$\int \frac{\cos \left(2\mathrm{x}\right) - 1}{\cos \left(2\mathrm{x}\right) + 1}\mathrm{dx}$ is equal to

• $\tan \left(\mathrm{x}\right) - \mathrm{x} + \mathrm{c}$

• $\mathrm{x} + \tan \left(\mathrm{x}\right) + \mathrm{c}$

• $\mathrm{x} - \tan \left(\mathrm{x}\right) + \mathrm{c}$

• $- \mathrm{x} - \cot \left(\mathrm{x}\right) + \mathrm{c}$

Suppose f is such that f( - x) = - f(x), for every real x and ${\int }_0^1\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} = 5, \mathrm{then} {\int }_{- 1}^0\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$ is equal to

• 10

• 5

• 0

• - 5

833.

If f(y) = e^y, g(y) = y, y > 0 and F(t) = ${\int }_0^{\mathrm{t}}\mathrm{f}\left(\mathrm{t} - \mathrm{y}\right) . \mathrm{g}\left(\mathrm{y}\right)\mathrm{dy}$, then 

• F(t) = 1 - e^{- t}(1 + t)

• F(t) = e^t - (1 + t)

• F(t) = te^t

• F(t) = te^{- t}

Let f(x) be a function satisfying f'(x) = f(x) with f(0) = 1 and g(x) be a function that satisfies f(x) + g(x) = x². Then, the value of the integeral ${\int }_0^1\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}$ is

• $\mathrm{e} - \frac{{\mathrm{e}}^2}{2} - \frac{5}{2}$

• $\mathrm{e} + \frac{{\mathrm{e}}^2}{2} - \frac{3}{2}$

• $\mathrm{e} - \frac{{\mathrm{e}}^2}{2} - \frac{3}{2}$

• $\mathrm{e} + \frac{{\mathrm{e}}^2}{2} + \frac{5}{2}$

${\int }_{- 1}^0\frac{\mathrm{dx}}{{\mathrm{x}}^2 + 2\mathrm{x} + 2}$ is equal to

• 0

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{- \mathrm{\pi }}{4}$

The value of f(x) is given only at x = 0, $\frac{1}{3}, \frac{2}{3}$ wich of the following can be used to evaluate ${\int }_0^1\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$ approximately

• Trapezoidal rule

• Simpson's rule

• Trapezoidal as well as Simpson's rule

• None of the above

If f(x) = $\frac{{\mathrm{e}}^{\mathrm{x}}}{1 + {\mathrm{e}}^{\mathrm{x}}}, {\mathrm{I}}_1 = {\int }_{\mathrm{f}\left(- \mathrm{a}\right)}^{\mathrm{f}\left(\mathrm{a}\right)}\mathrm{xg}\left\{\mathrm{x}\left(1 - \mathrm{x}\right)\right\}\mathrm{dx}$ and ${\mathrm{I}}_2 = {\int }_{\mathrm{f}\left(- \mathrm{a}\right)}^{\mathrm{f}\left(\mathrm{a}\right)}\mathrm{g}\left\{\mathrm{x}\left(1 - \mathrm{x}\right)\right\}\mathrm{dx}$, the value of $\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}$ is

• 2

• - 3

• - 1

• 1

${\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$ is equal to

• ${\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{a} - \mathrm{x}\right)\mathrm{dx}$

• ${\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x} - \mathrm{a}\right)\mathrm{dx}$

• ${\int }_0^{\mathrm{a}}\mathrm{f}\left(2\mathrm{a} - \mathrm{x}\right)\mathrm{dx}$

• ${\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x} + 2\mathrm{a}\right)\mathrm{dx}$

$\int {\sin }^4\left(\mathrm{x}\right)\mathrm{dx}$ is equal to

• $\frac{1}{8}\left[3\mathrm{x} + \frac{\sin \left(4\mathrm{x}\right)}{4} - \frac{4\sin \left(2\mathrm{x}\right)}{2}\right] + \mathrm{c}$

• $\frac{1}{8}\left[3\mathrm{x} + \frac{\sin \left(4\mathrm{x}\right)}{4} + \frac{4\sin \left(2\mathrm{x}\right)}{2}\right] + \mathrm{c}$

• $\frac{1}{4}\left[3\mathrm{x} + \frac{\sin \left(4\mathrm{x}\right)}{4} - \frac{4\cos \left(2\mathrm{x}\right)}{2}\right] + \mathrm{c}$

• $\frac{1}{8}\left[3\mathrm{x} + \frac{\sin \left(4\mathrm{x}\right)}{4} + \frac{4\cos \left(4\mathrm{x}\right)}{2}\right] + \mathrm{c}$

$\int \log \left(\mathrm{x}\right)\mathrm{dx}$ is equal to

• $\mathrm{x} + \mathrm{xlog}\left(\mathrm{x}\right) + \mathrm{c}$

• $\mathrm{xlog}\left(\mathrm{x}\right) - \mathrm{x} +\hspace{0.17em}\mathrm{c}$

• ${\mathrm{x}}^2\log \left(\mathrm{x}\right) +\hspace{0.17em}\mathrm{c}$

• $\frac{1}{\mathrm{x}}\log \left(\mathrm{x}\right) + \mathrm{xc}$