MCQ
$\int\cos(\log_\text{e}.\text{x})\text{dx}$ is equal to:
  • A
    $\frac{1}{2}\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$
  • $\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$
  • C
    $\frac{1}{2}\text{x}[\cos(\log_\text{e}\text{x})-\sin(\log_\text{e}\text{x})]$
  • D
    $\text{x}[\cos(\log_\text{e}\text{x})-\sin(\log_\text{e}\text{x})]$

Answer

Correct option: B.
$\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$
$\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Consider the system of equations

$ x-2 y+3 z=-1 $ ; $ -x+y-2 z=k $ ; $ x-3 y+4 z=1$

$STATEMENT -1$ : The system of equations has no solution for $\mathrm{k} \neq 3$. and

$STATEMENT - 2$ : The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & \mathrm{k} \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $\mathrm{k} \neq 3$.

If $\triangle_1=\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix},\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix},$ then:
Let $\mathrm{F}:[3,5] \rightarrow \mathrm{R}$ be a twice differentiable function on $(3,5)$ such that $\mathrm{F}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}}$ $\int_{3}^{x}\left(3 t^{2}+2 t+4 F^{\prime}(t)\right) \,d t$

If $F^{\prime}(4)=\frac{\alpha e^{\beta}-224}{\left(e^{\beta}-4\right)^{2}}$, then $\alpha+\beta$ is equal to $....$

The value of $\sum\limits_{n = 1}^N {{U_n},} $ if ${U_n} = \left| {\,\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}\,} \right|$ is
If $\left|\begin{array}{lll}\alpha & 3 & 4 \\ 1 & 2 & 1 \\ 1 & 4 & 1\end{array}\right|=0,$ then the value of $\alpha$ is
$\int\frac{\text{x+sin x}}{1+\text{cos x}}\ dx$ is equal to:
The differential equation $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\text{x}^{2}$ has the general solution :
Let the image of the point $P (1,2,3)$ in the line $L : \frac{ x -6}{3}=\frac{ y -1}{2}=\frac{ z -2}{3}$ be $Q .$ let $R (\alpha, \beta, \gamma)$ be a point that divides internally the line segment $PQ$ in the ratio $1: 3$. Then the value of $22(\alpha+\beta+\gamma)$ is equal to
If two constraints do not intersect in the positive quadrant of the graph, then.
If $y = (A + Bx) e^{mx} + (m - 1)^{-2} e^x$  then $\frac{{{d^2}y}}{{d{x^2}}}$ $- 2m \frac{{dy}}{{dx}} \,\,+ m^2y$ is equal to :