Question
The value of $$\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^3}\,x}}{{\sin \,x + \cos \,x}}dx} $$ is :
A.
$$\frac{{\pi - 2}}{8}$$
B.
$$\frac{{\pi - 1}}{4}$$
C.
$$\frac{{\pi - 2}}{4}$$
D.
$$\frac{{\pi - 1}}{2}$$
Answer :
$$\frac{{\pi - 1}}{4}$$
Solution :
$${\text{Let, }}I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^3}\,x}}{{\sin \,x + \cos \,x}}} .....(1)$$
Use the property $$\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} $$
$$ \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}x\,dx}}{{\sin \,x + \cos \,x\,}}} .....(2)$$
Adding equations (1) and (2), we get
$$\eqalign{
& 2I = \int\limits_0^{\frac{\pi }{2}} {\left( {1 - \frac{1}{2}\sin \left( {2x} \right)} \right)dx} \cr
& \Rightarrow I = \frac{1}{2}\left[ {x + \frac{1}{4}\cos \,2x} \right]_0^{\frac{\pi }{2}} \cr
& \Rightarrow I = \frac{{\pi - 1}}{4} \cr} $$