{2}}}{x} + C$
$- \frac{\sqrt{x^{2} - 1}}{x} + C$

694.

If $\int \sin^{- 1} (x) \cos^{- 1} (x) dx = f^{- 1} (x) [\frac{π}{2} x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}]$

$\frac{π}{2} \sqrt{1 - x^{2}} + 2 x + C$ , then

f(x) = sin(x)
f(x) = cos(x)
f(x) = tan(x)
None of these

695.
If $\int_{0}^{π} xf (\sin^{2} (x) + \sec^{2} (x)) dx = k \int_{0}^{\frac{π}{2}} f (\sin^{2} (x) + \sec^{2} (x)) dx$ , then the value of k is

$\frac{π}{2}$

$π$

$- \frac{π}{2}$

None of these

$π$

$We have, \int_{0}^{π} xf (\sin^{2} (x) + \sec^{2} (x)) dx = k \int_{0}^{\frac{π}{2}} f (\sin^{2} (x) + \sec^{2} (x)) dx Let I = \int_{0}^{π} xf (\sin^{2} (x) + \sec^{2} (x)) dx . . . (i) = \int_{0}^{π} (π - x) f (\sin^{2} (π - x) + \sec^{2} (π - x)) dx = \int_{0}^{π} (π - x) f (\sin^{2} (x) + \sec^{2} (x)) . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = π \int_{0}^{π} f (\sin^{2} (x) + \sec^{2} (x)) \Rightarrow 2 I = 2 π \int_{0}^{\frac{π}{2}} f (\sin^{2} (x) + \sec^{2} (x)) \Rightarrow I = π \int_{0}^{\frac{π}{2}} f (\sin^{2} (x) + \sec^{2} (x))$

On comparing with given integral, we get $k = π$ .

696.

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

697.

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

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

698.

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

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

699.

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

700.

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

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

Previous Year Papers

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Multiple Choice Questions

695. If ∫0πxfsin2x + sec2xdx = k∫0π2fsin2x + sec2xdx, then the value of k is π2 π - π2 None of these

695.
If $\int_{0}^{π} xf (\sin^{2} (x) + \sec^{2} (x)) dx = k \int_{0}^{\frac{π}{2}} f (\sin^{2} (x) + \sec^{2} (x)) dx$ , then the value of k is

$\frac{π}{2}$

$π$

$- \frac{π}{2}$

None of these