If I1 = ∫sin-1xdx and I2 = ∫sin-11 - x2dx, then
I1 = I2
I2 = π2I1
I1 + I2 = π2x
I1 + I2 = π2
∫sinθ + cosθsin2θdθ is equal to
logcosθ - sinθ + sin2θ + c
logsinθ - cosθ + sin2θ + c
sin-1sinθ - cosθ + c
sin-1sinθ + cosθ + c
∫π6π3dx1 + tanx is equal to
π12
π2
π6
π4
A.
∫π6π3dx1 + tanx= ∫π6π3cosxsinx + cosxdx ...(i)= ∫π6π3cosπ2 - xsinπ2 - x + cosπ2 - xdx⇒ I = ∫π6π3sinxcosx + sinxdx ...(ii)On adding Eqs. (i) and (ii), we get 2I = ∫π6π31dx = xπ6π3 = π3 - π6 = π6⇒ I = π12
If f is a continuous function, then
∫- 22f(x)dx = ∫02f(x) - f(- x)dx
∫- 352f(x)dx = ∫- 610fx - 1dx
∫- 35fxdx = ∫- 44fx - 1dx
∫- 35fxdx = ∫- 26fx - 1dx
∫1 + sinx1 + cosxdx is equal to
xtanx2 + c
log1 + cosx + c
cotx2 + c
logx + sinx + c
∫cos3x . elogsinxdx is equal to
- sin4x4 + c
- cos4x4 + c
esinx4 + c
None of the above
The value of ∫0π2cos3x + 12cosx - 1dx is
2
1
12
0
The value of ∫01tan-12x - 11 + x - x2dx is
- 1
By the application of Simpson's one - third rule for numerical integration, with two subintervals, the value of ∫01dx1 + x is
1736
1725
2536
1724
∫xe - 1 + ex - 1xe + exdx is equal to
logxe + ex + c
elogxe + ex + c
1elogxe + ex + c