If ∫0∞log1 + x21 + x2dx = k∫01log1 + x1 + x2dx, then k is equal to
4
8
π
2π
∫0xsintdt, where x ∈ 2nπ, 2n + 1π, n ∈ N, is equal to
4n - 1 - cos(x)
4n - sin(x)
4n - cos(x)
4n + 1 - cos(x)
Let I1 = ∫01exdx1 + x and I2 = ∫01x2dxex32 - x3 Then, I1I2 is equal to
13e
3e
e3
∫52525 - x23x4dx is equal to
π3
2π3
π6
5π6
A.
I = ∫525 25 - x23x4dxLet x = 5sinθ ⇒ dx = 5cosθdθ∴ I = ∫π6π2 25 - 25sin2θ3x4sin4θ . 5cosθdθ
= ∫π6π253cos3θ . 5cosθ54sin4θ = ∫π6π2cot2θcsc2θ - 1dθ = ∫π6π2cot2θcsc2θdθ - ∫π6π2cot2θ = ∫π6π2cot2θcsc2θdθ - ∫π6π2csc2θ - 1dθ = - cot3θ3 + cotθ + θπ6π2 = - 0 + 0 + π2 - - 333 + 3 + π6 = π3
∫dxcosx + 3sinx equals
12logtanx2 + π12 + C
13logtanx2 - π12 + C
logtanx2 + π6 + C
12logtanx2 - π6 + C
∫0π2sin2x . logtanxdx
0
2
7
If ∆x = 1cosx1 - cosx1 + sinxcosx1 + sinx + cosxsinxsinx1, then ∫0π4∆xdx is equal to
14
12
- 14
∫02πsinx + sinxdx is equal to
1
The value of ∫02x2dx, where [.] is the greatest integer function, is
2 - 2
2 + 2
2 - 1
If l (m,n) = ∫01tm1 + tndt, then the expression for l (m, n) in terms of l (m + 1, n + 1) is
2nm + 1 - nm + 1 . l m + 1, n - 1
nm + 1 . l m + 1, n - 1
2nm + 1 + nm + 1 . l m + 1, n - 1
mn + 1 . l m + 1, n - 1