Question
The value of $$I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\left( {\sin \,x + \cos \,x} \right)}^2}}}{{\sqrt {1 + \sin \,2x} }}dx} ,$$ is-
A.
$$3$$
B.
$$1$$
C.
$$2$$
D.
$$0$$
Answer :
$$2$$
Solution :
$$\eqalign{
& I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\left( {\sin \,x + \cos \,x} \right)}^2}}}{{\sqrt {1 + \sin \,2x} }}dx} \cr
& {\text{We know }}\left[ {{{\left( {\sin \,x + \cos \,x} \right)}^2} = 1 + \sin \,2x} \right],{\text{ so}} \cr
& I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\left( {\sin \,x + \cos \,x} \right)}^2}}}{{\left( {\sin \,x + \cos \,x} \right)}}dx} \cr
& {\text{or }}\,I = \int\limits_0^{\frac{\pi }{2}} {\left( {\sin \,x + \cos \,x} \right)dx} \cr
& \left[ {\because \,\sin \,x + \cos \,x > 0\,\,if\,0 < x < \frac{x}{2}} \right] \cr
& {\text{or }}\,I = \left[ { - \cos \,x + \sin \,x} \right]_0^{\frac{\pi }{2}} = 2 \cr} $$