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Mere Teacher · Previous Year Questions
JEE Main Mathematics · Indefinite Integration · Set 8 (Q211-240)
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211.
For n∈N, If In=∫sinxsinnxdx=n−12sin(n−1)x+In−2 and ∫0πsinxsinnxdx=2kπ then k=
(1)(−1)n−1
(2)1−(−1)n
(3)(−1)n
(4)(−1)n+1
212.
The integral∫4x2−4x+6(2x−1)cos(2x−1)2+5dx is equal to (where c is a constant of integration)
(1)21sin(2x−1)2+5+c
(2)21cos(2x+1)2+5+c
(3)21cos(2x−1)2+5+c
(4)21sin(2x+1)2+5+c
213.
∫cosec(x−a)cosecxdx=
(1)sina1log[sin(x−a)cosecx]+C
(2)sina1log[sin(x−a)sinx]+C
(3)sina−1log|sinxcosec(x−a)|+C
(4)sina−1log[sin(x−a)sinx]+C
214.
∫sinx+cosxdx is equal to
(1)21log(tan(2x+8π))+C
(2)21log(tan(2x+4π))+C
(3)21log(tan(4x+2π))+C
(4)21log(tan(8x+2π))+C
215.
The integral∫(xsinx+cosxx)2dxis equal to (where C is a constant of integration)
(1)tanx−xsinx+cosxxsecx+C
(2)secx−xsinx+cosxxtanx+C
(3)tanx+xsinx+cosxxsecx+C
(4)secx+xsinx+cosxxtanx+C
216.
∫1+x1−xdx= (where C is a constant of integration.)
(1)21−x+cos−1x+x(1−x)
(2)−21−x+cos−1x+x(1−x)
(3)−21−x+cos−1x−x(1−x)
(4)−21−x−cos−1x+x(1−x)
217.
∫0π/43+sin2xsinx+cosxdx=
(1)log3
(2)41log3
(3)2log3
(4)21log3
218.
Let α∈(0,π/2)be fixed. If the integral∫tanx−tanαtanx+tanαdx=A(x)cos2α+B(x)sin2α+C Where C is a constant of integration, then the functions A(x) and B(x) are respectively:
(1)x+α and loge|sin(x+α)|
(2)x−α and loge|sin(x−α)|
(3)x-α and loge|cos(x−α)|
(4)x+α and loge|sin(x−α)
219.
If (x2+1)3x4+24x2+28=x2+1Ax+B+(x2+1)2Cx+D+(x2+1)3Ex+F then the value of A+B+C+D+E+F=
(1)21
(2)22
(3)28
(4)29
220.
∫sinxsin4xdx= (Where C is a constant of integration.)
(1)3sin3x+4sinx+C
(2)31sin3x−32sinx+C
(3)32sin3x+2sinx+C
(4)32sin3x−2sinx+C
221.
∫π/6π/3sin3x+cos3xsin3xdx=
(1)6π
(2)2π
(3)3π
(4)12π
222.
Let ∫3+tanx2−tanxdx=21(αx+loge|βsinx+γcosx|)+C, where C is the constant of integration. Then α+βγ is equal to :
(1)7
(2)4
(3)1
(4)3
223.
limn→∞(n+m1+n+2m1+n+3m1+…+n+nm1)=
(1)mloge(m)
(2)1+mloge(1+m)
(3)mloge(1+m)
(4)1−mloge(1+m)
224.
∫(x+1)3(x−1)exdx=
(1)x+1ex+C
(2)(x+1)2ex+C
(3)(x+1)3ex+C
(4)(x+1)x⋅ex+C
225.
There exist θ such that a>|secθ|, then ∫1+acosxdx=