If $f(x)=\int\limits_{0}^{\sin^2x}\sin^{-1}\sqrt{t}\,dt$ and $f(x)=\int\limits_{0}^{\cos^2x}\cos^{-1}\sqrt{t}\,dt$, then the value of f(x)+ g(x) is
Answer & explanation
Correct answer: option 2
$f(x)+g'(x)=\sin^{-1}(\sin x)2\sin x \cos x -\cos^{-1}(\cos x)2\sin x \cos x$
$= x\sin 2x − x\sin 2x = 0$
for all x ∈ R
Hence f(x) + g(x) = constant = c (say)
Putting $x=\frac{π}{4},\,c=\int\limits_0^{1/2}\sin^{-1}\sqrt{t}\,dt+\int\limits_0^{1/2}\cos^{-1}\sqrt{t}\,dt=\int\limits_0^{1/2}\frac{π}{2}\,dt=\frac{π}{4}$
Hence $f(x)+g(x)=\frac{π}{4}$