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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.
| 51. |
Evaluate `int((sinx+cosx)dx)/(sqrt(3+sin2x)).` |
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Answer» `int((sinx+cosx)dx)/(sqrt(3+sin2x))=int((sinx+cosx)dx)/sqrt(4-(sinx-cosx)^(2))` `=int(dt)/(sqrt(4-t^(2))) " " ("Putting " sinx-cosx=t)` `="sin"^(-1)(t)/(2)+c` `="sin"^(-1)(sinx-cosx)/(2)+c` |
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| 52. |
Evaluate `int(dx)/(2+sinx+cosx).` |
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Answer» `I=int(dx)/(2+sinx+cosx)` `=int(dx)/(2+(2tanx//2)/(1+tan^(2)x//2)+(1-tan^(2)x//2)/(1+tan^(2)x//2))` `=int(1+tan^(2)x//2)/(tan^(2)x//2+2tanx//2+3)dx` `=int(sec^(2)x//2)/(tan^(2)x//2+2tanx//2+3)dx` `" put " "tan"(x)/(2)=t . :. (1)/(2)"sec"^(2)(x)/(2)dx=dt` `:. I=int(2dt)/(t^(2)+2t+3)` `=2int(dt)/((t+1)^(2)+2)` `=2int(1)/(sqrt(2))"tan"(t+1)/(sqrt(2))+c` `=sqrt(2)"tan"("tan"(x)/(2)+2)/(sqrt(2))+c` |
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| 53. |
` int sqrt(1+sinx)dx " is equal to "`A. ` -2sqrt(1-sinx)+C`B. `sin(x//2)+cos(x//2)+C`C. `cos(x//2)-sin(x//2)+C`D. `2sqrt(1-sinx)+C` |
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Answer» Correct Answer - A `I=int(sqrt(1+sinx)sqrt(1-sinx))/(sqrt(1-sinx))dx` `=int(cosx)/(sqrt(1-sinx))dx=-2sqrt(1-sinx)+C` |
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| 54. |
Evaluate: `int1/((x-3)sqrt(x+1)) dx` |
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Answer» Let `I=int(1)/((x-3)sqrt(x+1))dx` `" Let " x+1=t^(2) " or " dx=2t dt` `:. I=int(1)/((t^(2)-1-3))(2t)/(sqrt(t^(2)))dt` `=2int(dt)/(t^(2)-2^(2))=2xx(1)/(2(2))log|(t-2)/(t+2)|+C` `=(1)/(2)log|(sqrt(x+1)-2)/(sqrt(x+1)+2)|+C` |
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| 55. |
Evaluate `int(3sinx+2cosx)/(3cosx+2sinx)dx.` |
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Answer» We have `I=int(3sinx+2cosx)/(3cosx+2sinx)dx` Let `3sinx+2cosx=mu(d)/(dx)(3cosx+2sinx)+lambda(3cosx+2sinx)` `=mu(-3sinx+2cosx)+lambda(3cosx+2sinx)` Comparing the coefficients of `sin x` and `cosx` on both sides, we get `-3mu+2lambda=3 " and " 2mu+3lambda=2` ` " or " lambda=(12)/(13) " and " mu=-(5)/(13) ` `:. I=int(mu(-3sinx+2cosx)+lambda(3cosx+2sinx))/(3cosx+2sinx)dx` `=lambda int 1dx+mu int(-3sinx+2cosx)/(3cosx+2sinx)dx` `=lambda x +mu int(dt)/(t), " where " t=3cosx+2sinx` `=lambda x +mu log|t|+C` `=(12)/(13)x+(-5)/(13)log|3cosx+2sinx|+C` |
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| 56. |
Evaluate: `int((3sinx-2)cosx)/(5-cos^2x-4sinx) dx`A. `cosx+(1)/(2)cos2x-(1)/(3)cos3x+C`B. `cosx-(1)/(2)cos2x-(1)/(3)cos3x+C`C. `cosx+(1)/(2)cos2x+(1)/(3)cos3x+C`D. `cosx-(1)/(2)cos2x+(1)/(3)cos3x+C` |
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Answer» Correct Answer - B `I=int 4sinx "cos"(x)/(2)"cos"(3x)/(2)dx ` `=int 2sinx(cos2x+cosx)dx` `=int(sin 3x-sinx+sin2x)dx` `=cosx-(1)/(3)cos 3x-(1)/(2)cos2x+C` |
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| 57. |
`int(e^(cotx))/(sin^(2)x)("2 ln cosec x"+sin2x)dx`A. `2e^(cotx)ln |sinx|+c`B. `2e^(tanx)ln|sinx|+c`C. `2e^(cotx)ln|cosx|+c`D. `2e^(tanx)ln |cosx|+c` |
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Answer» Correct Answer - A `t=cotxrArrdt=-"cosec"^(2)xdx` `I=int(e^(cotx))/(sin^(2)x)(ln" cosec"^(2)x+(2tanx)/(1+tan^(2)x))dx` `=-inte^(t)(ln(1+t^(2))+((2)/(t))/(1+(1)/t^(2)))dt` `=-e^(t)ln|1+t^(2)|+c=-2e^(cotx)ln|"cosec x"|+c` `=2e^(cotx)ln|sinx|+c` |
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| 58. |
`int e^(x)((2 tanx)/(1+tanx)+cot^(2)(x+(pi)/(4)))dx` is equal toA. `e^(x)tan((pi)/(4)-x)+c`B. `e^(x)tan(x-(pi)/(4))+c`C. `e^(x)tan((3pi)/(4)-x)+c`D. none of these |
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Answer» Correct Answer - B Putting `x^(2)=t,` `int e^(x)((2 tanx)/(1+tanx)+tan^(2)(x-(pi)/(4)))dx` `=int e^(x)(tan(x-(pi)/(4))+sec^(2)(x-(pi)/(4)))dx` `=e^(x)tan(x-(pi)/(4))+C` |
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| 59. |
The value of `int("cosec x")/(cos^(2)(1+logtan.(x)/(2)))dx` isA. `-tan(1+logtan.(x)/(2))+c`B. `sec^(2)(1+logtan.(x)/(2))+c`C. `tan(1+log tan.(x)/(2))+c`D. `sin^(2)(1+logtan.(x)/(2))+c` |
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Answer» Correct Answer - C `I=int("cosec x")/((cos^2)(1+log tan.(x)/(2)))dx` `"Let "1+log tan.(x)/(2)=t` `therefore" "(1)/(sin x)dx=dt` `therefore" "I=int(dt)/(cos^(2)t)` `" "=tant+c` `" "=tan(1+log tan.(x)/(2))+c` |
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| 60. |
`int ((x+2)/(x+4))^2 e^x dx` is equal toA. `e^(x)((x)/(x+4))+c`B. `e^(x)((x+2)/(x+4))+c`C. `e^(x)((x-2)/(x+4))+c`D. `((2x e^(2))/(x+4))+c` |
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Answer» Correct Answer - A `I=int((x+2)/(x+4))^(2)e^(x) dx=int e^(x)[(x^(2)+4x+4)/((x+4)^(2))]dx` `=int e^(x)[(x(x+4))/((x+4)^(2))+(4)/((x+4)^(2))]dx` `=int e^(x)[(x)/(x+4)+(4)/((x+4)^(2))]dx` `=e^(x) ((x)/(x+4))+C` |
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| 61. |
If `int(dx)/(sqrt(sin^(3)xcos^(5)x))=a sqrt(cot x)+bsqrt(tan^(3)x)+c,` then |
| Answer» Correct Answer - 10 | |
| 62. |
`int(cosec^2x-2005)/cos^[2005]x.dx`A. `(cotx)/((cosx)^(2005))+c`B. `(tan x)/((cosx)^(2005))+c`C. `(-(tan x))/((cosx)^(2005))+c`D. none of these |
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Answer» Correct Answer - D `int("cosec"^(2)x-2005)/(cos^(2005)x)dx` `=int(cotx)^(-2005)"cosec"^(2)xdx-2005 int (dx)/(cos^(2005)x)` `=(cosx)^(-2005)(-cot x)-int (-2005)(cosx)^(-2006)(-sinx)(-cotx)dx-2005int(dx)/(cos^(2005)x)` `= -(cotx)/((cosx)^(2005))+C` |
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| 63. |
Evaluate: `inte^tantheta(sectheta-sintheta)d theta`A. `-e^(tan theta)sintheta+C`B. `e^(tan theta)sintheta+C`C. `e^(tan theta)sectheta+C`D. `e^(tan theta)costheta+C` |
| Answer» Correct Answer - d | |
| 64. |
What is `int(dx)/(x(x^7+1))` equal to?A. `(1)/(2) ln |(x^(7) -1)/(x^(7) + 1)| + c`B. `(1)/(7) ln |(x^(7) + 1)/(x^(7))| +_c`C. `ln |(x^(7) -1)/(7x)| +c`D. `(1)/(7) ln |(x^(7))/(x^(7) + 1)| + c` |
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Answer» Correct Answer - D `int (dx)/(x(x^(7) + 1)) = int (x^(6))/(x^(7)(x^(7) + 1)).dx` Let `x^(7) = t` `rArr 7x^(6).dx = dt rArr x^(6) dx = (dt)/(7)` Then `int (x^(6) dx)/(x^(7) (x^(7)+ 1)) = (1)/(7) int (dt)/(t(t + 1))` `= (1)/(7) [int (1)/(t) dt - int (1)/(t + 1) dt]` `= (1)/(7)[l n |t| - l n |t + 1|] + c` `= (1)/(7) l n |(t)/(t + 1)| + c` `= (1)/(7) l n |(x^(7))/(x^(7) + 1)| + c` |
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| 65. |
Evaluate `inte^(3logx)(x^(4)+1)^(-1)dx` |
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Answer» Correct Answer - `(1)/(4) log(x^(4)+1)+C` `inte^(3logx)(x^(4)+1)^(-1)dx=inte^(logx^(3))(dx)/(x^(4)+1)=int(x^(3)dx)/(x^(4)+1)` `=(1)/(4)int(4x^(3))/(x^(4)+1)dx` `=(1)/(4) log(x^(4)+1)+C` |
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| 66. |
`int((x^4-1)dx)/(x^2sqrt(x^4+x^2+1))`A. `sqrt((x^(4) + x^(2) + 1)/(x)) + c`B. `sqrt(x^(4) + 2 - (1)/(x^(2))) + c`C. `sqrt(x^(2) + (1)/(x^(2)) + 1) + c`D. `sqrt((x^(4) - x^(2) + 1)/(x)) + c` |
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Answer» Correct Answer - C Take option (a) `I_(1) = sqrt((x^(4) + x^(3) + 1)/(x)) + C` `(dI_(1))/(dx) = (d)/(dx) [(x^(3) + x^(2) + x^(-1))^(1//2) + C]` `(dI_(2))/(dx) = (1)/(2) (x^(3) + x^(2) + x^(-1))^(-1//2) (3x^(2) + 2x - x^(-2))` `= (1)/(2) [(3x^(2) + 2x - (1)/(x^(2)))/(sqrt(x^(3) + x^(2) + (1)/(x)))]` `(dI_(2))/(dx) = (1)/(2) [(3x^(4) + 2x^(3) - 1)/((x^(2))/(sqrtx) sqrt(x^(4) + x^(3) + 1))]` Take option (b) : `I_(2) = sqrt(x^(4) + 2 - (1)/(x^(2))) + C` `(dI_(2))/(dx) = (1)/(2) [x^(4) + 2 - x^(-2)]^(-1//2) [4x^(3) + 0 + 2x^(-3)]` `= (1)/(2) [(4x^(3) + (2)/(x^(3)))/(sqrt(x^(4) + 2 - (1)/(x^(2))))] = (2x^(6) + 1)/((x^(3))/(x) sqrt(x^(6) + 2x^(2) - 1))` `(dI_(2))/(dx) = (2x^(6) + 1)/(x^(2) sqrt(x^(6) + 2x - 1))` Take option (c) `I_(3) = sqrt(x^(2) + x^(-2) + 1) + C` `(dI_(3))/(dx) = (1)/(2) [x^(2) + x^(-2) + 1]^(-1//2) [2x - 2x^(-3) + 0]` `= (1)/(2) [(2x - (2)/(x^(3)))/(sqrt(x^(2) + (1)/(x^(2)) + 1))] = (1)/(2) [(2(x^(4) -1))/(x^(3) sqrt((x^(4) + 1 + x^(2))/(x^(2))))]` `(dI_(3))/(dx) = (x^(4) -1)/(x^(2) sqrt(x^(4) + x^(2) + 1))` |
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| 67. |
`int(e^(sinx))/(cos^2x)(xcos^3x-sinx) dx`A. `(x + sec x) e^(sin x) + c`B. `(x - sec x) e^(sin x) + c`C. `(x + tan x) e^(sin x) + c`D. `(x - tan x)e^(sin x) + c` |
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Answer» Correct Answer - B Let us differentiate all the options one by one to get the expression in the question whose integral is to be found. Here `xe^(sin x)` is the common term in all the options. So let us differentiate it first Let `l = xe^(sin x)` `rArr (dl)/(dx) = e^(sin x) [ x cos x + 1]` `rArr (dl)/(dx) = (e^(sin x))/(cos^(2)x) [x cos^(3) x + cos^(2) x]` Let `m = sec xe^(sin x)` `rArr (dm)/(dx) = sec xe^(sin x). cos x + e^(sin x) sec x tan x` `rArr (dm)/(dx) = e^(sin x) [1 + (sin x)/(cos^(2) x)]` `rArr (dm)/(dx) = (e^(sin x))/(cos^(2)x) [cos^(2) x + sin x]` Differentiation of option (a) is `= (e^(sin x))/(cos^(2) x) [x cos^(3) x + cos^(2) x + cos^(2) x + sin x]` `= (e^(sin x))/(cos^(2)x) [x cos^(3) x + 2 cos^(2) x + sin x]` Differentiation of option (b) is `= (e^(sin x))/(cos^(2) x) [x cos^(3) x + cos^(2) x - cos^(2) x - sin x]` `= (e^(sin x))/(cos^(2)x) [x cos^(3) x - sin x]` `:.` Option (b) is correct |
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| 68. |
`inte^(tanx)(sinx-secx)dx` is equal toA. `e^(tanx)cosx +C`B. `e^(tanx)sinx +C`C. `-e^(tanx)cosx +C`D. `e^(tanx)sec x +C` |
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Answer» Correct Answer - C `I=inte^(tanx)(sinx-secx)dx` `=int sinx e^(tanx)dx-int secx e^(tanx)dx` `=-e^(tanx)cosx+int cos x e^(tanx)sec^(2)x dx-int sec x e^(tanx) dx` `= -cosx e^(tanx) +C` |
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| 69. |
Evaluate: `inte^(x)(1-sinx)/(1+sinx)dx` |
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Answer» Given integral `=inte^(x)(1-2sinx/2cosx/2)/(2sin^(2)x/2)dx` `=inte^(x)(1/2"cosec")^(2)x/2-cosx/2dx=-e^(x)cotx/2+C` |
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| 70. |
`int(sin^(6)x + cos^(6)x)/(sin^(2)xcos^(2)x)dx` |
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Answer» Correct Answer - `secx-"cosec"x+C` `int(sin^(3)x+cos^(3)x)/(sin^(2)xcos^(2)x)dx=int(sin^(3)x)/(sin^(2)xcos^(2)x)dx+int(cos^(3)x)/(sin^(2)xcos^(2)x)dx` `=int(sinx)/(cos^(2)x)dx+int(cosx)/(sin^(2)x)dx` `=int tanxsecxdx+int cotx"cosec"xdx` `=secx-"cosec"x+C` |
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| 71. |
if `int 1/(1+sinx)dx = tan(x/2+a) + C` thenA. `a=-pi/4, C in R`B. `a=pi/4, C in R`C. `a=(5pi)/(4), C in R`D. `a=pi/3, C in R` |
| Answer» Correct Answer - A | |
| 72. |
If `inte^(sinx).(xcos^(3)x-sinx)/(cos^(2)x)dx=e^(sinx)f(x)+C`, such that f(0)`=-1` then `pi/3-f(pi/3)` is equal to: |
| Answer» Correct Answer - 2 | |
| 73. |
if `int (x-1)^2/(x^4+x^2+1)dx=1/sqrta tan^-1 ((x^2-1)/(xsqrt3))-b/sqrta tan^-1((2x^2+1)/3)+c` then `a^2+b^2` is |
| Answer» Correct Answer - 13 | |
| 74. |
If `f(x) =sqrt(x-1),g(x)=e^(x)` and `intlog(x)dx=Alog(x)+Btan^(-1)(logx)+C` then `A^(3)+B^(2)` equals: |
| Answer» Correct Answer - 12 | |
| 75. |
If `int(2sin2phi-cosphi)/(60cos^(2)phi-4sinphi)dphi=p" ln "|sin^(2)phi-4sinphi+5|+qtan^(-1)(sinphi-r)+C` then p+q+r equal to: |
| Answer» Correct Answer - 11 | |
| 76. |
The integral `intsqrt(cotx)e^(sqrt(sinx))sqrt(cosx)dx` equalsA. `(sqrt(tanx)e^(sqrt(sinx)))/(sqrt(cosx))+C`B. `2e^(sqrt(sinx))+C`C. `-(1)/(2)e^(sqrtsinx)+C`D. `(sqrt(cotx)e^(sqrt(sinx)))/(2sqrt(cosx))+C` |
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Answer» Correct Answer - B `e^(sqrt(sinx))=trArr(e^(sqrt(sinx)))/(2sqrtsinx)cosxdx=dt` `rArr" "e^(sqrt(sinx))sqrt(cotx)sqrt(cosx)dx=2dt` Hence `int2dt=2t+C=2e^(sqrt(sinx))+C` |
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| 77. |
Evaluate `int(sqrt(tanx))/(sinx cosx)dx` |
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Answer» Correct Answer - `2 sqrt(tanx)+C` `int(sqrt(tanx))/(sinx cosx)dx=int(sqrt(tanx)(1)/("cos"^(2)x))/((sinx cosx)/("cos"^(2)x))dx` `=int(sqrt(tanx)sec^(2)x)/(tanx)dx=int(tan x)^(-1//2)sec^(2)x dx` `=((tan x)^(1//2))/(1//2)+C=2 sqrt(tanx)+C` |
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| 78. |
Evaluate:`int1/((x^2-4)sqrt(x+1))dx` |
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Answer» Correct Answer - `(1)/(4sqrt(3))log|(sqrt(x+1)-sqrt(3))/(sqrt(x+1)+sqrt(3))|-(1)/(2)"tan"^(-1)sqrt(x+1)+C` `"Let " I=int(1)/((x^(2)-4)sqrt(x+1))dx` `"Putting " x+1=t^(2) " and " dx=2tdt, " we get " ` `I=int(2tdt)/([(t^(2)-1)^(2)-4]sqrt(t^(2)))` `=2int(dt)/((t^(2)-1-2)(t^(2)-1+2))` `=2int(dt)/((t^(2)-3)(t^(2)+1))` `=(2)/(4)int((1)/(t^(2)-3)-(1)/(t^(2)+1))dt` `=(1)/(4sqrt(3))log|(t-sqrt(3))/(t+sqrt(3))|-(1)/(2)"tan"^(-1)t+C, " where " t=sqrt(x+1)` |
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| 79. |
Evaluate:`int1/((x+1)sqrt(x^2+x+1))dx` |
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Answer» Correct Answer - `-log|(1)/(x+1)-(1)/(2)+(sqrt(x^(2)+x+1))/(x+1)|+C` `I=int(1)/((x+1)sqrt(x^(2)+x+1))dx` ` "Let " x+1=(1)/(t) " or " dx=-(1)/(t^(2))dt ` `:. I=int(1)/((1)/(t)sqrt(((1)/(t)-1)^(2)+((1)/(t)-1)+1))(-(1)/(t^(2)))dt` `=-int(dt)/(sqrt((1-t)^(2)+(t-t^(2))+t^(2)))` `=-int(dt)/(sqrt(t^(2)-t+1))` ` =-int(dt)/(sqrt((t-(1)/(2))^(2)+(3)/(4)))` `=-log|(t-(1)/(2))+sqrt(t^(2)-t+1)|+C` `=-log|(1)/(x+1)-(1)/(2)+(sqrt(x^(2)+x+1))/(x+1)|+C` |
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| 80. |
Evaluate:`int1/((x+1)sqrt(x^2-1))dx` |
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Answer» Correct Answer - `sqrt((x-1)/(x+1))+C` `" Let " I=int(1)/((x+1)sqrt(x^(2)-1))dx` ` " Putting " x+1=(1)/(t) " and " dx=-(1)/(t^(2))dt, " we get " ` ` I=int(1)/((1)/(t)sqrt(((1)/(t)-1)^(2)-1))(-(1)/(t^(2)))dt ` `=-int(dt)/(sqrt(1-2t))=-int(1-2t)^(-1//2)dt ` `=-((1-2t)^(1//2))/((-2)((1)/(2)))+C=sqrt(1-2t)+C ` `=sqrt(1-(2)/(x+1))+C=sqrt((x-1)/(x+1))+C` |
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| 81. |
Evaluate:`int1/(3+sin2x)dx` |
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Answer» Correct Answer - `(1)/(sqrt(8)) "tan"^(-1) (3tanx+1)/(sqrt(8))+c` `I=int(dx)/(3+sin2x)` `=int(dx)/(3+sinx cosx)` ` =int(sec^(2)xdx)/(3sec^(2)x+2tanx)" "("Dividing Nr. and Dr. by " cos^(2)x)` ` =int(sec^(2)xdx)/(3tan^(2)x+2tanx+3) ` ` =int(dt)/(3t^(2)+2t+3)" " ("Putting " tanx=t)` `=(1)/(3) int (dt)/(t^(2)+(2)/(3)t+1)` ` = (1)/(3) int (dt)/((t+(1)/(3))^(2)+(8)/(9))` ` = (1)/(3)* (1)/(sqrt(8)//3) "tan"^(-1) ((t+(1)/(3)))/((sqrt(8))/(3))+c ` ` =(1)/(sqrt(8)) "tan"^(-1) (3tanx+1)/(sqrt(8))+c` |
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| 82. |
`int(sqrt((cosx)/(x))-sqrt((x)/(cosx))sinx)dx` equalsA. `-sqrt(xcosx)+C`B. `sqrt(xsinx)+C`C. `2sqrt(xcos x)+C`D. `C-2sqrt(xcosx)` |
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Answer» Correct Answer - C `I=int(cosx-x sin x)/(sqrt(xcosx))dx` Put `xcosx=t^(2)` `rArr" "(cosx-x sin x)dx=2t dt` `rArr" "I=int(2tdt)/(t)=2t+C=2sqrt(x cos x)+C` |
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| 83. |
Evaluate the following: i) `int(xcosx-sinx)/(xsinx)` dx, ii) `int((x/(x+1)-In(x+1))/(x(In(x+1))))dx` |
| Answer» i) In`(sinx)/(x)+C` , ii) In`("ln"(x+1))/(x)+C` | |
| 84. |
Evaluate:`int1/(x^4-1)dx` |
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Answer» Correct Answer - `(1)/(4)log|(x-1)/(x+1)|-(1)/(2)"tan"^(-1)x+C` `int(1)/(x^(4)-1)dx=int(1)/((x^(2)+1)(x^(2)-1))dx` `=(1)/(2)int((1)/(x^(2)-1)-(1)/(x^(2)+1))dx ` `=(1)/(4)log|(x-1)/(x+1)|-(1)/(2)"tan"^(-1)x+C` |
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| 85. |
If `int(sqrt(4+x^2))/(x^6)dx = ((a+x^2)^(3/2).(x^2-b))/(120x^5)+C` then `a+b` equals to |
| Answer» Correct Answer - 10 | |
| 86. |
Integrate with respect to x: i) `x ln x` |
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Answer» i) `x^(2)/2lnx-x^(2)/4+C`, ii) `x^(2)/4-x/4sin2x-1/8cos2x+C` iii) `x^(2)/2tan^(-1)-x/2+1/2tan^(-1)x+C`, iv) `x(lnx-1)+C`, v) `(secxtanx)/(2) + 1/2ln |sec x + tanx|+C` , vi) `(x^(2)-1)e^(x^(2))+C`, vii) `xsin^(-1)sqrt(x)+(sqrt(x)sqrt(1-x))/(2)-1/2sin^(-1)sqrt(x)+C`, viii) `xtan^(-1)x-1/2ln(1+x^(2))-(tan^(-1)x)^(2)/(2)+C` ix) `e^(x)/2(sinx-cosx)+C`, x) `e^(x)tanx+C` |
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| 87. |
Integrate with respect to `x`: i) `1/(x^(2)+4)` |
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Answer» i) `1/2tan^(-1)x2+C` ii) `1/sqrt(5)tan^(-1)x/sqrt(5)+C`, iii) `1/2tan^(-1)(x+1)/(2)+C`, iv) `ln|x^(2)+3x+4|-4/sqrt(7)tan^(-1)(2x+3)/sqrt(7)+C`, v) `x-arctanx+"ln"sqrt(1+x^(2))/(x)+C`, vi) `ln|x+sqrt(x^(2)-4)|+C`, vii) `x/2sqrt(x^(2)+4+2ln|x+sqrt(x^(2)+4)|+C)` viii) `(x+1)/(2)sqrt(x^(2)+2x+5+2ln|x+1+sqrt(x^(2)+2x+5)|+C` , ix) `-(1-x-x^(2))^(3//2)/(3)-3/8(2x+1)sqrt(1-x-x^(2)) -15/16sin^(-1)(2x+1)/sqrt(5)+C`, X) `2/15(a^(3)+x^(3))^(5//2)-(2a^(3))/(9)(a^(3)+x^(3))^(3//4)+C` |
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| 88. |
Let F(x) be the primitive of `(3x+2)/sqrt(x-9)`w.r.t.x. If `F(10)=60` then the value of `F(13)` |
| Answer» Correct Answer - 12 | |
| 89. |
If `y=int1/(1+x^2)^(3/2)dx` and `y=0` when `x=0` , then value of y when `x=1` isA. `sqrt(2/3)`B. `sqrt(2)`C. `3sqrt(2)`D. `1/sqrt(2)` |
| Answer» Correct Answer - D | |
| 90. |
Find the antiderivative of `f(x) = In (In x) + (In x)^-2` whose graph passes through `(e, e)`. |
| Answer» `y=x[ln(lnx)-1/(lnx)]+2e` | |
| 91. |
The value of `int1/(cos^(4)x+sin^(5)x)`dx is equal toA. `tan^(-1)(tanx+cotx)+C`B. `-tan^(-1)(tanx+cotx)+C`C. `tan^(-1)(tanx-cotx)+C`D. `-tan^(-1)(tanx-cotx)+C` |
| Answer» Correct Answer - c | |
| 92. |
`int(2^x)/(sqrt(1-4^x))dx=ksin^(- 1)2^x+c`, then k =A. `ln 2`B. `1/2` ln 2C. `1/2`D. `1/(ln 2)` |
| Answer» Correct Answer - D | |
| 93. |
The value of `int(cos2x)/(sinx+cosx)^(2)` dx is equal toA. `-1/(sinx+cosx)+C`B. `ln(sinx+cosx)+C`C. `ln(sinx-cosx)+C`D. `ln(sinx+cosx)^(2)+C` |
| Answer» Correct Answer - B | |
| 94. |
`"If " f(x)=sqrt(x),g(x)=e^(x)-1, " and " int fog(x)dx=A fog(x)+B tan^(-1)(fog(x))+C, " then " A+B " is equal to "-.` |
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Answer» ` fog(x)=sqrt(e^(x)-1)` ` :. I=int sqrt(e^(x)-1)dx` `=int(2t^(2))/(t^(2)+1)dt, " where " sqrt(e^(x)-1)=t` `=2t-2tan^(-1)t+C` `=2sqrt(e^(x)-1)-2tan^(-1)(sqrt(e^(x)-1))+C` `=2fog(x)-2tan^(-1)(fog(x))+C` ` :. A+B=2+(-2)=0` |
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| 95. |
If `I_(n)=int cos^(n)x dx`. Prove that `I_(n)=(1)/(n)(cos^(n-1)x sinx)+((n-1)/(n))I_(n-2)`. |
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Answer» `I_(n)=int cos^(n)x dx` `=cos^(n-1)x intcosx dx+(n-1)int(sin^(2)x)cos^(n-2)x dx` `=(cos^(n-1) x sinx )+(n-1) int cos^(n-2)x(1-cos^(2)x)dx` `=(cos^(n-1) x sinx )+(n-1) int [cos^(n-2)x-cos^(n)x]dx` or `I_(n)+(n-1) I_(n)=(cos^(n-1)x sinx)+(n-1)(I_(n-2))` or `I_(n)=(1)/(n)(cos^(n-1) x sin x)+((n-1)/(n))I_(n-2)` |
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| 96. |
The value of `int(x e^ln(sinx)-cosx)dx` is equal toA. `xcosx+C`B. `sinx-xcosx+C`C. `-e^(ln x)cosx+C`D. `sinx+xcosx+C` |
| Answer» Correct Answer - C | |
| 97. |
Consider the following statements: `S_(1)`: The antiderivative of every function is an odd function. `S_(2)`: Primitive of `(3x^(4)-1)/(x^(4)-x+1)^(2)` w.r.t is `x/(x^(4)+x+1)` +C `S_(3)`: `int1/(sqrt(sinxcosx))dx = -2/sqrt(tanx)+C` `S_(4):` The value of `int(sqrt(a+x)/(a-x)-sqrt(a-x)/(a+x))` dx is equal to `-2sqrt(a^(2)-x^(2))+C` State, in order whether `S_(1),S_(2),S_(3),S_(4)` are true or falseA. FFTTB. TTTTC. FFFFD. TFTF |
| Answer» Correct Answer - a | |
| 98. |
If `I_(n)=int(sinx+cosx)^(n)`dx, snd `I_(n)=1/n(sinx+cosx)^(n-1)(sinx-cosx)+(2k)/(n) I_(n-2)` then k=A. (n+1)B. `(n-1)`C. `(2n+1)`D. `(2n-1)` |
| Answer» Correct Answer - b | |
| 99. |
`int(x+sqrt(x+1))/(x+2)` dx is equal to:A. `(x+1)-2sqrt(x+1)+2"ln"|x+2|=2tan^(-1)sqrt(x+1)+C`B. `(x+1)+2sqrt(x+2)-2"ln"|x+2|-2tan^(-1)sqrt(x+2)+C`C. `(x+1)+2sqrt(x+1)-2"ln"|x+2|-2tan^(-1)sqrt(x+1)+C`D. `(x=1)+2sqrt(x+2)-2"ln"|x+1|+2tan^(-1)sqrt(x+2)+C` |
| Answer» Correct Answer - c | |
| 100. |
If `int (xtan^(-1)x)/sqrt(1+x^2) dx = sqrt(1+x^2)f(x)+Aln|x+sqrt(x^2+1)|+c` thenA. `f(x) = tan^(-1)x, A=-1`B. `f(x) = tan^(-1)x,A=1`C. `f(x)=2tan^(-1)x, A=-1`D. `(x) = 2tan^(-1)x,A=1` |
| Answer» Correct Answer - a | |