Question
If $$f\left( x \right) = \int_0^{\sin \,x} {{{\cos }^{ - 1}}t\,dt} + \int_0^{\cos \,x} {{{\sin }^{ - 1}}t\,dt,\,0 < x < \frac{\pi }{2},} $$ then $$f\left( {\frac{\pi }{4}} \right)$$ is :
A.
$$\frac{\pi }{{\sqrt 2 }}$$
B.
$$1 + \frac{\pi }{{2\sqrt 2 }}$$
C.
1
D.
none of these
Answer :
$$1 + \frac{\pi }{{2\sqrt 2 }}$$
Solution :
$$\eqalign{
& f\left( x \right) = \int_0^{\sin \,x} {\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}t} \right)\,dt} + \int_0^{\cos \,x} {\left( {\frac{\pi }{2} - {{\cos }^{ - 1}}t} \right)\,dt} \cr
& = \frac{\pi }{2}\left[ t \right]_0^{\sin \,x} - \int_0^{\sin \,x} {{{\sin }^{ - 1}}} t\,dt + \frac{\pi }{2}\left[ t \right]_0^{\cos \,x} - \int_0^{\cos \,x} {{{\cos }^{ - 1}}t\,dt} \cr
& = \frac{\pi }{2}\left( {\sin \,x + \cos \,x} \right) - \int_0^{\sin \,x} {\sin {\,^{ - 1}}t\,dt} - \int_0^{\cos \,x} {{{\cos }^{ - 1}}t\,dt} \cr
& \int_0^{\sin \,x} {{{\sin }^{ - 1}}t\,dt} = \int_0^x {\theta \cos \,\theta \,d\theta \,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{putting }}t = \sin \,\theta } \right)} \cr
& = \left[ {\theta \sin \,\theta } \right]_0^x - \int_0^x {\sin \,\theta \,d\theta } \cr
& = x\sin \,x + \left[ {\cos \,\theta } \right]_0^x \cr
& = x\sin \,x + \cos \,x - 1 \cr
& \int_0^{\cos \,x} {{{\cos }^{ - 1}}t\,dt} = \int_0^x { - \phi \sin \,\phi \,d\phi \,\,\,\,\,\,\,\,\,\left( {{\text{putting }}t = \cos \,\phi } \right)} \cr
& = \left[ {\phi \cos \,\phi } \right]_0^x - \int_0^x {\cos \phi \,d\phi } \cr
& = x\cos \,x - \left[ {\sin \,\phi } \right]_0^x \cr
& = x\cos \,x - \sin \,x \cr
& \therefore f\left( x \right) = \frac{\pi }{2}\left( {\sin \,x + \cos \,x} \right) - x\sin \,x - \cos \,x + 1 - x\cos \,x + \sin \,x \cr
& \therefore f\left( {\frac{\pi }{4}} \right) = \frac{\pi }{2}.\sqrt 2 - \frac{\pi }{4}.\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} + 1 - \frac{\pi }{4}.\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = 1 + \frac{\pi }{{2\sqrt 2 }} \cr} $$