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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The value of `int_0^oox/((1+x)(x^2+1))dx` isA. `2pi`B. `pi/4`C. `pi/16`D. `pi/32` |
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Answer» Correct Answer - B Let `l = int_(0)^(oo) (x dx)/((1+x)(x^(2)+1))` By Partial fraction, `(x)/((1+x)(x^(2)+1))=(A)/((1+x))+(Bx+C)/((x^(2)+1))` `rArr x=A(x^(2)+1)+(1+x)(Bx+C)` `rArr x=A(x^(2)+1)+(Bx+Bx^(2)+C+Cx)` `rArr x =(A+B)x^(2)+(B+C)x +(A+C)` On comparing both sodes, we get `A+B=0, B+C=1, A+C=0` ...(i) On adding all these equations, we get `A+B+C=(1)/(2)` ...(ii) `therefore A=(1)/(2)-1=-(1)/(2),C=(1)/(2)` and `B=(1)/(2)` Then, `l=int_(0)^(oo){(-1)/(2(1+x))+(1)/(2)((x+1))/((x^(2)+1))}dx` `=-(1)/(2)int_(0)^(oo)(dx)/(1+x)+(1)/(2)int_(0)^(oo)(x)/(x^(2)+1)dx +(1)/(2)int_(0)^(oo)(dx)/(1+x^(2))` `=-(1)/(2)[log(1+x)]_(0)^(oo)+(1)/(4)[log(x^(2)+1)]_(0)^(oo)+(1)/(2)xx(pi)/(2)` `=-(1)/(2)lim_(x to oo)log (1+x)+(1)/(4) lim_(x to oo)log (1+x^(2))+(pi)/(4)` `= lim_(x to oo)log [((1+x^(2))^(1//4))/((1+x)^(1//2))]+(pi)/(4)` `= lim_(x to oo)log [(sqrt(x)((1)/(x^(2))+1)^(1//4))/(sqrt(x)((1)/(x)+1)^(1//2))]+(pi)/(4)` `= log. ((0+1)^(1//4))/((0+1)^(1//2))+(pi)/(4)` `= log (1) +(pi)/(4)=0+(pi)/(4)=(pi)/(4)` |
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| 2. |
` int _(0)^(a) sqrt(a^(2) - x^(2))` dx is equal toA. `pia^(2)`B. `1/2 pia^(2)`C. `1/3 pia^(2)`D. `1/4 pia^(2)` |
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Answer» Correct Answer - D `int_(0)^(a)sqrt(a^(2)-x^(2))dx=[(x)/(2)sqrt(a^(2)-x^(2))+(a^(2))/(2)sin^(-1)((x)/(a))]_(0)^(a)` `=[0+(a^(2))/(2)sin^(-1)(1)-0-(a^(2))/(2)sin^(-1)(0)]` `(a^(2))/(2).(pi)/(2)-0=(a^(2)pi)/(4)` |
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| 3. |
` int _(0)^(pi//3) (cos x + sin x)/(sqrt(1+sin 2x))dx ` is equal toA. `(4pi)/(3)`B. `(2pi)/3`C. `pi`D. `pi/3` |
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Answer» Correct Answer - D Let `l=int_(0)^(pi//3)(cosx+sinx)/(sqrt(1+2sinxcosx))dx` `=int_(0)^(pi//3)(cosx+sinx)/(sqrt(cos^(2)x+sin^(2)x+2sinxcosx))dx` `=int_(0)^(pi//3)(cosx+sinx)/(sqrt((cosx+sinx)^(2)))dx` `=int_(0)^(pi//3)(cosx+sinx)/(cosx+sinx)dx=int_(0)^(pi//3)1dx=(pi)/(3)` |
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| 4. |
The value of ` int _(0)^(pi//2) ((sin x + cos x)^(2))/(sqrt(1+sin 2x) dx` is |
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Answer» Correct Answer - C Let `l=int_(0)^(pi//2)((sinx+cosx)^(2))/(sqrt(sin^(2)x+cos^(2)x+2sinxcosx))dx` `=int_(0)^(pi//2)((sinx+cosx)^(2))/(sqrt((sinx+cosx)^(2)))dx` `=int_(0)^(pi//2)(sinx+cosx)dx` `=[-cosx+sinx]_(0)^(pi//2)=-0+1+1-0=2` |
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| 5. |
` int _(-3pi//2)^(-pi//2) [ ( x + pi)^(3) + cos^(2) x ] ` dx is equalt toA. `((pi^(4))/32)+(pi/2)`B. ` (pi/2)`C. ` (pi/4)-1`D. ` (pi^(4))/32` |
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Answer» Correct Answer - B Let ` l = int_(-3pi//2)^(-pi//2)[(x+pi)^(3)+cos^(2)x]dx` … (i) `rArrl = int_(-3pi//2)^(-pi//2)` `[(-(pi)/(2)-(3pi)/(2)-x+pi)^(3)+cos^(2)(-(pi)/(2)-(3pi)/(2)-x)]dx` `rArr l = int_(-3pi//2)^(-pi//2)[-(x+pi)^(3)+cos^(2)x]dx` … (ii) On adding Eqs . (i) and (ii) , we get `2l=int_(-3pi//2)^(-pi//2)2cos^(2)x dx=int_(-3pi//2)^(-pi//2)(1+cos2x)dx` `=[x+(sin2x)/(2)]_(-3pi//2)^(-pi//2)` `=[-(pi)/(2)+(sin(-pi))/(2)-(-3pi)/(2)+(sin(-3pi))/(2)]` `rArrl=(pi)/(2)` |
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| 6. |
The value of ` int _(0)^(pi) (dx)/(5+4 cos x ) ` isA. `2pi`B. `(3pi)/2`C. `(5pi)/4`D. `pi/3` |
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Answer» Correct Answer - D Put tan `(x)/(2)=t` `therefore " " dx=(2dt)/(1+t^(2))` and `cos x = (1-t^(2))/(1+t^(2))` `therefore " " int_(0)^(pi)(dx)/(5+4 cos x)=int_(0)^(oo)(2dt)/(9+t^(2))` `= [(2)/(3)tan^(-1)((t)/(3))]_(0)^(oo)=(2)/(3)(tan^(-1)oo-0)=(pi)/(3)`. |
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| 7. |
If `int _(0)^(pi//2) sin^(6) dx = (5pi)/32 ,` then the value of ` int _(-pi)^(pi) (sin ^(6) x + cos^(6)x)dx` isA. `(5pi)/8`B. `(5pi)/16`C. `(5pi)/2`D. `(5pi)/4` |
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Answer» Correct Answer - D Let `int_(-pi)^(pi)(sin^(6)x+cos^(6)x)dx` `=2int_(0)^(pi)(sin^(6)x+cos^(6)x)dx` `=4int_(0)^(pi//2)sin^(6)x dx+4int_(0)^(pi//2)cos^(6)((pi)/(2)-x)dx` `8int_(0)^(pi//2)sin^(6)xdx=8xx(5pi)/(32)=(5pi)/(4)" [given]"` |
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| 8. |
The value of `int_0^1 (x^4+1)/(x^2+1)dx` isA. `(1)/(6)(3pi-4)`B. `(1)/(6)(3-4pi)`C. `(1)/(6)(3pi+4)`D. `(1)/(6)(3+4pi)` |
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Answer» Correct Answer - A `int_(0)^(1)(x^(4)+1)/(x^(2)+1)dx=int_(0)^(1)(x^(4)-1)/(x^(2)+1)dx+2int_(0)^(1)(dx)/(1+x^(2))` `int_(0)^(1)(x^(2)-1)dx+2int_(0)^(1)(dx)/(1+x^(2))` `=[(x^(3))/(3)-x]_(0)^(1)+[2tan^(-1)x]_(0)^(1)` `=-(2)/(3)+(pi)/(2)=(3pi-4)/(6)` |
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| 9. |
`l=int_(-2)^(1)(tan^(-1)x+"cot"^(-1)(1)/(x))dx` is equal toA. ` (5pi)/2+ 4 tan ^(-1) 2 - In 5/2`B. ` (5pi)/2 - 4 tan^(-1) 2 + In 5/2`C. ` (5pi)/2 - 3 tan ^(-1) 2 - In 5/2`D. ` (5pi)/2 - 3 tan ^(-1) 2 + 5/2` |
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Answer» Correct Answer - B `l=int_(-2)^(1)(tan^(-1)+cot^(-1).(1)/(x))dx` `=int_(-2)^(0)(tan^(-1)x+pi+tan^(-1)x)dx+int_(0)^(1)(tan^(-1)x+tan^(-1)x)dx` `=int_(-2)^(0)2tan^(-1)xdx+int_(-2)^(0)pidx+int_(0)^(1)2tan^(1)xdx` `=2[x tan^(-1)x-(1)/(2)log(1+x^(2))]_(-2)^(0)+[pix]_(-2)^(0)+2[x tan^(-1)x-(1)/(2)log(1+x^(2))]_(0)^(1)` `=(5pi)/(2)-4tan^(-1)2+log.(5)/(2)` |
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| 10. |
`int_0^pi(x dx)/(a^2cos^2x+b^2sin^2x)`A. `pi/(2ab)`B. `pi/(ab)`C. `(pi^(2))/(2ab)`D. `(pi^(2))/(ab)` |
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Answer» Correct Answer - C Let `l=int_(0)^(pi)(xdx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)" ...(i)"` `rArr" "l=int_(0)^(pi)((pi-x)dx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)" ...(ii)"` On adding Eqs. (i) and (ii), we get `2l=2piint_(0)^(pi//2)(dx)/(a^(2)cos^(2)x+b^(2)sin 2x)` `rArr" "2l=2piint_(0)^(pi//2)(sec^(2)x)/(a^(2)+b^(2)tan^(@)x)` Now, put `tanx=t rArr dx=(dt)/(sec^(2)x)` `therefore" "2l=2piint_(0)^(oo)(dt)/(a^(2)+b^(2)t^(2))` `=(1)/(b^(2))xx2piint_(0)^(oo)(dt)/((a^(2))/(b^(2))+t^(2))` `=[(2pi)/(b^(2)).(b)/(a)tan^(-1).(bt)/(a)]_(0)^(oo)` `=(2pi)/(ab)[tan^(-1)oo-tan^(-1)0]=(2pi)/(ab)xx(pi)/(2)` `rArr" "l=(pi^(2))/(2ab)` |
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| 11. |
If `overset(a)underset(0)int(1)/(1+4x^(2))dx=(pi)/(8)`, then a equalsA. `(1)/(4)`B. `-(1)/(2)`C. `(3)/(2)`D. `(1)/(2)` |
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Answer» Correct Answer - D Given , ` int _(0)^(p) (dx)/(1+4x^(2))= pi/8` , ` LHS = int _(0)^(p) (dx)/(1+4x^(2)) = 1/4 int _(0)^(p) (dx)/(1/4+x^(2)) = 1/4 xx (1)/(1/2) [ tan^(-1)(x/(1/2))]_(0)^(p)` ` rArr 1/2 (tan^(-1) 2p ) = pi/8` ` rArr tan^(-1) 2p = pi/4 rArr 2p = 1` ` rArr p = 1/2` |
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| 12. |
`overset(-pi//4)underset(-pi//4)int e^(-x)sin x" dx"` isA. `-(1)/(2)e^((-pi)/(2))`B. `-(sqrt2)/(2)e^((-pi)/(4))`C. `-sqrt2(e^((-pi)/(4))+e^((-pi)/(4)))`D. 0 |
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Answer» Correct Answer - A `int_(-pi//4)^(pi//2)e^(-x)sin xdx=[(e^(-x))/(2)(-sin x -cos x)]_((-pi)/(4))^((pi)/(2))` `=(1)/(2)[e^(-x)(-sin x -cosx)]_(-pi//4)^(pi//2)` `=(1)/(2)[e^((-pi)/(2))(-1-0)-{e^((pi)/(2))((1)/(sqrt2)-(1)/(sqrt2))}]` `-(e^((-pi)/(2)))/(2)` |
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| 13. |
The value of `int_(0)^(oo) (dx)/((x^(2)+4)(x^(2)+9) )` isA. `(pi)/(60)`B. `(pi)/(20)`C. `(pi)/(40)`D. `(pi)/(80)` |
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Answer» Correct Answer - A Let ` l = int _(0)^(oo)(dx)/((x^(2)+ 4 )(x^(2) +9))` ` = 1/5(int _(0)^(oo) 1/((x^(2)+4)) dx - int _(0)^(oo) 1/((x^(2)+9)) dx )` , ` = 1/5 [ [ 1/2 tan^(-1). x/2 ]_(0)^(oo) - [1/3 tan^(-1) x/3 ]_(0)^(oo) ]` ` = 1/5 [ (1/2 .pi/2 -0) - (1/3 . pi/2 -0 )] = 1/5 (pi/4 - pi/6) = pi/60` |
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| 14. |
`int_(0)^(1//sqrt(2))(sin^(-1)x)/((1-x^(2))^(3//2))dx=?`A. `(1)/(2)(pi-log2)`B. `((pi)/(2)-2log2)`C. `((pi)/(4)-(1)/(2)log2)`D. none of these |
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Answer» Put `x=sint` and `dx=cost dt` `[x=0impliessint=0impliest=0]` and `[x=(1)/(sqrt(2))impliessint=(1)/(sqrt(2))impliest=(pi)/(4)]` `:. I=int_(0)^(pi//4)(tcost)/(cos^(3)t)dt=int_(0)^(pi//4)t sec^(2)t=[t tant]_(0)^(pi//4)-int_(0)^(pi//4)tant dt` `=(pi)/(4)+[logcost]_(0)^(pi//4)=(pi)/(4)+log.(1)/(sqrt(2))=((pi)/(4)-(1)/(2)log2)` |
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| 15. |
`int_(0)^(pi)|cosx|dx=?`A. `2`B. `(3)/(2)`C. `1`D. `0` |
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Answer» `I=int_(0)^(pi//2)cosxdx+int_(pi//2)^(pi)-cosxdx`. `=[sinx]_(0)^(pi//2)-[sinx]_(pi//2)^(pi)=1-(0-1)=2`. |
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| 16. |
The value of `alpha`, which satisfy `int_(pi/2) ^(alpha) sin x dx = sin2alpha (alpha in [0,2pi]` are equalA. `pi/2`B. `(3pi)/2`C. `(7pi)/6`D. All of these |
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Answer» Correct Answer - D ` int_(pi//2)^(alpha) sin x dx = [ -cos x ] _(pi//2)^(alpha) = - cos alpha = sin 2 alpha ` [ Given] ` :. - cos alpha = 2 sin alpha cos alpha ` ` rArr cos alpha (2 sin alpha +1)=0` ` rArr cos alpha = 0 and sin alpha = -1/2 ` Since , `alpha (0,2pi)` Hence , ` alpha = pi/2 , (3pi)/2, (7pi)/6` |
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| 17. |
`int_(pi/3)^(pi/2) sqrt(1+cosx)/(1-cosx)^(5/2) dx`A. `(3)/(64)`B. `-(3)/(64)`C. `(-3)/(64)`D. `-(1)/(128)` |
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Answer» Correct Answer - B ` int _(pi//3) ^(pi//2) (sqrt(1+cos x))/((1-cos x)^(5//2) ) dx ` ` int _(pi//3)^(pi//2) (sqrt(2 )cos. x/2)/(2^(5//2) sin^(5). x/2) dx = 1/4 int _(pi//3)^(pi//2) (cos.x/2)/(sin^(5).x/2)dx` Put `sin. x/2 = t` ` rArr = 1/2 cos. x/2 dx= dt` Also at ` x= pi/3 , t = sin. pi/6 = 1/2 ` and at ` x = pi/2 , t = sin. pi/4 = 1/(sqrt(2))` So integral ` = 1/2 int _(1//2) ^(1//sqrt(2))(dt)/(t^(5) ) = 1/2 [ (-1)/(4t^(4))]_(1//2)^(-1//sqrt(2))` ` = (-1)/8 [ 1/4 - 1/16 ] = (-3)/128` |
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| 18. |
`int_(pi)^(2pi)|sinx|dx=?`A. `0`B. `1`C. `2`D. none of these |
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Answer» `pi le x lt 2pi implies sin x le 0 implies |sinx|=-sinx` `:.I=int_(pi)^(2pi)-sinxdx=[cosx]_(pi)^(2pi)=2`. |
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| 19. |
`int_(0)^(2pi)|sinx|dx=?`A. `2`B. `4`C. `1`D. none of these |
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Answer» `I=int_(0)^(pi)sinxdx+int_(pi)^(2pi)-sinxdx`. `=[-cosx]_(0)^(pi)+[cosx]_(pi)^(2pi)=(2+2)=4`. |
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| 20. |
Let f and g be continuous functions on [ 0 , a] such that `f(x)=f(x)=f(a-x)andg(x)+g(a-x)=4`, then `int_(0)^(a) f(x)g(x) `dx is equal toA. `(a)/(2)`B. `(a)/(2)int_(0)^(a)f(x)dx`C. `int_(0)^(a)f(x)dx`D. `a int_(0)^(a)f(x)dx` |
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Answer» Correct Answer - B Let ` l = int_(0)^(a) f(x) . g(x) dx` ` = int_(0)^(a) f(a-x)g(a-x)dx= int_(0)^(a) f(x) {a-g(x)}dx` ` = a int _(0)^(a) f(x) dx - int _(0)^(a) f(x).g(x) dx` ` = a int _(0)^(a) f(x) dx-1 or l = a/2 int_(0)^(2) f(x) dx` |
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| 21. |
The value of ` l = int_(0)^(1) | x- 1/2| dx` isA. `(1)/(3)`B. `(1)/(4)`C. `(1)/(8)`D. 2 |
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Answer» Correct Answer - C Let ` l = int _(0)^(pi) x f (sin x) dx ` ` rarr l = int _(0)^(pi) (pi-x) f{sin (pi-x)}dx ` , ` = piu int _(0)^(pi) f(sin x) dx -x int_(0)^(pi) f(sin x) dx ` ` rArr l = pi int_(0)^(pi) f(sin x) dx -l` ` rArr 2l = 2pi int _(0)^(pi//2) f(sin x) dx ` ` rArr l = pi int_(0)^(pi//2) f(sin x) dx ` |
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| 22. |
`int_(0)^(pi)(1)/(1+sin x)dx` is equal toA. 1B. 2C. `-1`D. `-2` |
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Answer» Correct Answer - B Let ` l = int_(0)^(pi) 1/(1+sin x) dx` ` = int _(0)^(pi) 1/(1+(2tan.x/2)/(1+tan^(2).x/2))dx = int _(0)^(pi) (sec^(2)x/2)/((1+tan.x/2)^(2))dx ` Put ` tan x/2 = t` ` rArr 1/2 sec^(2). x/2 dx = dt` , Also ,at x = 0 , t=0 at `x = pi , t=alpha` ` :. l = int _(0)^(oo) (2dt)/(1+t)^(2) = [ 2/(1+t)]_(0)^(oo) = 2 ` |
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| 23. |
`int_0^(pi)sin^(2m)x cos^(2m+1)x dx=0` |
| Answer» Apply `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx` to get `I=-I`. | |
| 24. |
Show that `int_0^(pi/2) log(sin 2x) dx=-pi/2 (log2)`. |
| Answer» Putting `2x=t`, we get `I=(1)/(2)*int_(0)^(pi)log(sint)dt=int_(0)^(pi//2)lo(sint)dt`. | |
| 25. |
The value of `overset(pi//2)underset(-pi//2)int (x^(2)+x cosx+tan^(5)x+1)dx` is equal to |
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Answer» Correct Answer - C let ` l = int _(-pi//2)^(pi//2) (x^(3) + x cos x tan ^(5) x + 1) dx` Since ,`x^(3),xcos x and tan^(5)x` are odd functions , therefore `l = int _(-pi//2)^(pi//2) 1dx = [x] _(-pi//2)^(pi//2) = pi ` |
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| 26. |
Let `[x]` denote the greatest integer less than or equal to `x`. Then, `int_(0)^(1.5)[x]dx=?`A. `(1)/(2)`B. `(3)/(2)`C. `2`D. `3` |
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Answer» `{0 le x lt 1implies[x]=0}` and `{1 le x lt 1.5}implies[x]=1` `:. I=int_(0)^(1)[x]dx+int_(1)^(1.5)[x]dx=int_(0)^(1)0dx+int_(0)^(1.5)1*dx` `=0+[x]_(1)^(1.5)=(1.5-1)=0.5=(1)/(2)`. |
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| 27. |
` int _(0)^(1) xe^(x^(2))dx` is equal toA. `(1)/(2)(e+1)`B. `e-1`C. `(1)/(2)(e-1)`D. `(e^(2))/(2)` |
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Answer» Correct Answer - C Let ` I = int _(0)^(1) xe^(x^(2)) dx` Put ` x^(2) = t rArr 2x = (dt)/(dx)` ` rArr dx = (dt)/(2x)` ` :. l = int _(0)^(1) xe^(1) (dt)/(2x) = 1/2 int _(0)^(1) e^(t) dt ` ` = 1/2 [ e^(t)] _(0)^(1) = 1/2 [e^(1) - e^(0) ] = 1/2 [ e-1 ] ` |
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| 28. |
Evaluate : `(i) int_(0)^(2)e^(x//2)dx` `(ii) int_(2)^(4)(x)/((x^(2)+1))dx` `(iii) int_(0)^(1)cos^(-1)xdx` |
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Answer» `(i)` Put `(x)/(2)=t` so that `dx=2dt`. Also, `(x=0impliest=0)` and `(x=2=t=1)`. `:.int_(0)^(2)e^(x//2)dx=2int_(0)^(1)e^(t)dt=2[e^(t)]_(0)^(1)=2(e-1)`. `(ii)` Put `(x^(2)+1)=t` so that `xdx=(1)/(2)dt`. Also, `(x=2impliest=5)` and `(x=4impliest=17)`. `:.int_(2)^(4)(x)/((x^(2)+1))dx=(1)/(2)int_(5)^(17)(dt)/(t)=(1)/(2)[log|t|]_(5)^(17)=(1)/(2)(log17-log5)`. `(iii)` Put `x=cost` so that `dx=-sint dt`. Also, `(x=0impliest=(pi)/(2))` and `(x=1impliest=0)`. `:.int_(0)^(1)cos^(-1)xdx=-int_(pi//2)^(0)cos^(-1)(cost)sintdt=int_(0)^(pi//2)tsintdt` `=[t(-cost)]_(0)^(pi//2)-int_(0)^(pi//2)1*(-cost)dt` [integrating by parts] `=[sint]_(0)^(pi//2)=1`. `(iv)` Let `(2x+3)-=A*(d)/(dx)(5x^(2)+1)+B`. Then, `(2x+3)-=(10x)A+B`. Comparing the coefficients of like powers of `x`, we get `10A=2` or `A=(1)/(5)` and `B=3`. `:.(2x+3)=(1)/(5)(10x)+3`. So, `int_(0)^(1)((2x+3))/((5x^(2)+1))dx=int_(0)^(1)((1)/(5)(10x)+3)/((5x^(2)+1))dx` `=(1)/(5)(10x)/((5x^(2)+1))dx+3int_(0)^(1)(dx)/((5x^(2)+1))` `=(1)/(5)[log|5x^(2)+1|]_(0)^(1)+(3)/(5)int_(0)^(1)(dx)/(x^(2)+((1)/(sqrt(5)))^(2))` `=(1)/(5)log6+(3)/(5)*sqrt(5)[tan^(-1)(x)/((1//sqrt(5)))]_(0)^(1)` `=(1)/(5)log6+(3)/(sqrt(5))(tan^(-1)sqrt(5))`. |
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| 29. |
Evaluate : `(i) int_(1)^(3)(cos(logx))/(x)dx` `(ii) int_(0)^(pi//2)sqrt(cos theta)sin^(3)theta d theta` `(iii) int_(0)^(pi//2)(cosx)/((1+sinx)(2+sinx))dx` |
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Answer» `(i)` Put `logx=t` so that `(1)/(x)dx=dt`. Also, `(x=1impliest=log1=0)` and `(x=3impliest=log3)`. `:. Int_(1)^(3)(cos(logx))/(x)dx=int_(0)^(log3)costdt=[sint]_(0)^(log3)=sin(log3)`. `(ii)` Put `cos theta=t` so that `sin theta d theta=-dt`. Also, `(theta=0impliest=1)` and `(theta=(pi)/(2)impliest=0)` `:.int_(0)^(pi//2)sqrt(costheta)sin^(3)d theta=int_(0)^(pi//2)sqrt(costheta)*(1-cos^(2)theta)sin theta d theta` `=-int_(1)^(0)sqrt(t)(1-t^(2))dt=int_(0)^(1)(t^(1//2)-t^(5//2))dt` `=[(2)/(3)t^(3//2)-(2)/(7)t^(7//2)]_(0)^(1)=((2)/(3)-(2)/(7))=(8)/(21)`. `(iii)` Put `sinx=t` so that `cosx dx=dt`. Also, `(x=0impliest=0)` and `(x=(pi)/(2)impliest=1)`. `:.int_(0)^(pi//2)(cosx)/((1+sinx)(2+sinx))dx` `=int_(0)^(1)(dt)/((1+t)(2+t))` `=int_(0)^(1)[(1)/((1+t))-(1)/((2+t))]dt` [ by partial fractions] `=int_(0)^(1)(dt)/((1+t))-int_(0)^(1)(dt)/((2+t))` `=[log|1+t|]_(0)^(1)-[log|2+t|]_(0)^(1)` `=[(log2-log1)-(log3-log2)]=(2log2)-(log3)`. `(iv) int_(0)^(pi//2)(dx)/((1-2sinx))=int_(0)^(pi//2)(dx)/(1-2{(2tan(x//2))/(1+tan^(2)(x//2))})` `=int_(0)^(pi//2)(sec^(2)(x//2))/([1+tan^(2)(x//2)-4tan(x//2)])dx` `=2int_(0)^(1)(dt)/((1+t^(2)-4t))`, where `tan.(x)/(2)=t` `[{:(x=0impliest=0),(x=(pi)/(2)impliest=1):}]` `=2int_(0)^(1)(dt)/((t-2)^(2)-(sqrt(3))^(2))=2*(1)/(2sqrt(3))[log|(t-2sqrt(3))/(t-2+sqrt(3))|]_(0)^(1)` `=(1)/(sqrt(3))[log((sqrt(3)+1)/(sqrt(3)-1))-log((sqrt(3)+2)/(sqrt(3)-2))]`. `(v) int_(0)^(pi//2)(dx)/((3+2cosx))=int_(0)^(pi//2)(dx)/(3+2*[(1-tan^(2)(x//2))/(1+tan^(2)(x//2))])` `=int_(0)^(pi//2)(sec^(2)(x//2))/(tan^(2)(x//2)+5)dx` `=2int_(0)^(1)(dt)/(t^(2)+(sqrt(5))^(2))`, where `tan.(x)/(2)=t` `[{:(x=0impliest=0),(x=(pi)/(2)impliest=1):}]` `=2*(1)/(sqrt(5))[tan^(-1).(t)/(sqrt(5))]_(0)^(1)=(2)/(sqrt(5))tan^(-1).(1)/(sqrt(5))`. `(vi) int_(0)^(pi//2)(dx)/((4sin^(2)x+5cos^(2)x))=int_(0)^(pi//2)(sec^(2)x)/((4tan^(2)x+5))dx`. [dividing num. and denom. by `cos^(2)x`] `=int_(0)^(oo)(dt)/((4t^(2)+5))`, where `tanx=t` `[{:(x=0impliest=0),(x=(pi)/(2)impliest=oo):}]` `=(1)/(4)int_(0)^(oo)(dt)/(t^(2)+((sqrt(5))/(2))^(2))=(1)/(4)*(2)/(sqrt(5))[tan^(-1).(2)/(sqrt(5))]_(0)^(oo)` `=(1)/(2sqrt(5))[tan^(-1)(oo)-tan^(-1)(0)]` `=(1)/(2sqrt(5))((pi)/(2)-0)=(pi)/(4sqrt(5))`. |
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| 30. |
Evaluate: `int_0^1(x t a n^(-1)x)/((1+x^2)^(3//2))dx` |
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Answer» `(i)` Put `x=tantheta` so that `dx=sec^(2)theta d theta`. Clearly, `x=0impliestheta=0` and `x=1impliestheta=(pi//4)`. `:.int_(0)^(1)(xtan^(-1)x)/((1+x^(2))^(3//2))dx=int_(0)^(pi//4)(theta tan theta)/(sec^(3)theta)*sec^(2)theta d theta=int_(0)^(pi//4)theta sin d theta` `=[-thetacostheta]_(0)^(pi//4)-int_(0)^(pi//4)(-costheta)d theta`[integrating by parts] `=[-thetacostheta]_(0)^(pi//4)+[sintheta]_(0)^(pi//4)=-(pi)/(4)cos.(pi)/(4)+sin.(pi)/(4)` `=((-pi)/(4sqrt(2))+(1)/(sqrt(2)))=(4-pi)/(4sqrt(2))=(sqrt(2)(4-pi))/(8)`. `(ii)` Put `x=sin theta` so that `dx=costheta d theta` Clearly, `(x=0impliestheta=0)` and `(x=(1)/(sqrt(2))impliestheta=(pi)/(4))`. `:.int_(0)^(1//sqrt(2))(sin^(-1)x)/((1-x^(2))^(3//2))dx=int_(0)^(pi//4)(theta)/(cos^(3)theta)*costhetad theta=int_(0)^(pi//4)theta sec^(2)theta d theta` `=[theta tan theta]_(0)^(pi//4)-int_(0)^(pi//4)1*tan theta d theta`[integrating by parts] `=(pi)/(4)+[log(costheta)]_(0)^(pi//4)=(pi)/(4)+log(cos.(pi)/(4))` `=(pi)/(4)+log((1)/(sqrt(2)))=((pi)/(4)-(1)/(2)log2)`. |
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| 31. |
Evaluate `int_(0)^(pi//2)(x)/((sinx+cosx))dx`. |
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Answer» Let `I=int_(0)^(pi//2)(x)/((sinx+cosx))dx`……..`(i)` Then, `I=int_(0)^(pi//2)(((pi)/(2)-x))/(sin[(pi//2)-x]+cos[(pi//2)-x])dx` or `I=(((pi)/(2)-x))/((cosx+sinx))dx=int_(0)^(pi//2)(((pi)/(2)-x))/((sinx+cosx))dx`.......`(ii)` Adding `(i)` and `(ii)`, we get `2I=(pi)/(2)int_(0)^(pi//2)(dx)/((sinx+cosx))` `:.I=(pi)/(4)int_(0)^(pi//2)(dx)/((sinx+cosx))=(pi)/(4)int_(0)^(pi//2)(dx)/([(2tan(x//2))/(1+tan^(2)(x//2))+(1-tam^(2)(x//2))/(1+tan^(2)(x//2))])` `=(pi)/(4)int_(0)^(pi//2)(sec^(2)(x//2))/(1-tan^(2)(x//2)+2tan(x//2))dx` `=(pi)/(4)int_(0)^(1)(2dt)/((1-t^(2)+2t))`, where `t=tan.(x)/(2)` [`x=0impliest=0` and `x=(pi)/(2)impliest=1`] `=(pi)/(2)int_(0)^(1)(dt)/([(sqrt(2))^(2)-(t-1)^(2)])dt` `=(pi)/(2)*(1)/(2sqrt(2))log|(sqrt(2)+(t-1))/(sqrt(2)-(t-1))|_(0)^(1)=(pi)/(4sqrt(2))log|(sqrt(2)+1)/(sqrt(2)-1)|`. |
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| 32. |
`int_(0)^(pi//2)sqrt(1+cos2x)dx=?`A. `sqrt(2)`B. `(3)/(2)`C. `sqrt(3)`D. `2` |
| Answer» `I=int_(0)^(pi//2)sqrt(2cos^(2)x)dx=sqrt(2)int_(0)^(pi//2)cosxdx=sqrt(2)[sinx]_(0)^(pi//2)=sqrt(2)`. | |
| 33. |
Evaluate : `(i) int_(0)^(pi//2)xcosxdx` `(ii) int_(0)^(pi)cos2xlog sinx dx` `(iii) int_(1)^(2)(logx)/(x^(2))dx` `(iv) int_(0)^(pi//6)(2+3x^(2))cos3x dx` |
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Answer» `(i)` Integrating by parts, we get ` int_(0)^(pi//2)xcosxdx=[xsinx]_(0)^(pi//2)-int_(0)^(pi//2)1*sinxdx` `=(pi)/(2)+[cosx]_(0)^(pi//2)=((pi)/(2)-1)`. `(ii)` Integrating by parts, taking `log(sinx)` as the first function, we get ` int_(0)^(pi)cos2xlog sinx dx` `=[(logsinx)*(sin2x)/(2)]_(0)^(pi)-int_(0)^(pi)(cotx*(sin2x)/(2))dx` `=0-int_(0)^(pi)(cosx)/(sinx)*(2sinxcosx)/(2)dx=-int_(0)^(pi)cos^(2)xdx` `=-(1)/(2)int_(0)^(pi)2cos^(2)xdx=-(1)/(2)int_(0)^(pi)(1+cos2x)dx` `=-(1)/(2)*[x+(sin2x)/(2)]_(0)^(pi)=-(pi)/(2)`. `(iii)` Integrating by parts, taking `(logx)` as the first function, we get ` int_(1)^(2)(logx)/(x^(2))dx=int_(1)^(2)(logx)*x^(-2)dx` `=[(logx)(-(1)/(x))]_(1)^(2)-int_(1)^(2)(1)/(x)*(-(1)/(x))dx` `=[-(log2)/(2)+(log1)/(1)]+int_(1)^(2)(dx)/(x^(2))` `=(-log2)/(2)-[(1)/(x)]_(1)^(2)=(-log2)/(2)-{(1)/(2)-1}=((1-log2)/(2))`. `(iv) int_(0)^(pi//6)(2+3x^(2))cos3x dx` `=2int_(0)^(pi//6)cos3xdx+3int_(0)^(pi//6)x^(2)cos3xdx` `=2[(sin3x)/(3)]_(0)^(pi//6)+3{[x^(2)((sin3x)/(3))]_(0)^(pi//6)-int_(0)^(pi//6)2x((sin3x)/(3))dx}` [integrating by parts] `=(2)/(3)+(pi^(2))/(36)-2int_(0)^(pi//6)xsin3xdx` `=(2)/(3)+(pi^(2))/(36)-2{[x((-cos3x)/(3))]_(0)^(pi//6)-int_(0)^(pi//6)1*((-cos3x)/(3))dx}` [integrating by parts] `=(2)/(3)+(pi^(2))/(36)+(2)/(3)[xcos3x]_(0)^(pi//6)-(2)/(3)*[(sin3x)/(3)]_(0)^(pi//6)` `=(2)/(3)+(pi^(2))/(36)-(2)/(9)((pi^(2))/(36)+(4)/(9))=(1)/(36)(pi^(2)+16)`. |
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| 34. |
The value of ` int _(0)^(1) (x^(4) +1)/(x^(2)+1)dx ` isA. `1/6(3-4pi)`B. `1/6 (3pi+4)`C. `1/6(3+4pi)`D. `1/6(3pi-4)` |
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Answer» Correct Answer - D Let `l=int_(0)^(1)(x^(4)+1)/(x^(2)+1)dx=int_(0)^(1)(x^(4)+1-1+1)/(x^(2)+1)dx` `=int_(0)^(1)[(x^(4)-1)/(x^(2)+1)+(2)/(x^(2)+1)]dx` `=int_(0)^(1)(x^(2)-1+(2)/(x^(2)+1))dx` `=[(x^(3))/(3)-x+2tan^(-1)x]_(0)^(1)=[(1)/(3)-1+2tan^(-1)(1)-0]` `=-(2)/(3)+2*(pi)/(4)=(3pi-4)/(6)` |
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| 35. |
`int_(0)^(pi//2)(cosx)/((1+sin^(2)x))dx=?`A. `(pi)/(2)`B. `(pi)/(4)`C. `pi`D. none of these |
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Answer» Put `sinx=t` and `cosx dx=dt`. `[x=0impliest=0]` and `[x=(pi)/(2)impliest=1]`. `:.I=int_(0)^(1)(dt)/((1+t^(2)))=[tan^(-1)t]_(0)^(1)=(pi)/(4)` |
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| 36. |
`int_0^1(dx)/(1+x+x^2)`A. `(pi)/(sqrt(3))`B. `(pi)/(3)`C. `(pi)/(3sqrt(3))`D. none of these |
| Answer» `I=int_(0)^(1)(dx)/({(x^(2)+x+(1)/(4))+(3)/(4)})=int_(0)^(1)(dx)/({(x+(1)/(2))^(2)+((sqrt(3))/(2))^(2)})=[(2)/(sqrt(3))tan^(-1).((x+(1)/(2)))/(((sqrt(3))/(2)))]_(0)^(1)=(pi)/(3sqrt(3))` | |
| 37. |
`int_(0)^(pi//2)xcosxdx` `=?`A. `(pi)/(2)`B. `((pi)/(2)-1)`C. `((pi)/(2)+1)`D. none of these |
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Answer» Integrating by parts, we get : `I=int_(0)^(pi//2)xcosxdx=[xsinx]_(0)^(pi//2)-int_(0)^(pi//2)(1*sinx)dx=(pi)/(2)+[cosx]_(0)^(pi//2)-((pi)/(2)-1)`. |
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| 38. |
`int_(0)^(1)x^(3)sqrt(1+3x^(4))dx` |
| Answer» Correct Answer - Put `(1+3x^(4))=t`. | |
| 39. |
`int_0^9(dx)/(1+sqrt(x))` |
| Answer» Correct Answer - Put `x=t^(2)`. | |
| 40. |
`int_0^9(dx)/(1+sqrt(x))`A. `(3-2log2)`B. `(3+2log2)`C. `(6-2log4)`D. `(6+2log4)` |
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Answer» Put `x=t^(2)` and `dx=2t dt`. `[x=0impliest=0]` and `[x=9impliest^(2)=9impliest=3]` `:.I=int_(0)^(3)(2t)/((1+t))dt=int_(0)^(3){2-(2)/((1+t))}dt=[2t-2log(1+t)]_(0)^(3)=(6-2log4)`. |
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| 41. |
Evaluate:`int_0^1x(tan^(-1)x)^2dx` |
| Answer» Integrate by parth with `x` as the second function. | |
| 42. |
Evaluate:`int_0^1xsqrt((1-x^2)/(1+x^2))dx` |
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Answer» Put `x^(2)=t` and `x dx=(1)/(2)dt`. Then, `[x=0impliest=0]` and `[x=1impliest=1]`. |
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| 43. |
`int_(0)^(pi//2)sin^(2)xdx=?` |
| Answer» `sin^(2)x=(1)/(2)(1-cos2x)`. | |
| 44. |
`int_(-pi/2)^(pi/2)cosx dx`A. `0`B. `2`C. `-1`D. none of these |
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Answer» Correct Answer - B `f(x)=cosximpliesf(-x)=cos(-x)=cosx=f(x)`. `:.f(x)` is an even function. `:.I=2int_(0)^(pi//2)cosxdx=2[sinx]_(0)^(pi//2)=(2xx1)=2` |
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| 45. |
By using the properties of definiteintegrals, evaluate the integrals`int0a(sqrt(x))/(sqrt(x)+sqrt(a-x))dx`A. `(a)/(2)`B. `2a`C. `(2a)/(3)`D. `(sqrt(a))/(2)` |
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Answer» `I=int_(0)^(a)(sqrt(x))/((sqrt(x)+sqrt(a-x)))dx` and `I=int_(0)^(-a)(sqrt(a-x))/({sqrt(a-x)+sqrt(a-(a-x))})dx` `:. 2I=int_(a)^(0)((sqrt(x)+sqrt(a-x)))/((sqrt(x)-sqrt(a-x)))dx=int_(0)^(a)dx=[x]_(0)^(a)=aimpliesI=(a)/(2)`. |
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| 46. |
Evaluate the following definite integral: `int_1^4(x^2+x)/(sqrt(2x+1))dx` |
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Answer» Integrating by parts, taking `(x^(2)+x)` as the first function and `(1)/(sqrt(2x+1))` as the second function, we get `int_(2)^(4)((x^(2)+x))/(sqrt(2x+1))dx=[(x^(2)+x)*sqrt(2x+1)]_(2)^(4)-int_(2)^(4)(2x+1)*sqrt(2x+1)dx` `=(60-6sqrt(5))-int_(2)^(4)(2x+1)^(3//2)dx` `=(60-6sqrt(5))-(1)/(5)*[(2x+1)^(5//2)]_(2)^(4)` `=(60-6sqrt(5))-((243)/(5)-5sqrt(5))` `=((57)/(5)-sqrt(5))=sqrt((57-5sqrt(5))/(5))`. |
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| 47. |
`lim_(nto oo)+(1)/(sqrt(n^(2)+n))+(1)/(sqrt(n^(2)+2n))+...(1)/(sqrt(n^(2)+(n-1)n))` is equal toA. `2+2sqrt(2)`B. `2sqrt(2)-2`C. `2sqrt(2)`D. 2 |
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Answer» Correct Answer - B Given , `lim_(ntoinfty)[(1)/(n)+(1)/(sqrt(n^(2)+n))+..+(1)/(sqrt(n^(2)+(n-1)n))]` `=lim_(n to infty)[(1)/(n)+(1)/(nsqrt(1+(1)/(n)))+..+(1)/(nsqrt(1+((n-1))/(n)))]` `=(1)/(n)lim_(ntoinfty)[1+(1)/(sqrt(1)+(1)/(n))+..+(1)/(sqrt(1+((n-1))/(n)))]` `-lim_(ntoinfty)sum_(r=0)^(n-1)((1)/(sqrt(1+(r)/(n))))(1)/(n)` `=int_(0)^(1)(dx)/(sqrt(1+x))=[2sqrt(1+x)]_(0)^(1)=2sqrt(2)-2` |
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| 48. |
`lim_(n->oo) (1^p+2^p+3^p+...........+n^p)/n^(p+1)`A. `1/(p+1)`B. `1/(1-p)`C. ` 1/p - 1/(p-1)`D. `1/(p+2)` |
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Answer» Correct Answer - A Given , `lim_(ntoinfty)(1^(p)+2^(p)+3^(p)+...+n^(p))/(n^(p+1))=lim_(ntoinfty)sum_(r=1)^(n)[(r^(p))/(n^(p+1))]` `=lim_(ntoinfty)(1)/(n)sum_(r=1)^(n)((r)/(n))^(p)=int_(0)^(1)x^(p)dx=[(x^(p+1))/(p+1)]_(0)^(1)=(1)/(p+1)` |
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| 49. |
Evaluate the following limit: `lim_(nto oo)[(n!)/(n^(n))]^(1//n)`A. eB. `1/e`C. `pi/4`D. `4/pi` |
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Answer» Correct Answer - B Let `P=underset(nrarroo)(lim)[(n!)/(n^(n))]^(1//n)=underset(nrarroo)(lim)[(1)/(n).(2)/(n).(3)/(n).(4)/(n).(n)/(n)]^(1//n)` `therefore" "logP=underset(nrarroo)(lim)(1)/(n)[log.(1)/(n)+log.(2)/(n)+…+log.(n)/(n)]` `rArr" "logP=underset(nrarroo)(lim)sum_(=1)^(n)(1)/(n)log(r)/(n)` `rArr" "logP=int_(0)^(1)log x dx=x [xlogx-x]_(0)^(1)=-1` `rArr" "P=e^(-x)=(1)/(e)` |
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| 50. |
` int _(-pi//2)^(pi//2) (dx)/(1+cosx)` is equal to |
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Answer» Correct Answer - C Let `l=int_(-pi//2)^(pi//2)(dx)/(1+cosx)=int_(-pi//2)^(pi//2)(dx)/(2cos^(2).(x)/(2))` `=int_(0)^(pi//2)sec^(2).(x)/(2)dx=[2tan.(x)/(2)]_(0)^(pi//2)=2` |
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