∫0πxdxa2cos2x + b2sin2xdx is equal to
π2ab
πab
π22ab
The differential equation for which sin-1(x) + sin-1(y) = c is given by
1 - x2dy + 1 - y2dx = 0
1 - x2dx + 1 - y2dy = 0
1 - x2dx - 1 - y2dy = 0
1 - x2dy - 1 - y2dx = 0
∫ex1 + sinx1 + cosxdx is equal to
exsec2x2 + c
extanx2 + c
exsecx2 + c
extanx + c
∫1 + sinx4dx is equal to
8sinx8 + cosx8 + C
8sinx8 - cosx8 + C
8cosx8 - sinx8 + C
18sinx8 - cosx8 + C
∫0∞xdx1 + x1 + x2 is equal to
π2
0
1
π4
D.
∫0π2xdx1 + x1 + x2Put x = tanθ⇒ dxdθ = sec2θ= ∫0π2tanθsec2θdθ1 + tanθ1 + tan2θ∵ For limit, x = 0 ⇒ θ = 0, x = ∞ ⇒ θ = π2= ∫0π2tanθsec2θdθ1 + tanθsec2θ= ∫π2tanθ1 + tanθ = ∫0π2sinθcosθ1 + sinθcosθdθ= ∫0π2sinθcosθ + sinθdθLet I = ∫0π2sinθcosθ + sinθdθ ...i
I = ∫0π2sinπ2 - θcosπ2 - θ + sinπ2 - θdθ ∵ ∫0afxdx = ∫0afa - xdx I = ∫0π2cosθsinθ + cosθdθ ...iiAdding Eqs. (i) and (ii), we get 2I = ∫0π2sinθcosθ + sinθdθ + ∫0π2cosθsinθ + cosθdθ = ∫0π2sinθ + cosθcosθ + sinθdθ = ∫0π2dθ⇒ 2I = θ0π2⇒ 2I = π2⇒ I = π4
If In = ∫logxndx, then In + nIn - 1 is equal to
xlogxn
nlogxn
logxn - 1
The area included between the parabolas x2 = 4y and y2 = 4x is (in square units)
43
13
163
83