Property 1
Trending Questions
Q. For n>0, ∫2π0x sin2nxsin2nx+cos2nxdx=
- π2
- 4π2
- π2
- π
Q. ∫ba√[x−ab−x]dx=
- 2π(b−a)3
- π(b+a)2
- π(b−a)2
- π(b−a)3
Q. ∫10 tan−1x1+x2dx=
- π28
- π216
- π24
- π232
Q. Let f be a differentiable function from R to R such that |f(x)−f(y)|≤2|x−y|3/2, for all x, y∈R. If f(0)=1, then 1∫0f2(x)dx is equal to :
- 0
- 1
- 12
- 2
Q. ∫π20 cos x(1+sin x)(2+sin x)dx= [UPSEAT 1999]
- log43
- log13
- log34
- None of these
Q. ∫π20 3 sec x+5 cosec xsec x+cosec x dx=
- π
- 2 π
- 3 π
- π2
Q. Let F(x) = f(x) + f(1x), f(x)=∫x1logt1+tdt. Then F(e) equals
- 12
- \N
- 1
- 2
Q. ∫π2π4 cos θ cosec2θ dθ=
- √2−1
- 1−√2
- √2+1
- None of these
Q. ∫1√20sin−1x(1−x2)32dx=
- π4+12log 2
- π4−12log 2
- π2+log 2
- π2−log 2
Q. If I=∫baf(g(x)).g′(x)dx then on substituting g(x) = t where the equation g(x) = t is continuous in the interval [a, b] , I will be equal to -
- I=∫baf(t).dt
- I=∫g(b)g(a)f(t).dt
- I=∫g(a)g(b)f(t).dt
- None of these
Q. Evaluate ∫π/20sin−1(cosx)dx
- 0
- π24
- π28
- π2
Q. ∫π20√cos θsin3θ d θ=
- 2021
- 821
- −2021
- −821
Q.
∫π4−π4ex.sec2xdxe2x−1is equal to
0
2
e
2e
Q. ∫a0x4dx(a2+x2)4=
- 116 a3(π4−13)
- 116 a3(π4+13)
- 116a3(π4−13)
- 116a3(π4+13)
Q. ∫−π2−3π2[(x+π)3+cos2(x+π)]dx is equal to
- (π432)+(π2)
- π2
- −1
- π432
Q. The value of the integral I = ∫10x(1−x)n dx is
- 1n+1
- 1n+2
- 1n+1−1n+2
- 1n+1+1n+2
Q. Let ddxF(x)=(esin xx), x>0.If ∫413xesin x3dx=F(k)−F(1), then one of the possible values of k is
- 15
- 16
- 63
- 64
Q.
The value of ∫1−1(2|x|−|x|3)dx is
0
3/4
3/2
3
Q. ∫120x sin−1x√1−x2dx=
- 12+√3π12
- 12−√3π12
- 12−√3π12
- None of these
Q. ∫π40 sin x+cos x9+16 sin 2xdx=
- 120log 3
- log 3
- 120log 5
- None of these
Q. ∫π0tan xsec x+cos xdx=
- π3
- π4
- π2
- 2π