cos−1√t dt Differentiating on both sides by Leibnitz rule, f ' (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 ⇒ f (x) = Constant Now, we check the constant value of this integration ondifferent value of x. (i) At (x=
Ï€
4
) f (
Ï€
4
) =
1
2
∫
0
sin−1√t dt +
1
2
∫
0
cos−1√t dt =
1
2
∫
0
(si−√t + cos−√t) dt =
1
2
∫
0
Ï€
2
dt =
Ï€
2
(
1
2
−0) =
Ï€
4
(ii) At (x = 0) f (0) = 0 +
1
∫
0
cos−1√t dt Let t = cos2θ , dt = - sin 2θ sθ = -
0
∫
Ï€
2
θ . sin 2θ dθ = -
Ï€
2
∫
0
θ . sin 2θ dθ (Since
b
∫
a
f (x) dx =
a
∫
b
f (x) dx) = [−θ
cos2θ
2
+
1
4
sin2θ]0
Ï€
2
= [−
Ï€
2.
1
2
(−1)+0] =
Ï€
4
(iii) At (x=
Ï€
2
) f (
Ï€
2
) =
1
∫
0
sin−1√t dt + 0 Let t = sin2θ , dt = sin 2θ dθ =