Question
The value of $$\int_0^{{{\sin }^2}x} {{{\sin }^{ - 1}}\sqrt t } \,dt + \int_0^{{{\cos }^2}x} {{{\cos }^{ - 1}}\sqrt t } \,dt{\text{ is :}}$$
A.
$$\pi $$
B.
$$\frac{\pi }{2}$$
C.
$$\frac{\pi }{4}$$
D.
$$1$$
Answer :
$$\frac{\pi }{4}$$
Solution :
$$\eqalign{
& {\text{Let }}{I_1} = \int_0^{{{\sin }^2}x} {{{\sin }^{ - 1}}\sqrt t } \,dt \cr
& {\text{Put }}t = {\sin ^2}u \Rightarrow dt = 2\,\sin \,u\,\cos \,u\,du \Rightarrow dt = \sin \,2u\,du \cr
& \therefore \,{I_1} = \int_0^x {u\,\sin \,2u\,du} \cr
& {\text{Let }}{I_2} = \int_0^{{{\cos }^2}x} {{{\cos }^{ - 1}}\sqrt t } \,dt \cr
& {\text{Put }}t = {\cos ^2}v \Rightarrow dt = - 2\,\cos \,v\,\sin \,v\,dv \Rightarrow dt = - \sin \,2v\,dv \cr
& \therefore \,{I_2} = \int_{\frac{\pi }{2}}^x {v\left( { - \sin \,2v} \right)\,dv} \cr
& = - \int_{\frac{\pi }{2}}^x {v\,\sin \,2v\,dv} \cr
& = - \int_{\frac{\pi }{2}}^x {u\,\sin \,2u\,du\,\,\,\,\,\,\,\,\left[ {{\text{change of variable}}} \right]} \cr
& \therefore \,I = {I_1} + {I_2} \cr
& = \int_0^x {u\,\sin \,2u\,du} - \int_{\frac{\pi }{2}}^x {u\,\sin \,2u\,du} \cr
& = \int_0^{\frac{\pi }{2}} {u\,\sin \,2u\,du} + \int_{\frac{\pi }{2}}^x {u\,\sin \,2u\,du} - \int_{\frac{\pi }{2}}^x {u\,\sin \,2u\,du} \cr
& = \int_0^{\frac{\pi }{2}} {u\,\sin \,2u\,du} \cr
& = \frac{\pi }{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Integrate by parts}}} \right] \cr} $$