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Integrals

Multiple Choice Questions

401.

If $\int_{0}^{\infty} \frac{\log (1 + x^{2})}{1 + x^{2}} d x = k \int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} dx$ , then k is equal to

4
8
$π$
2 $π$

402.

$\int_{0}^{x} |\sin (t)| dt, where x \in [2 nπ, (2 n + 1) π], n \in N$ , is equal to

4n - 1 - cos(x)
4n - sin(x)
4n - cos(x)
4n + 1 - cos(x)

403.

Let I₁ = $\int_{0}^{1} \frac{e^{x} dx}{1 + x}$ and I₂ = $\int_{0}^{1} \frac{x^{2} dx}{e^{x^{3}} (2 - x^{3})}$ Then, $\frac{I_{1}}{I_{2}}$ is equal to

$\frac{1}{3 e}$
3e
$\frac{e}{3}$
$\frac{3}{e}$

404.

$\int_{\frac{5}{2}}^{5} \frac{\sqrt{{(25 - x^{2})}^{3}}}{x^{4}} dx$ is equal to

$\frac{π}{3}$
$\frac{2 π}{3}$
$\frac{π}{6}$
$\frac{5 π}{6}$

405.

$\int \frac{dx}{\cos (x) + \sqrt{3} \sin (x)}$ equals

$\frac{1}{2} \log (\tan (\frac{x}{2} + \frac{π}{12})) + C$
$\frac{1}{3} \log (\tan (\frac{x}{2} - \frac{π}{12})) + C$
$\log (\tan (\frac{x}{2} + \frac{π}{6})) + C$
$\frac{1}{2} \log (\tan (\frac{x}{2} - \frac{π}{6})) + C$

406.
$\int_{0}^{\frac{π}{2}} \sin (2 x) . \log (\tan (x)) dx$

0

2

4

7

$Consider, I = \int_{0}^{\frac{π}{2}} (\log (\tan (x))) . \sin (2 x) dx . . . (i) I = \int_{0}^{\frac{π}{2}} (\log (\tan (\frac{π}{2} - x))) . \sin (2 (\frac{π}{2} - x)) dx$

$[∵ \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx]$

$\Rightarrow$ $I = \int_{0}^{\frac{π}{2}} \log (\cot (x)) \sin (2 x) dx$ ...(ii)

$[∵ \sin (π - 2 x) = \sin (2 x)]$

On adding Eqs. (i) and (ii), we get

$2 I = \int_{0}^{\frac{π}{2}} \log (\tan (x)) . \sin (2 x) dx + \int_{0}^{\frac{π}{2}} \log (\cot (x)) . \sin (2 x) dx = \int_{0}^{\frac{π}{2}} \sin (2 x) \log (\tan (x) . \cot (x)) dx [∵ \log (m) + \log (n) = \log (m . n)] = \int_{0}^{\frac{π}{2}} \sin (2 x) \log (1) dx \Rightarrow I = 0 [∵ \log (1) = 0] ∴ \int_{0}^{\frac{π}{2}} \sin (2 x) \log (\tan (x)) dx = 0$

407.

If $∆ x = |\begin{matrix} 1 & \cos (x) & 1 - \cos (x) \\ 1 + \sin (x) & \cos (x) & 1 + \sin (x) + \cos (x) \\ \sin (x) & \sin (x) & 1 \end{matrix}|$ , then $\int_{0}^{\frac{π}{4}} ∆ (x) dx$ is equal to