The value of ∫π4π2exlogsinx + cotxdx&nbs

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Integrals

Multiple Choice Questions

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641.
The value of $\int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} (\log (\sin (x)) + \cot (x)) dx$ is

$e^{\frac{π}{4}} \log (2)$

$- e^{\frac{π}{4}} \log (2)$

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

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

C.

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

$Let I = \int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} (\log (\sin (x)) + \cot (x)) dx \Rightarrow I = \int_{\frac{π}{4}}^{\frac{π}{2}} + \int_{\frac{π}{4}}^{\frac{π}{2}} = \int_{\frac{π}{4}}^{\frac{π}{2}} + - \int_{\frac{π}{4}}^{\frac{π}{2}} = e^{\frac{π}{2}} \log (\sin (\frac{π}{2})) - e^{\frac{π}{4}} \log (\sin (\frac{π}{4})) = e^{- \frac{π}{2}} \log (\frac{1}{\sqrt{2}}) = \frac{1}{2} e^{\frac{π}{4}} \log (2)$

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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

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

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645.

The value of $\int_{0}^{x} {xsin}^{3} (x) dx$ is

$\frac{4 π}{3}$
$\frac{2 π}{3}$
0
None of these

646.

Which of the following is true ?

$\int_{0}^{1} e^{x} dx = e$
$\int_{0}^{1} 2^{x} dx = \log (2)$
$\int_{0}^{1} \sqrt{x} dx = \frac{2}{3}$
$\int_{0}^{1} xdx = \frac{1}{3}$

647.

$\int [\sin (\log (x)) + \cos (\log (x))] dx$ is equal to

$xcos (\log (x)) + c$
$\cos (\log (x)) + c$
$xsin (\log (x)) + c$
$\sin (\log (x)) + c$

648.

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

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

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649.

$\int_{5}^{10} \frac{1}{(x - 1) (x - 2)} dx$

$\log (\frac{27}{32})$
$\log (\frac{32}{27})$
$\log (\frac{8}{9})$
$\log (\frac{3}{4})$

650.

$\int xlog (x) dx$ is equal to

$\frac{x^{2}}{4} (2 \log (x) - 1) + c$
$\frac{x^{2}}{2} (2 \log (x) - 1)$
$\frac{x^{2}}{4} (2 \log (x) + 1) + c$
$\frac{x^{2}}{2} (2 \log (x) + 1)$

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