Question

What is the value of the definite integration \[\int_0^\pi {\sqrt {\dfrac{{1 + \cos 2x}}{2}} dx} \]?
A. 0
B. 2
C. 1
D. -1

Answer 1

Hint: First we will apply double angle formula of cosine to simplify \[\sqrt {\dfrac{{1 + \cos 2x}}{2}} \]. Then we find the interval where cosine function positive and where it is negative. According to that we will rewrite the given integration as a sum of two integrations. Then solve the integration to get solution of the given integration.



Formula Used:Double angle formula:
\[1 + \cos 2x = 2{\cos ^2}x\]
Integration formula:
\[\int {\cos xdx} = \sin x + c\]



Complete step by step solution:The given definite integration is \[\int_0^\pi {\sqrt {\dfrac{{1 + \cos 2x}}{2}} dx} \].
Assume that, \[I = \int_0^\pi {\sqrt {\dfrac{{1 + \cos 2x}}{2}} dx} \]
Now applying double angle formula:
\[ \Rightarrow I = \int_0^\pi {\sqrt {\dfrac{{2{{\cos }^2}x}}{2}} dx} \]
Cancel out 2 from denominator and numerator:
\[ \Rightarrow I = \int_0^\pi {\left| {\cos x} \right|dx} \]
We know that in the first quadrant all trigonometry ratios are positive and in the second quadrant sine and cosec are positive.
This implies \[\left| {\cos x} \right| = \left\{ {\begin{array}{*{20}{c}}{\cos x}&{0 < x < \dfrac{\pi }{2}}\\{ - \cos x}&{\dfrac{\pi }{2} < x < \pi }\end{array}} \right.\]
Now rewrite the integration as a sum of two integrations. The limit of the first integration is 0 to \[\dfrac{\pi }{2}\] and limit of the second integration is \[\dfrac{\pi }{2}\] to \[\pi \].
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\cos xdx} + \int_{\dfrac{\pi }{2}}^\pi { - \cos xdx} \]
\[ \Rightarrow I = \left[ {\sin x} \right]_0^{\dfrac{\pi }{2}} + \left[ { - \sin x} \right]_{\dfrac{\pi }{2}}^\pi \]
\[ \Rightarrow I = \left[ {\sin \dfrac{\pi }{2} - \sin 0} \right] + \left[ { - \sin \pi + \sin \dfrac{\pi }{2}} \right]\]
\[ \Rightarrow I = \left[ {1 - 0} \right] + \left[ { - 0 + 1} \right]\]
\[ \Rightarrow I = 2\]



Option ‘B’ is correct



Note: Students often do mistake to integrate \[I = \int_0^\pi {\left| {\cos x} \right|dx} \]. They do not consider the cosine as absolute function. They wrote \[I = \int_0^\pi {\sqrt {\dfrac{{2{{\cos }^2}x}}{2}} dx} = \int_0^\pi {\cos xdx} \]. But \[\cos x\] is not over the interval \[\left[ {0,\pi } \right]\]. Thus we have to break the limit of the integration.