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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

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

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

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

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

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

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

$Let I = \int \sin^{- 1} (\sqrt{\frac{x}{a + x}}) dx Put x = {atan}^{2} (θ) \Rightarrow dx = 2 atan (θ) \sec^{2} (θ) dθ Now, I = \int \sin^{- 1} (\sqrt{\frac{{atan}^{2} (θ)}{a + {atan}^{2} (θ)}} . 2 atan (θ) \sec^{2} (θ)) dθ = \int \sin^{- 1} (\sqrt{\frac{\tan^{2} (θ)}{\sec^{2} (θ)}} 2 atan (θ) \sec^{2} (θ) dθ) [∵ 1 + \tan^{2} (θ) = \sec^{2} (θ)] = \int \sin^{- 1} (\sin (θ)) 2 atan (θ) \sec^{2} (θ) dθ = 2 a \int \underset{I}{θ} \underset{II}{(\tan (θ) \sec^{2} (θ)) dθ} [∵ \sin^{- 1} (\sin (θ)) = θ] = 2 a [θ \int (\tan (θ) \sec^{2} (θ)) dθ - \int \{\frac{d}{dθ} (θ) \int \tan (θ) \sec^{2} (θ) dθ\} dθ] = 2 a [θ \frac{\sec^{2} (θ)}{2} - \int 1 \frac{\sec^{2} (θ)}{2} dθ] + C [∵ \int \tan (θ) \sec^{2} (θ) = \int zdz = \frac{z^{2}}{2} = \frac{\sec^{2} (θ)}{2}] = a [θ \sec^{2} (θ) - \tan (θ)] + C = a [\tan^{- 1} (\sqrt{\frac{x}{a}} . \frac{(a + x)}{a} - \sqrt{\frac{x}{a}})] + C [∵ 1 + \tan^{2} (θ) = \sec^{2} (θ) and \sec^{2} (θ) = 1 + \frac{x}{a}]$