\(\frac{sin^{-1}\sqrt{x}-cos^{-1}\sqrt{x}}{sin^{-1}\sqrt{x}+cos^{-1}\sqrt{x}}\),\(x∈[0,1]\)
Let I=∫sin-1√x-cos-1√x/sin-1√x+cos-1√xdx
It is known that,sin-1√x+cos-1√x=π/2
⇒I=∫(π/2-cos-1√x)-cos-1√x/π/2dx
=2π/π∫(/2-2cos-1√x)dx
=2π/π./2∫1.dx4//π∫cos-1√xdx
=x-4/π∫cos-1√xdx...(1)
Let I1=∫cos-1√xdx
Also,let √x=t⇒dt=2t dt
⇒I1=2∫cos-1t.tdt
=[2cos-1t.t2/2-∫-1/√1-t2.t2/2dt]
=t2cos-1t+∫t2/√1-t2dt
=t2cos-1t-∫1-t2-1/√1-t2dt
=t2cos-1t-∫√1-t2dt+∫1/√1-t2dt
=t2cos-1t-t/2√1-t2-1/2sin-1t+sin-1t
=t2cos-1t-t/2√1-t2+1/2sin-1t
From equation(1),we obtain
I=x-4/π[t2cost-1/t√1-t2+1/2sin-1t]
=x-4/π[xcos-1√x-√x/2√1-x+1/2sin-1√x]
=x-4π/π[x(/2-sin-1√x)-√xx2/2+/2sin-1√x]
=x-2x+4x/πsin-1√x+2/π√x-x2-2/πsin-1√x
=-x+2/π[(2x-1)sin-1√x]+2/π√x-x2+C
=2(2x-1)/πsin-1√x+2/π√x-x2-x+C
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Find the area of the region bounded by the curve y2=x and the lines x=1,x=4 and the x-axis
Find the area of the region bounded by y2=9x, x=2, x=4 and the x-axis in the first quadrant.
Find the area of the region bounded by x2=4y,y=2,y=4 and the x-axis in the first quadrant.
Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)