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Integrals

Multiple Choice Questions

641.

The value of $\int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} (\log (\sin (x)) + \cot (x)) dx$ is

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

642.

Considering four sub-intervals, the value of $\int_{0}^{1} \frac{1}{1 + x} dx$ by Trapezoidal rule, is

0.6870
0.6677
0.6977
0.5970

643.

By Simpson's rule, the value of $\int_{1}^{2} \frac{dx}{x}$ dividing the interval (1, 2) into four equal parts, is

0.6932
0.6753
0.6692
7.1324

644.
The value of $\int xsin (x) \sec^{3} (x) dx$ is

$\frac{1}{2} [\sec^{2} (x) - \tan (x)] + c$

$\frac{1}{2} [{xsec}^{2} (x) - \tan (x)] + c$

$\frac{1}{2} [{xsec}^{2} (x) + \tan (x)] + c$

$\frac{1}{2} [\sec^{2} (x) + \tan (x)] + c$

$\frac{1}{2} [{xsec}^{2} (x) - \tan (x)] + c$

$\int xsin (x) \sec^{3} (x) dx = \int xsin (x) \frac{1}{\cos^{3} (x)} dx \int xtan (x) . \sec^{2} (x) dx = dt Put \tan (x) = t \Rightarrow \sec^{2} (x) dx = dt and x = \tan^{- 1} (t) Then, it reduces to \int \tan^{- 1} (t) . t dt = \frac{t^{2}}{2} \tan^{- 1} (t) - \int \frac{t^{2}}{2 (1 + t^{2})} dt = \frac{{xtan}^{2} (x)}{2} - \frac{1}{2} t + \frac{1}{2} \tan^{- 1} (t) + c = \frac{x (\sec^{2} (x) - 1)}{2} - \frac{1}{2} \tan (x) + \frac{1}{2} x + c = \frac{1}{2} [\sec^{2} (x) - \tan (x)] + c$