MCQ
$\int_{\,0}^{\,\pi } {\sqrt {\frac{{1 + \cos 2x}}{2}} \,dx} $ is equal to
  • A
    $0$
  • $2$
  • C
    $1$
  • D
    $ - 1$

Answer

Correct option: B.
$2$
b
(b) $I = \int_0^\pi {\sqrt {\frac{{1 + \cos 2x}}{2}} dx = \int_0^{\,\pi } {|\cos x|\,dx} } $

$I = \int_{\,0}^{\,\pi /2} {\cos x\,dx} - \int_{\,\pi /2}^{\,\pi } {\cos x\,dx} $

$= [\sin x]_0^{\pi /2} - [\sin x]_{\pi /2}^\pi $

$I = \left[ {\sin \frac{\pi }{2} - \sin 0} \right] - \left[ {\sin \pi - \sin \frac{\pi }{2}} \right] $

$=1+ 1 = 2.$

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

If $x = {e^{y + {e^{y + ....t{\rm{o}}\,\,\infty }}}}$, $x > 0,$ then ${{dy} \over {dx}}$ is
Let the ellipse, $E_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$ and $\mathrm{E}_{2}: \frac{\mathrm{x}^{2}}{\mathrm{~A}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~B}^{2}}=1, \mathrm{~A}<\mathrm{B}$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$, and the distance between the foci of $E_{1}$ be 4. If $E_{1}$ and $E_{2}$ meet at $A, B, C$ and $D$, then the area of the quadrilateral ABCD equals:
If $A = \left( {\begin{array}{*{20}{c}}1&{ - 2}&1\\2&1&3\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}2&1\\3&2\\1&1\end{array}} \right)$, then ${(AB)^T}$ is equal to
The slope of the line touching both the parabolas ${y^2} = 4x$ and ${x^2} = - 32y$, is
If $\frac{{dy}}{{dx}} + \frac{3}{{{{\cos }^2}\,x}}\,y = \frac{1}{{{{\cos }^2}\,x}},$ $x \in \left( {\frac{{ - \pi }}{3},\frac{\pi }{3}} \right)$ and $y\left( {\frac{\pi }{4}} \right) = \frac{4}{3}$, then $y\left( { - \frac{\pi }{4}} \right)$ equals
If $\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-a x\right)=b$, then the ordered pair $(a, b)$ is:
If $f(x) = \left\{ \begin{array}{l}1 + x,\;{\rm{when\,\, }}x \le 2\\5 - x,\,{\rm{when \,\,}}\,x \le 3\end{array} \right.,$ then
A six faced fair dice is thrown until $2$ comes, then the probability that $2$ comes in even number of trials is (dice having six faces numbered $1, 2, 3, 4, 5$ and $6$)
If $a_n=\frac{-2}{4 n^2-16 n+15}$, then $a_1+a_2+\ldots \ldots+a_{25}$ is equal to :
If $P(n,r) = 1680$ and $C(n,r) = 70$, then $69n + r! = $