MCQ
The value of $\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ is:
  • A
    $2\cos\sqrt{\text{x}}+\text{C}$
  • B
    $\sqrt{\frac{\cos\text{x}}{\text{x}}}+\text{C}$
  • C
    $\sin\sqrt{\text{x}}+\text{C}$
  • $2\sin\sqrt{\text{x}}+\text{C}$

Answer

Correct option: D.
$2\sin\sqrt{\text{x}}+\text{C}$
$\text{I}=\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$Put $\sqrt{\text{x}}=\text{t}$
$\frac{1}{2\sqrt{\text{x}}}\text{ dx}=\text{dt}$
$\frac{1}{\sqrt{\text{x}}}\text{ dx}=2\text{dt}$
$\text{I}=\int\cos\text{t }2\text{ dt}$
$\text{I}=2\sin\text{t}+\text{C}$
$\text{I}=2\sin\sqrt{\text{x}}+\text{C}$

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

$\tan^{-1}1+\cos^{-1}\Big(\frac{-1}{2}\Big)+\sin^{-1}\Big(\frac{-1}{2}\Big)$
If $f$ is a real-valued differentiable function satisfying $|f(x)-f(y)| \leq(x-y)^2, x, y \in R$ and $f(0)=0$, then $f(1)$ equals
Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{a\sin \left( {x + b} \right)}&{x \ge 0}\\
{6{x^7} - x + 1}&{x < 0}
\end{array}} \right.$ is differentiable for all real $x$. If $a \in R$ and $b \in \left[ {0,2\pi } \right]$ , then number of ordered pair $(s)$ of $(a, b)$ is
$\int {x{e^{{x^2}}}} dx = $
The domain of $f(x) = \frac{1}{{\sqrt {{{\log }_{\frac{\pi }{4}}}({{\sin }^{ - 1}}x) - 1} }}$,is
Let $y = y\left( x \right)$ be the solutions of the differential equation, $\left( {{x^2} + 1} \right)^2\,\frac{{dy}}{{dx}} + 2x\left( {{x^2} + 1} \right)\,y = 1$ such that $y\left( 0 \right) = 0$. If $\sqrt a y\left( 1 \right) = \frac{\pi }{{32}}$, then the value of $‘a’$ is
If order of A + B is n × n, then the order of AB is:
Let $A = \left[ {\begin{array}{*{20}{c}}4&6&{ - 1}\\3&0&2\\1&{ - 2}&5\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}2&4\\0&1\\{ - 1}&2\end{array}} \right]$ and $C = [3\,\,1\,\,2]$. The expression which is not defined is
$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ equal to
The area bounded by the parabolas $y=x^2$ and $y=1-x^2$ equals