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Integrals

Multiple Choice Questions

211.

$\int {(27 e^{9 x} + e^{12 x})}^{\frac{1}{3}} dx$ is equal to

$(\frac{1}{4}) {(27 + e^{3 x})}^{\frac{1}{3}} + C$
$(\frac{1}{4}) {(27 + e^{3 x})}^{\frac{2}{3}} + C$
$(\frac{1}{3}) {(27 + e^{3 x})}^{\frac{4}{3}} + C$
$(\frac{1}{4}) {(27 + e^{3 x})}^{\frac{4}{3}} + C$

212.

$\int \frac{4 dx}{x^{2} \sqrt{4 - 9 x^{2}}}$ is equal to

$\sqrt{4 - 9 x^{2}} + C$
$\frac{- 2}{3} \sqrt{4 - 9 x^{2}} + C$
$\frac{- \sqrt{4 - 9 x^{2}}}{x} + C$
$\frac{2}{3} \frac{\sqrt{4 - 9 x^{2}}}{x} + C$

213.

$\int e^{- x} \csc (x) (1 + \cot (x)) dx$ is equal to

$e^{- x} \csc (x) + C$
$- e^{- x} \csc (x) + C$
- $e^{- x} (\csc (x) + \cot (x)) + C$
- $e^{- x} (\csc (x) - \tan (x)) + C$

214.

$\int \frac{\sin (x) + \cos (x)}{e^{- x} + \sin (x)} dx$ is equal to

$\log |1 - e^{x} \sin (x)| + C$
$\log |1 + e^{x} \sin (x)| + C$
$\log |1 + e^{x} \sin (x)| + C$
$\log |1 - e^{- x} \sin (x)| + C$

215.

The solution of $\int_{\sqrt{2}}^{x} \frac{dt}{\sqrt{t^{2} - 1}} = \frac{π}{12}$ is

216.

Let f(x) = x² - 2. If $\int_{3}^{6} f (x) dx = 3 f (c)$ for some c $\in$ (3, 6), then the value of c is equal to

$\sqrt{12}$
$\sqrt{21}$
$\sqrt{19}$
$\sqrt{17}$

217.

$\int_{0}^{\frac{π}{2}} \frac{\sin (2 x)}{1 + 2 \cos^{2} (x)} dx$ is equal to

$\frac{1}{2} \log (2)$
$\log (2)$
$\frac{1}{2} \log (3)$
$\log (3)$

218.
$\int_{0}^{\frac{π}{2}} \frac{dx}{1 + \tan^{3} (x)}$ is equal to

1

$π$

$\frac{π}{2}$

$\frac{π}{4}$

$\frac{π}{4}$

$Let I = \int_{0}^{\frac{π}{2}} \frac{dx}{1 + \tan^{3} (x)} \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos^{3} (x)}{\sin^{3} (x) + \cos^{3} (x)} dx . . . (i) \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos^{3} (\frac{π}{2} - x)}{\sin^{3} (\frac{π}{2} - x) + \cos^{3} (\frac{π}{2} - x)} dx [∵ \int_{0}^{a} f (x) = \int_{0}^{a} f (a - x) dx] \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin^{3} (x)}{\cos^{3} (x) + \sin^{3} (x)} On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{3} (x) + \cos^{3} (x)}{\sin^{3} (x) + \cos^{3} (x)} d x = \int_{0}^{\frac{π}{2}} (1) dx = {[x]}_{0}^{\frac{π}{2}} \Rightarrow 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4}$