Integration Using Substitution
Trending Questions
Q. The integral π/4∫π/6dxsin2x(tan5x+cot5x) equals :
- 110[π4−tan−1(19√3)]
- 15[π4−tan−1(13√3)]
- π40
- 120[tan−1(19√3)]
Q. Let f be a real-valued function and satisfies f(x)+f(y)=1x+1y ∀ x, y∈R−{0}. If 3∫2⎛⎝3(f(x))5−f(x)1−(f(x))4⎞⎠dx=12ln(2α3β), then
- α≤β
- α+β=17
- tan−1√α−β=sec−1(α5)
- (α−β) is not a prime number
Q. The value of ∫π015+4 cos xdx is
- π
- π2
- π3
- π4
Q. For x>0, let f(x)=x∫1logt1+tdt. Then f(x)+f(1x) is equal to
- 14logx2
- 14(logx)2
- logx
- 12(logx)2
Q. Let f(x)=sin−1x and g(x)=x2−x−22x2−x−6. If g(2)=limx→2g(x), then the domain of the function fog is :
- (−∞, −1]∪[2, ∞)
- (−∞, −2]∪[−43, ∞)
- (−∞, −2]∪[−1, ∞)
- (−∞, −2]∪[−32, ∞)
Q. If k∫ln21√ex−1 dx=π6, then k=lnp. The value of p+1 is
Q. For x>0, let f(x)=x∫1logt1+tdt. Then f(x)+f(1x) is equal to
- 14logx2
- 14(logx)2
- logx
- 12(logx)2
Q. If y=x3∫x21lntdt (where x∈R+−{1}), then the value of dydx is
- x(x−1)(lnx)−1
- x(x−1)(lnx)2
- x(x−1)(lnx)
- x(x−1)(lnx)−2
Q. Let f:[0, 27]→[13, 6] be a differentiable function such that f′(x)<0 ∀ x∈Df. If 27∫0xf′(x)dx=λ−33∫0x2f(x3)dx, then the minimum value of λ is
(Note: Df denotes the domain of the function)
(Note: Df denotes the domain of the function)