Concept:

• \({\sin ^2}{\rm{x}} + {\cos ^2}{\rm{x}} = 1\)
• \(\sin 2{\rm{x}} = 2\sin {\rm{x}}\cos {\rm{x}}\)

Calculation:

Let, \({\rm{I}} = \mathop \smallint \limits_0^{1} \sqrt {1 + \sin \frac{{\rm{x}}}{2}} {\rm{dx}}\)

\(= \mathop \smallint \limits_0^1 \sqrt {{{\sin }^2}\left( {\frac{{\rm{x}}}{4}} \right) + {{\cos }^2}\left( {\frac{{\rm{x}}}{4}} \right) + \sin \frac{{\rm{x}}}{2}} {\rm{dx}}\)                  ( \({\sin ^2}\left( {\frac{{\rm{x}}}{4}} \right) + {\cos ^2}\left( {\frac{{\rm{x}}}{4}} \right)\) = 1)

\(= \mathop \smallint \limits_0^1 \sqrt {{{\sin }^2}\left( {\frac{{\rm{x}}}{4}} \right) + {{\cos }^2}\left( {\frac{{\rm{x}}}{4}} \right) + 2\sin \left( {\frac{{\rm{x}}}{4}} \right)\cos \left( {\frac{{\rm{x}}}{4}} \right)} {\rm{dx}}\)             \(\left( {\sin \frac{{\rm{x}}}{2} = 2\sin \left( {\frac{{\rm{x}}}{4}} \right)\cos \left( {\frac{{\rm{x}}}{4}} \right)} \right)\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^1 \sqrt {{{\left( {\sin \left( {\frac{{\rm{x}}}{4}} \right) + \cos \left( {\frac{{\rm{x}}}{4}} \right)} \right)}^2}} {\rm{dx}}\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^1\left| {\sin \left( {\frac{{\rm{x}}}{4}} \right) + \cos \left( {\frac{{\rm{x}}}{4}} \right)} \right|{\rm{dx}}\)

Let \(\frac{{\rm{x}}}{4} = {\rm{t}}\)

Differentiating with respect to x, we get

\(\Rightarrow \frac{{{\rm{dx}}}}{4} = {\rm{dt}}\)                      

\(\Rightarrow {\rm{dx}} = 4{\rm{dt}}\)

\(\therefore {\rm{I}} = 4\mathop \smallint \limits_0^1 \left| {\sin \left( {\rm{t}} \right) + \cos \left( {\rm{t}} \right)} \right|{\rm{dt}}\)

\(\Rightarrow {\rm{I}} = 4\left[ { - \cos {\rm{t}} + \sin {\rm{t}}} \right]_0^1\)

\(\rm = 4\left[ { - \cos {\rm{\frac x 4}} + \sin {\rm{\frac x 4}}} \right]_0^1\)

\(= 4\left[ {\left( { - \cos \left( {\frac{{\rm{1 }}}{4}} \right) + \sin \left( {\frac{{\rm{1 }}}{4}} \right)} \right) - \left( { - \cos 0 + \sin 0} \right)} \right]\)

\(= 4\left[ { 1 - \cos \left( {\frac{{\rm{1 }}}{4}} \right) + \sin \left( {\frac{{\rm{1 }}}{4}} \right)} \right]\)

Hence, option (4) is correct.