Integral Calculus Question 143
Question: The value of $ \int_0^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}dt}+\int_0^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}dt} $ is
Options:
A) $ \pi $
B) $ \frac{\pi }{2} $
C) $ \frac{\pi }{4} $
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let $ I_1=\int_0^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}dt} $ Put $ t={{\sin }^{2}}u\Rightarrow dt=2\sin u\cos udu $
$ \Rightarrow dt=\sin 2udu $
$ \therefore ,I_1=\int_0^{x}{u\sin 2udu} $ Let $ I_2=\int_0^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t},dt} $ Put $ t={{\cos }^{2}}v\Rightarrow dt=-2\cos vsinvdv $
$ \Rightarrow dt=-\sin 2,vdv $
$ \therefore I_2=\int_{\frac{\pi }{2}}^{x}{v}(-sin2v)dv=-\int_{\frac{\pi }{2}}^{x}{v\sin 2,vdv} $ $ =-\int_{\frac{\pi }{2}}^{x}{u\sin 2udu} $ [change of variable]
$ \therefore I=I_1+I_2=\int_0^{x}{u\sin 2udu-\int_{\frac{\pi }{2}}^{x}{u\sin 2,udu}} $ $ =\int\limits_0^{\frac{\pi }{2}}{u\sin 2udu+\int\limits_{\frac{\pi }{2}}^{x}{u\sin 2udu-\int\limits_{\frac{\pi }{2}}^{x}{u\sin 2udu}}} $ $ =\int\limits_0^{\frac{\pi }{2}}{u\sin 2udu}=\frac{\pi }{4} $ [Integrate by parts]
BETA
