∫1 - x1 + xdx is equal to
cos-1x + 1 - xx - 2 + c
cos-1x - 1 - xx - 2 + c
cos-1x + 1 - xx + 2 + c
None of the above
A.
Let I = ∫1 - x1 + xdxPut x = cos2θ and dx = - 2sinθcosθdθ∴ I = ∫1 - cosθ1 + cosθ- 2sinθcosθdθ = - 2∫2sin2θ22cos2θ2 . sinθ2cosθ2cosθdθ = - 4∫sin2θ2cosθdθ = - 2∫1 - cosθcosθdθ = - 2∫cosθ - 1 + cos2θ2dθ = θ + 12sin2θ - 2sinθ = θ + sinθcosθ - 2sinθ = cos-1x + 1 - xx - 2 + c
∫logx + 1 + x21 + x2dx is equal to
12logx + 1 + x22 + c
logx + 1 + x22 + c
logx + 1 + x2 + c
∫sin8x - cos8x1 - 2sin2xcos2xdx is equal to
sin2x + c
- 12sin2x + c
12sin2x + c
- sin2x + c
∫dxsinx + sin2x is equal to
16log1 - cosx + 12loglog1 + cosx - 23log1 + 2cosx + c
6log1 - cosx + 2loglog1 + cosx - 23log1 + 2cosx + c
6log1 - cosx + 12loglog1 + cosx + 23log1 + 2cosx + c
∫fxg''x - f''xgxdx is equal to
fxg'x
f'(x)g(x) - f(x)g'(x)
f(x)g'(x) - f'(x)g(x)
f(x)g'(x) + f'(x)g(x)
Correct value of ∫0πsin4xdx is
8π3
2π3
4π3
3π8
∫0π2logsinxdx is equal to
- π2log2
πlog12
- π2log12
log2
∫0πcos3xdx is equal to
- 1
0
1π
1
Suppose f is such that f(- x) = - f(x), for every x and ∫01fxdx = 5, then ∫- 10ftdt is equal to
10
5
- 5
Withthehelp of Trapezoidal rule fornumerical integration and the following table
the value of ∫01fxdx is
0.35342
0.34375
0.34457
0.33334