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Integrals

Multiple Choice Questions

841.

Intersection point of f₁(x) = $\int_{2}^{x} (2 t - 5) dt and f_{2} (x) = \int_{0}^{x} 2 tdt$ is

$(\frac{6}{5}, \frac{36}{25})$
$(\frac{2}{3}, \frac{4}{9})$
$(\frac{1}{3}, \frac{1}{9})$
$(\frac{1}{5}, \frac{1}{25})$

842.

The value of $\lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} e^{\frac{r}{n}}$ is

e
e - 1
1 - e
1 + e

843.

$\int_{0}^{2} \sqrt{\frac{2 + x}{2 - x}} dx$ is equal to

$π + 2$
$π + \frac{3}{2}$
$π + 1$
None of these

844.

$\int e^{x} \sin (e^{x}) dx$ is equal to

$- \cos (e^{x}) + c$
$\cos (e^{x}) + c$
$- \csc (e^{x}) + c$
None of these

845.

$\int e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) dx$ is equal to

$\frac{- e^{x}}{x^{2}} + c$
$\frac{e^{x}}{x^{2}} + c$
$\frac{e^{x}}{x} + c$
$\frac{- e^{x}}{x} + c$

846.

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

$\cos^{- 1} (\sqrt{x}) + \sqrt{1 - x} (\sqrt{x} - 2) + c$
$\cos^{- 1} (\sqrt{x}) - \sqrt{1 - x} (\sqrt{x} - 2) + c$
$\cos^{- 1} (\sqrt{x}) + \sqrt{1 - x} (\sqrt{x} + 2) + c$
None of the above

847.

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

$\frac{1}{2} {[\log (x + \sqrt{1 + x^{2}})]}^{2} + c$
$\log {(x + \sqrt{1 + x^{2}})}^{2} + c$
$\log (x + \sqrt{1 + x^{2}}) + c$
None of the above

848.

$\int \frac{\sin^{8} (x) - \cos^{8} (x)}{1 - 2 \sin^{2} (x) \cos^{2} (x)} dx$ is equal to

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

849.
$\int \frac{dx}{(\sin (x) + \sin (2 x))}$ is equal to

$\frac{1}{6} \log (1 - \cos (x)) + \frac{1}{2} \log (\log (1 + \cos (x))) - \frac{2}{3} \log (1 + 2 \cos (x)) + c$

$6 \log (1 - \cos (x)) + 2 \log (\log (1 + \cos (x))) - \frac{2}{3} \log (1 + 2 \cos (x)) + c$

$6 \log (1 - \cos (x)) + \frac{1}{2} \log (\log (1 + \cos (x))) + \frac{2}{3} \log (1 + 2 \cos (x)) + c$

None of the above

$\frac{1}{6} \log (1 - \cos (x)) + \frac{1}{2} \log (\log (1 + \cos (x))) - \frac{2}{3} \log (1 + 2 \cos (x)) + c$

$Let I = \int \frac{dx}{(\sin (x) + \sin (2 x))} = \int \frac{dx}{\sin (x) + 2 \sin (x) \cos (x)} = \int \frac{dx}{\sin (x) (1 + 2 \cos (x))} = \int \frac{\sin (x) dx}{(1 - \cos (x)) (1 + \cos (x)) (1 + 2 \cos (x))} Put \cos (x) = t and \sin (x) dx = - dt ∴ I = - \int \frac{dt}{(1 - t) (1 + t) (1 + 2 t)} = - \int \{\frac{1}{6 (1 - t)} - \frac{1}{2 (1 + t)} + \frac{4}{3 (1 + 2 t)}\} dt = - \{- \frac{1}{6} \log (1 - t) - \frac{1}{2} \log (1 + t) + \frac{4}{6} \log (1 + 2 t)\} + c = \frac{1}{6} \log (1 - \cos (x)) + \frac{1}{2} \log (\log (1 + \cos (x))) - \frac{2}{3} \log (1 + 2 \cos (x)) + c$