A) 2
B) 4
C) 6
D) 8
Correct Answer: D
Solution :
[d] \[I=\int\limits_{0}^{2\pi }{\sqrt{1+\sin \frac{x}{2}}.dx}\] \[=\int\limits_{0}^{2\pi }{\sqrt{{{\sin }^{2}}\frac{x}{4}+{{\cos }^{2}}\frac{x}{4}+2\sin \frac{x}{4}\cos \frac{x}{4}}.dx}\] \[=\int\limits_{0}^{2\pi }{\sqrt{{{\left( \sin \frac{x}{4}+\cos \frac{x}{4} \right)}^{2}}}.dx}\]\[=\int\limits_{0}^{2\pi }{\left( \sin \frac{x}{4}+\cos \frac{x}{4} \right).dx}\] \[=4\left( -\cos \frac{x}{4}+\sin \frac{x}{4} \right)_{0}^{2\pi }=4\left[ \left( -\cos \frac{\pi }{2}+\sin \frac{\pi }{2} \right)-(-\cos {{0}^{{}^\circ }}-{{\sin }^{{}^\circ }}) \right]\]\[=4\left( 1+1 \right)=4\times 2=8\] Hence, option [d] is correct.
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