InterviewSolution
Saved Bookmarks
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. |
`int (dx)/((4sin^(2)x+5cos^(2)x))=?` |
|
Answer» On dividing the numerator and denominator by `cos^(2)x` , we get `int(dx)/((4sin^(2)x+5cos^(2)x))=int(sec^(2)x)/((4tan^(2)x+5))dx` `=int(dt)/(4t^(2)+5)` [ putting tan x = t] `=(1)/(4)int(dt)/((t^(2)+(5)/(4)))=(1)/(4)*int(dt)/([t^(2)+(sqrt(5)/(2))^(2)])` `=(1)/(4)*(1)/((sqrt(5)/(2)))tan^(-1)(t)/((sqrt(5)/(2)))+C` `=(1)/(2sqrt(5))tan^(-1)((2t)/(sqrt(5)))+C=(1)/(2sqrt(5))tan^(-1)((2tanx)/(sqrt(5)))+C`. |
|
| 2. |
Evaluate `int(sqrt(tanx)+sqrt(cotx))dx`. |
|
Answer» We have `int(sqrt(tanx)+sqrt(cotx))dx` `=int(sqrt(tanx)+(1)/(sqrt(tanx)))dx=int((tanx+1))/(sqrt(tanx))dx` `=int((t^(2)+1))/(t)*(2t)/((1+t^(4)))dt " where tan" =t^(2)rArrx=tan^(-1)t^(2)rArrdx=(2t)/((1+t^(4)))dt` `=2int((t^(2)+1))/((t^(4)+1))dt=2int((1+(1)/(t^(2))))/((t^(2)+(1)/(t^(2))))dt` [ on dividing num . and denom by `t^(2)`] `=2int((1+(1)/(t^(2))))/((t-(1)/(t))^(2)+2)dt` `=2int(du)/((u^(2)+2))`, where ` (t-(1)/(t))=u and (1+(1)/(t^(2)))dt=du` `=2*(1)/(sqrt(2))"tan"^(-1)(u)/(sqrt(2))+C=sqrt(2)"tan"^(-1)((t-(1)/(t)))/(sqrt(2))+C` `[becauseu=(t-(1)/(t))]` `=sqrt(2)"tan"^(-1)((t^(2)-1))/((sqrt(2)t))+C=sqrt(2)tan^(-1)((tanx-1)/(sqrt(2tanx)))+C [ becauset^(2)=tanx]`. |
|
| 3. |
Evaluate `intsqrt(cotx)dx`. |
|
Answer» Put `cotx=t^(2) " so that " -" coscec "^(2)x dx=2tdtordx=(-2t)/((1+t^(4)))dt`. `thereforeintsqrt(cotx)dx=-int(2t^(2))/((t^(4)+1))dx` `=-int([(t^(2)+1)+(t^(2)-1)])/((t^(4)+1))dt=-int((t^(2)+1))/((t^(4)+1))dt-int((t^(2)-1))/((t^(4)+1))dt` `=- int((1+(1)/(t^(2))))/((t^(2)+(1)/(t^(2))))dt-int((1-(1)/(t^(2))))/((t^(2)+(1)/(t^(2))))dt` `=-int((1+(1)/(t^(2))))/([(t-(1)/(t))^(2)+2])dt-int((1-(1)/(t^(2))))/([(t+(1)/(t))^(2)-2])dt` `=-int(du)/([u^(2)+(sqrt(2))^(2)])-int(dv)/([v^(2)-(sqrt(2))^(2)])` `["putting"(t-(1)/(t))=u " in the 1 st integral , and "(t+(1)/(t))=v "in the 2nd"]` `=-(1)/(sqrt(2))tan^(-1)((u)/(sqrt(2)))-(1)/(2sqrt(2))log|(v-sqrt(2))/(v+sqrt(2))|+C` `=-(1)/(sqrt(2))tan^(-1)((t-(1)/(t))/(sqrt(2)))-(1)/(2sqrt(2))log|(t+(1)/(t)-sqrt(2))/(t+(1)/(t)+sqrt(2))|+C` `=-(1)/(sqrt(2))tan^(-1)((t^(2)-1)/(sqrt(2)t))-(1)/(2sqrt(2))log|(t^(2)-sqrt(2)t+1)/(t^(2)+sqrt(2)t+1)|+C` `=-(1)/(sqrt(2))tan^(-1)((cotx-1)/(sqrt(2cotx)))-(1)/(2sqrt2)log|(cotx-sqrt(2 cotx)+1)/(cotx+sqrt(2cotx)+1)|+C`. |
|
| 4. |
`int(x^(2))/((x^(2)+6x-3))dx` |
|
Answer» Correct Answer - `x-3log|x^(2)+6x-3|+(7sqrt(3))/(4)log|(x+3-2sqrt(3))/(x+3+2sqrt(3))|+C` On dividing `x^(2)` by `(x^(2)+6x-3)` , we get `I=int{1-(6x-3)/(x^(2)+6x-3)}dx=x-3int((2x-1))/((x^(2)+6x-3))dx`. Let `(2x-1)=A*(d)/(dx)(x^(2)+6x-3)+BrArr(2x-1)=A(2x+6)+B`. |
|
| 5. |
`int(dx)/((1+cos^(2)x))` |
|
Answer» Correct Answer - `(1)/(sqrt(2))tan^(-1)((tanx)/(sqrt(2)))+C` On dividing num and denom by `cos^(2)x` , we get `I=int(sec^(2)x)/((2+tan^(2)x))dx=int(dt)/((2+t^(2)))` , where tan x =t. |
|
| 6. |
`int(dx)/((2+sin^(2)x))` |
|
Answer» Correct Answer - `(1)/(sqrt(6))tan^(-1)((sqrt(3)tanx)/(sqrt(2)))+C` Divide num and denom by `cos^(2)x`. |
|
| 7. |
`int (dx)/((cos ^(2)x-3sin^(2)x))=?` |
|
Answer» Correct Answer - `(1)/(2sqrt(3))log|(1+sqrt(3)tanx)/(1-sqrt(3)tanx)|+C` Divide num and denom by `cos^(2)x` and put and x =t. |
|
| 8. |
Evaluate `int(x)/((x^(4)-x^(2)+1))dx`. |
|
Answer» Putting `x^(2)=t" and "2xdx=dt` we get `int(x)/((x^(4)-x^(2)+1))dx=(1)/(2)*int(dt)/((t^(2)-t+1))=(1)/(2)*int(dt)/({(t-(1)/(2))^(2)+((sqrt(3))/(2))^(2)})` `=(1)/(2)*(1)/(((sqrt(3))/(2)))" tan "^(-1)((t-(1)/(2)))/(((sqrt(3))/(2)))+C` `=(1)/(sqrt(3))*tan^(-1)((2t-1)/(sqrt(3)))+C=(1)/(sqrt(3))tan^(-1)((2x^(2)-1)/(sqrt(3)))+C`. |
|
| 9. |
Evaluate `int((2x+1))/((4-3x-x^(2)))dx`. |
|
Answer» Let `(2x+1)=A*(d)/(dx)(4-3x-x^(2))+B`. Then , `(2x+1)=A(-3-2x)+B` . . . (i) Comparing the coefficients of like terms , we get `(-2A=2and-3A+B=1)rArr(A=-1,B=-2)`. `thereforeint((2x+1))/((4-3x-x^(2)))dx = int{((-1)*(-3-2x)-2)/((4-3x-x^(2)))}dx` `=-int((3-2x))/((4-3x-x^(2)))dx-2int(dx)/((4-3x-x^(2)))` `=-log|4-3x-x^(2)|+2int(dx)/((4-3x-x^(2)))` `=-log|4-3x^(2)-x^(2)|+2int(dx)/((x+(2)/(2))^(2)-(4+(9)/(4)))` `=-log|4-3x-x^(2)|+2int(dx)/({(x+(3)/(2))^(2)-((5)/(2))^(2)})` `=-log|4-3x-x^(2)|+(2)/((2xx(5)/(2)))log|((x+(3)/(2))-(5)/(2))/((x+(3)/(2))+(5)/(2))|+C` `=-log|4-3x-x^(2)|+(2)/(5)log|(x-1)/(x+4)|+C`. |
|
| 10. |
Evaluate: `int1/(a^2sin^2x+b^2cos^2x) dx` |
|
Answer» Dividing the numerator and the denominator of the given integrand by `cos^(2)` x , we get `int(dx)/((a^(2)sin^(2)x+b^(2)cos^(2)x))=int(sec^(2)x)/(a^(2)tan^(2)x+b^(2))dx` `=int(dt)/((a^(2)t^(2)+b^(2)))` [ putting tan x = t] `=(1)/(a^(2))int(dt)/([t^(2)+((b)/(a))^(2)])=(1)/(a^(2))*(1)/((b/(a)))"tan"^(-1)(t)/((b/(a)))+C` `(1)/(ab)tan^(-1)((at)/(b))+C=(1)/(ab)tan^(-1)((atanx)/(b))+C`. |
|
| 11. |
Evaluate:`int(x^2+1)/(x^4+x^2+1)dx` |
|
Answer» We have `int((x^(2)-1))/((x^(4)+x^(2)+1))dx=int((1-(1)/(x^(2))))/((x^(2)+(1)/(x^(2))+1))dx` [ diving num . And denom . By `x^(2)` ] `=int((1-(1)/(x^(2))))/([(x+(1)/(x))^(2)-1])dx=int(dt)/((t^(2)-1))` `["Putting"(x+(1)/(x))=t and(1-(1)/(x^(2))dx=dt]` `=(1)/(2)log|(t-1)/(t+1)|+C=(1)/(2)log|(x+(1)/(x)-1)/(x+(1)/(x)+1)|+C` `=(1)/(2)log|(x^(2)-x+1)/(x^(2)+x+1)|+C`. |
|
| 12. |
`int1/(a^2cos^2 x+ b^2sin^2x)` dx |
|
Answer» Correct Answer - `(1)/(ab)tan^(-1)((b)/(a)tanx)+C` Divide num and denom by `cos^(2)x` and put and x =t. |
|
| 13. |
Evaluate:`int1/(2x^2+x-1)dx` |
|
Answer» We have `int(dx)/((2x^(2)+x-1))=(1)/(2)*int(dx)/((x^(2)+(1)/(2)x-(1)/(2)))` `=(1)/(2)int(dx)/([{x^(2)+(1)/(2)x+((1)/(2))^(2)}-(1)/(16)-(1)/(2)])=(1)/(2)int(dx)/([(x+(1)/(4))^(2)-((3)/(4))^(2)])` `=(1)/(2)*(1)/(2*(3)/(4))log|((x+(1)/(4))-(3)/(4))/((x+(1)/(4))+(3)/(4))|+C=(1)/(3)log|(2x-1)/(2(x+1))|+C` |
|
| 14. |
Evaluate: `int(x+1)/(sqrt(4+5x-x^2)) dx` |
|
Answer» Let `(x+1)=A*(d)/(dx)(4+5x-x^(2))+B. "Then", (x+1)=A(5-2x)+B`. Comparing the coefficients of like powers of x , we get `(-2A=1and5A+B=1)rArr(A=-(1)/(2)andB=(7)/(2))`. `therefore(x+1)=-(1)/(2)(5-2x)+(7)/(2)`. `thereforeint((x+1))/(sqrt(4+5x-x^(2)))dx=int(-(1)/(2)(5-2x)+(7)/(2))/(sqrt(4+5x-x^(2)))dx` `=-(1)/(2)int((5-2x))/(sqrt(4+5x-x^(2)))dx+(7)/(2)int(1)/(sqrt(4+5x-x^(2)))dx` `=-(1)/(2)int(1)/(sqrt(t))dt+(7)/(2)*int(1)/(sqrt(4-(x^(2)-5x)))dx " where "t=4+5x-x^(2)` `=-(1)/(2)*2sqrt(t)+(7)/(2)*int(dx)/(sqrt(4-(x^(2)-5x+(25)/(4))+(25)/(4)))` `=-sqrt(t)+(7)/(2)*int(dx)/(sqrt(((sqrt41)/(2))^(2)-(x-(5)/(2))^(2)))` `=-sqrt(4+5x-x^(2))+(7)/(2)*"sin"^(-1)((x-(5)/(2)))/(((sqrt(41))/(2)))+C` `=-sqrt(4+5x-x^(2))+(7)/(2)sin^(-1)((2x-5)/(sqrt(41)))+C`. |
|
| 15. |
Evaluate `int((3x+1))/((2x^(2)-2x+3))dx`. |
|
Answer» Let `(3x+1)=A*(d)/(dx)(2x^(2)-2x+3)+B`.Then , `(3x+1)=A(4x-2)+B` " " (i) Comparing the coefficients of like powers of x we get `(4A=3andB-2A=1)rArr(A=(3)/(4)andB=(5)/(2))`. `thereforeint((3x+1))/((2x^(2)-2x+3))dx=int(A*(4x-2)+B)/((2x^(2)-2x+3))` `=int((3)/(4)*(4x-2)+(5)/(2))/((2x^(2)-2x+3))dx=(3)/(4)*int((4x-2))/((2x^(2)-2x+3))dx+(5)/(2)int(dx)/(2(x^(2)-x+(3)/(2)))` `=(3)/(4)log|2x^(2)-2x+3|+(5)/(4)*int(dx)/({(x^(2)-x+(1)/(4))+(5)/(4)})` `=(3)/(4)log|2x^(2)-2x+3|+(5)/(4)*int(dx)/({(x-(1)/(2))^(2)+((sqrt(5))/(2))^(2)})` `=(3)/(4)log|2x^(2)-2x+3|+(5)/(4)*(1)/(((sqrt(5))/(2)))"tan"^(-1){((x-(1)/(2)))/(((sqrt(5))/(2)))}+C` `=(3)/(4)log|2x^(2)-2x+3|+(sqrt(5))/(2)tan^(-1)((2x-1)/(sqrt(5)))+C`. |
|
| 16. |
`int1/(3x^2+13x-10)dx` |
|
Answer» We have `(3x^(2)+13x-10)=3(x^(2)+(13)/(3)x-(10)/(3))` `=3{(x+(13)/(6))^(2)-(169)/(36)-(10)/(3)}=3{(x+(13)/(6))^(2)-(289)/(36)}` `=3{(x+(13)/(6))^(2)-((17)/(6))^(2)}`. `thereforeint(dx)/((3x^(2)+13x-10))=int(dx)/(3{(x+(13)/(6))^(2)-((17)/(6))^(2)})` `=(1)/(3)int(dt)/({t^(2)-((17)/(6))^(2)}) , "where" (x+(13)/(6))=t` `=(1)/(3)*(1)/((2xx(17)/(6)))log|(t-(17)/(6))/(t+(17)/(6))|+C[becauseint(dx)/((x^(2)-a^(2)))=(1)/(2a)log|(x-a)/(x+a)|]` `=(1)/(17)log|(6t-17)/(6t+17)|+C` `=(1)/(17)log|(6(x+(13)/(6))-17)/(6(x+(13)/(6))+17)|=(1)/(17)log|(6x-4)/(6x+30)|+C` `=(1)/(17){log(1)/(3)+log|(3x-2)/(x+5)|}+C` `=(1)/(17)log|(3x-2)/(x+5)|+k "where" (1)/(17)log(1)/(3)+C=k=` constant. |
|
| 17. |
Evaluate: `int(5x-2)/(1+2x+3x^2)dx` |
|
Answer» Let `(5x-2)=A*(d)/(dx)(1+2x+3x^(2))+B`. Then , `(5x-2)=A(6x+2)+B`. " " ...(i) Comparing the coefficients of like powers of x on both sides , we get `6A=5and2A+B=-2` This gives `A=(5)/(6)andB=(-11)/(3)`. `thereforeI=int({(5)/(6)(6x+2)-(11)/(3)})/((1+2x+3x^(2)))dx` `=(5)/(6)*int(6x+2)/((1+2x+3x^(2)))dx-(11)/(3)int(dx)/((3x^(2)+2x+1))` `=(5)/(6)log|1+2x+3x^(2)|-(11)/(9)*int(dx)/({(x+(1)/(3))^(2)+((1)/(3)-(1)/(9))})` `=(5)/(6)log|1+2x+3x^(2)|-(11)/(9)*int(dx)/({(x+(1)/(3))^(2)+((sqrt(2))/(3))^(2)})+C` `=(5)/(6)log|1+2x+3x^(2)|-(11)/(9)*(1)/(((sqrt(2))/(3)))tan^(-1){(x+(1)/(3))/(((sqrt(2))/(3)))}+C` `=(5)/(6)log|1+2x+3x^(2)|-(11)/(3sqrt(2))tan^(-1)((3x+1)/(sqrt(2)))+C`. |
|
| 18. |
Evaluate `int(dx)/((1+x-x^(2)))`. |
|
Answer» We have `int(dx)/((1+x-x^(2)))=-int(dx)/((x^(2)-x-1))` `=-int(dx)/({(x^(2)-x+(1)/(4))-(5)/(4)})=-int(dx)/({(x-(1)/(2))^(2)-((sqrt(5))/(2))^(2)})` `=int(dx)/({(sqrt(5)/(2))^(2)-(x-(1)/(2))^(2)})=int(dx)/({(sqrt(5)/(2))^(2)-u^(2)}) "where" (x-(1)/(2))=u` `=(1)/(2xx(sqrt(5)/(2)))*log|(sqrt(5)/(2)+u)/(sqrt(5)/(2)-u)|+C` `=(1)/(sqrt(5))log|(sqrt(5)+2u)/(sqrt(5)-2u)|+C=(1)/(sqrt(5))log|(sqrt(5)+2(x-(1)/(2)))/(sqrt(5)-2(x-(1)/(2)))|+C` `=(1)/(sqrt(5))log|(sqrt(5)+2x-1)/(sqrt(5)-2x+1)|+C=(1)/(sqrt(5))log|((sqrt(5)-1)+2x)/((sqrt(5)+1)-2x)|+C`. |
|
| 19. |
`int(dx)/(sqrt(15-8x^(2)))` |
|
Answer» We have `int(dx)/(sqrt(15-8x^(2)))=(1)/(sqrt(8))*int(dx)/(sqrt((15)/(8)-x^(2)))` `=(1)/(2sqrt(2))sin^(-1){(x)/(sqrt((15)/(8)))}+C=(1)/(2sqrt(2))sin^(-1)(sqrt((8)/(15))x)+C`. |
|
| 20. |
Evaluate `int(dx)/((2+cosx))`. |
|
Answer» We have `int(dx)/((2+cosx))=int(dx)/(1+(1+cosx))=int(dx)/(1+2cos^(2)(x//2))=int(sec^(2)(x//2)dx)/(sec^(2)(x//2)+2)` [ dividing the num . And denom . By `cos^(2)(x//2)`] `=int(sec^(2)(x//2))/(3+tan^(2)(x//2))dx=2int(dt)/(3+t^(2))`, where tan `(x//2)=` `=2*int(dt)/((sqrt(3))^(2)+t^(2))=2*(1)/(sqrt(3))"tan"^(-1)(t)/(sqrt(3))+C` `=(2)/(sqrt(3))tan^(-1)[(tan(x//2))/(sqrt(3))]+C`. |
|
| 21. |
Evaluate `int(dx)/((5-8x-x^(2)))`. |
|
Answer» We have `int(dx)/((5-8x-x^(2)))=-int(dx)/((x^(2)+8x-5))` `=-int(dx)/({(x^(2)+8x+16)-21})=-int(dx)/({(x+4)^(2)-(sqrt(21))^(2)})` `=int(dx)/({(sqrt(21))^(2)-(x+4)^(2)})=int(dt)/({(sqrt(21))^(2)-t^(2)}), "where" (x+4)=t` `=(1)/(2sqrt(21))*log|(sqrt(21)+t)/(sqrt(21)-t)|+C` `=(1)/(2sqrt(21))*log|(sqrt(21)+4+x)/(sqrt(21)-4-x)|+C`. |
|
| 22. |
Evaluate `int(cosx)/((sin^(2)x+4sinx+5))dx`. |
|
Answer» Putting `sinx=t " and "cosx dx=dt` we get `int(cosx)/((sin^(2)x+4sinx+5))dx=int(dt)/((t^(2)+4t+5))=int(dt)/({(t^(2)+4t+4)+1}}` `=int(dt)/({(t+2)^(2)+1^(2)}}=int(du)/((u^(2)+1)), "where "u=(t+2)` `=tan^(-1)u+C=tan^(-1)(t+2)+C` `=tan^(-1)(sinx+2)+C`. |
|
| 23. |
Evaluate : (i) `intsqrt(9-x^(2))dx` (ii) `intsqrt(1-4x^(2))dx` (iii) `intsqrt(16-9x^(2))dx` |
|
Answer» We know that `intsqrt(a^(2)-x^(2))dx=(x)/(2)sqrt(a^(2)-x^(2))+(a^(2))/(2)"sin"^(-1)(x)/(a)+C`. `therefore` (i) `intsqrt(9-x^(2))dx=intsqrt(3^(2)-x^(2))dx` `=(x)/(2)sqrt(9-x^(2))+(9)/(2)"sin"^(-1)(x)/(3)+C`. (ii) `sqrt(1-4x^(2))dx=2intsqrt(((1)/(4)-x^(2)))dx=2int{sqrt(((1)/(2))^(2)-x^(2))}dx` `=2[(x)/(2)sqrt((1)/(4)-x^(2))+(1)/(8)sin^(-1)((x)/((1//2)))]+C` `=(x)/(2)sqrt(1-4x^(2))+(1)/(4)sin^(-1)(2x)+C`. (iii) `intsqrt(16-9x^(2))dx=3int{sqrt(((16)/(9)-x^(2)))}dx=3int{sqrt(((4)/(3))^(2)-x^(2))}dx` `=3[(x)/(2)sqrt((16)/(9)-x^(2))+(8)/(9)"sin"^(-1)(x)/((4//3))]+C` `(x)/(2)sqrt(16-9x^(2))+(8)/(3)sin^(-1)((3x)/(4))+C`. |
|
| 24. |
Evaluate: `intsqrt(3-2x-2x^2) dx` |
|
Answer» We have `(3-2x-2x^(2))=2*{(3)/(2)-x-x^(2)}` `=2*[(3)/(2)-(x^(2)+x+(1)/(4))+(1)/(4)]` `=2*[(7)/(4)-(x+(1)/(2))^(2)]=2*{((sqrt(7))/(2))^(2)-(x+(1)/(2))^(2)}`. `thereforesqrt(3-2x-2x^(2))=sqrt(2)*sqrt((sqrt(7)/(2))^(2)-(x+(1)/(2))^(2))` `rArrintsqrt(3-2x-2x^(2))dx=sqrt(2)*intsqrt((sqrt(7)/(2))^(2)-(x+(1)/(2))^(2))dx` `=sqrt(2)*intsqrt((sqrt(7)/(2))^(2)-t^(2))dt, "where"(x+(1)/(2))=t` and dx = dt `=sqrt(2)*{(t)/(2)*sqrt(7)/(4)-t^(2)+(7)/(8)" sin"^(-1)(t)/((sqrt(7)//2))}+C` `[becauseintsqrt(a^(2)-t^(2))dt=(t)/(2)sqrt(a^(2)-t^(2))+(a^(2))/(2)*"sin"^(-1)(t)/(a)+C]` `=sqrt(2)*{(1)/(2)(x+(1)/(2))sqrt((7)/(4)-(x+(1)/(2))^(2))+(7)/(8)"sin"^(-1)((x+(1)/(2)))/((sqrt(7)//2))}+C` `=sqrt(2){(1)/(4)(2x+1)*sqrt((3)/(2)-x^(2))+(7)/(8)sin^(-1)((2x+1)/(sqrt(7)))}+C` `=(1)/(4)(2x+1)sqrt(3-2x-2x^(2))+(7)/(4sqrt(2))sin^(-1)((2x+1)/(sqrt(7)))+C`. Integrals of the form `int(px+q)sqrt((ax^(2)+bx+c))dx` Method Let `(px+q)=A*(d)/(dx)(ax^(2)+bx+c)+B`. Find A and B . Then , we get the integrand which is easily integrable. |
|
| 25. |
Evaluate:`intsqrt(2x^2+3x+4)dx` |
|
Answer» We have `(2x^(2)+3x+4)=2(x^(2)+(3)/(2)x+2)` `=2*{(x^(2)+(3)/(2)x+(9)/(16))+(2-(9)/(16))}=2*{(x+(3)/(4))^(2)+(sqrt(23)/(4))^(2)}`. `thereforesqrt(2x^(2)+3x+4)=sqrt(2)*sqrt((x+(3)/(4))^(2)+(sqrt(23)/(4))^(2))dx` `rArrintsqrt(2x^(2)+3x+4)dx=sqrt(2)*intsqrt((x+(3)/(4))^(2)+(sqrt(23)/(4))^(2))dx` `=sqrt(2)intsqrt(t^(2)+((sqrt(23))/(4))^(2))dt " where"(x+(3)/(4))=t` `=sqrt(2)*{(t)/(2)*sqrt(t^(2)+(23)/(16))+(23)/(32)log|t+sqrt(t^(2)+(23)/(16))|}+C` `[becauseintsqrt(t^(2)+a^(2))dt=(t)/(2)sqrt(t^(2)+a^(2))+(a^(2))/(2)log|t+sqrt(t^(2)+a^(2))|+C]` `=(sqrt(2))/(2)(x+(3)/(4))sqrt((x+(3)/(4))^(2)+(23)/(16))+(23sqrt(2))/(32)log|(x+(3)/(4))+sqrt(x^(2)+(3)/(2)x+2)|+C` `=((4x+3))/(4sqrt(2))*sqrt(x^(2)+(3)/(2)x+2)+(23sqrt(2))/(32)log|((4x+3))/(4)+(sqrt(2x^(2)+3x+4))/(sqrt(2))|+C` `=((4x+3)sqrt(2x^(2)+3x+4))/(8)+(23sqrt(2))/(32)log|((4x+3))/(4)+(sqrt(2x^(2)+3x+4))/(sqrt(2))|+C` |
|
| 26. |
`int(e^(x))/((e^(2x)+1))dx=?` |
|
Answer» Correct Answer - `tan^(-1)(e^(x))+C` Put `e^(x)= t and e^(x) dx= dt`. |
|
| 27. |
Evaluate : (i) `intsqrt(x^(2)-16)dx` (ii) `intsqrt(4x^(2)-5)dx` (iii) `intsqrt(17x^(2)-11)dx` |
|
Answer» We know that `intsqrt(x^2-a^(2))dx=(x)/(2)sqrt(x^(2)-a^(2))-(a^(2))/(2)log|x+sqrt(x^(2)-a^(2))|+C`. `therefore` (i) `sqrt(x^(2)-16)dx=intsqrt(x^(2)-4^(2))dx` `=(x)/(2)*sqrt(x^(2)-4^(2))-(16)/(2)log|x+sqrt(x^(2)-16)|+C` `=(x)/(2)*sqrt(x^(2)-16)-8log|x+sqrt(x^(2)-16)|+C`. (ii) `intsqrt(4x^(2)-5)dx=2intsqrt(x^(2)-(5)/(4))dx=2intsqrt(x^(2)-((sqrt(5))/(2))^(2))dx` `=2*[(x)/(2)sqrt(x^(2)-(5)/(4))-(5)/(8)-log|x+sqrt(x^(2)-(5)/(4))|]+C` `=xsqrt(x^(2)-(5)/(4))-(5)/(4)log|xsqrt(x^(2)-(5)/(4))|+C`. (iii) `intsqrt(17x^(2)-11)dx=sqrt(17)*intsqrt(x^(2)-(11)/(17))dx` `=sqrt(17)*{(x)/(2)sqrt(x^(2)-(11)/(17))-(11)/(34)log|x+sqrt(x^(2)-(11)/(17))|}+C` `=(x)/(2)sqrt(17x^(2)-11-)(11sqrt(17))/(34)log|x+sqrt(x^(2)-(11)/(17))|+C`. |
|
| 28. |
Evaluate `intsqrt(x^(2)+3x)dx`. |
|
Answer» We have `(x^(2)+3x)={x^(2)+3x+((3)/(2))^(2)-((3)/(2))^(2)}=(x+(3)/(2))^(2)-((3)/(2))^(2)`. `thereforeI=intsqrt(x^(2)+3x)dx` `=intsqrt((x+(3)/(2))^(2)-((3)/(2))^(2))dx=intsqrt(t^(2)-((3)/(2))^(2))dt "where"(x+(3)/(2))=t` `=(1)/(2)tsqrt(t^(2)-(9)/(4))-(9)/(8)log|t+sqrt(t^(2)-(9)/(4))|+C` `{"using"intsqrt(x^(2)-a^(2))dx=(x)/(2)sqrt(x^(2)-a^(2))-(a^(2))/(2)log|x+sqrt(x^(2)-a^(2))|+C}` `=(1)/(2)(x(3)/(2))sqrt(x^(2)+3x)-(9)/(8)log|(x+(3)/(2))+sqrt(x^(2)+3x)|+C`. |
|
| 29. |
`int(dx)/((1-6x-9x^2)` |
|
Answer» We have `int(dx)/((1-6x-9x^(2)))=-int(dx)/((9x^(2)+6x-1))=-(1)/(9)int(dx)/((x^(2)+(2)/(3)x-(1)/(9)))` `=-(1)/(9)*int(dx)/({(x^(2)+(2)/(3)x+(1)/(9))-(2)/(9)})` `=-(1)/(9)*int(dx)/({(x+(1)/(3))^(2)-((sqrt(2))/(3))^(2)})=(1)/(9)*int(dx)/({((sqrt(2))/(3))^(2)-(x+(1)/(3))^(2)})` `=(1)/(9)*int(dx)/({((sqrt(2))/(3))^(2)-t^(2)}),"where"(x+(1)/(3))=t` `=(1)/(9)*(1)/(2*(sqrt(2))/(3))log|(sqrt(2)/(3)+t)/(sqrt(2)/(3)-t)|+C=(1)/(6sqrt(2))log|(sqrt(2)+3t)/(sqrt(2)-3t)|+C` `=(1)/(6sqrt(2))log|(sqrt(2)+3(x+(1)/(3)))/(sqrt(2)-3(x+(1)/(3)))|+C` `=(1)/(6sqrt(2))log|(sqrt(2)+1+3x)/(sqrt(2)-1-3x)|+C`. |
|
| 30. |
Evaluate: `intsqrt(16 x^2+25) dx` |
|
Answer» We know that `intsqrt(x^(2)+a^(2))dx=(x)/(2)sqrt(x^(2)+a^(2))+(a^(2))/(2)log|x+sqrt(x^(2)+a^(2))|+C`. `thereforeintsqrt(16x^(2)+25)dx=4int{sqrt(x^(2)+(25)/(16))}dx=4int{sqrt(x^(2)+((5)/(4))^(2))}dx` `=4*{(x)/(2)sqrt(x^(2)+(25)/(16))+(25)/(32)log|x+sqrt(x^(2)+(25)/(16))|}+C` `=(x)/(2)*sqrt(16x^(2)+25)+(25)/(8)log|x+sqrt(x^(2)+(25)/(16))|+C`. |
|
| 31. |
Evaluate: `intcosx sqrt(4-sin^2x) dx` |
|
Answer» Putting sin x = t and cos x dx = dt , we get `I=intsqrt(4-t^(2))dt` `=(t)/(2)sqrt(4-t^(2))+(4)/(2)"sin"^(-1)(t)/(2)+C` `=(1)/(2)sinxsqrt(4-sin^(2)x)+2sin^(-1)((1)/(2)sinx)+C`. Integrals of the form `intsqrt((ax^(2)+bx+c))dx` Method Express `(ax^(2)+bx+c)" as a"[(x+alpha)^(2)+-beta^(2)]` and obtain an integral which can be evaluated easily. |
|
| 32. |
`int(x^2)/(9+4x^2)dx` |
|
Answer» Correct Answer - `(x)/(4)-(3)/(8)"tan"^(-1)((2x)/(3))+C` On dividing `x^(2)` by `(4x^(2)+9)`, we get ` (x^(2))/((9+4x^(2)))=(1)/(4)-((9//4))/(4x^(2)+9)`. `thereforeI=(1)/(4)dx-(9)/(19)*int(dx)/({x^(2)((3)/(2))^(2)})`. |
|
| 33. |
`int(sqrt(16+(logx)^(2)))/(x)dx=?` |
|
Answer» Putting log `x=t and (1)/(x)dx=dt` we get `I=intsqrt(16+t^(2))dt=intsqrt(4^(2)+t^(2))dt` `=(t)/(2)sqrt(16+t^(2))+(16)/(2)log|t+sqrt(16+t^(2))|+C` `=(1)/(2)logx*sqrt(16+(logx)^(2))+8 log|logx+sqrt(16+(logx)^(2))|+C`. |
|
| 34. |
`inte^(x)sqrt(e^(2x)+4)dx=?` |
|
Answer» Putting `e^(x)=t and e^(x)dx = dt` we get `I=intsqrt(t^(2)+4)dx` `=(t)/(2)*sqrt(t^(2)+4)+(4)/(2)log|t+sqrt(t^(2)+4)|+C` `=(1)/(2)e^(x)sqrt(e^(2x)+4)+2log|e^(x)+sqrt(e^(2x)+4)|+C`. |
|
| 35. |
Evaluate: `int(3x-2) sqrt(x^2+x+1) dx` |
|
Answer» Let `(3x-2)=A*(d)/(dx)(x^(2)+x+1)+B`. Then , `( 3x -2) =A (2x =1)+B`. Comparing the coefficients of like powers of x , we get `2A=3and A+B =-2. "So",A=(3)/(2)andB=(-7)/(2)`. `therefore(3x-2)=(3)/(2)(2x+1)-(7)/(2)*` So , `int(3x-2)sqrt(x^(2)+x+1)dx` `=int{(3)/(2)(2x+1)-(7)/(2)}sqrt(x^(2)+x+1)dx` `=(3)/(2)int(2x+1)sqrt(x^(2)+x+1)dx-(7)/(2)intsqrt(x^(2)+x+1)dx` `=(3)/(2)intsqrt(t)dt-(7)/(2)intsqrt((x^(2)+x+(1)/(4))+(3)/(4))dx`, where `x^(2)+x+1=t` in the 1st integral `=t^(3//2)-(7)/(2)intsqrt({(x+(1)/(2))^(2)+((sqrt(3))/(2))^(2)})dx` `=(x^(2)+x+1)^(3//2)-(7)/(2)*intsqrt(u^(2)+((sqrt(3))/(2))^(2))du, "where"u=(x+(1)/(2))` `=(x^(2)+x+1)^(3//2)-(7)/(2){(u)/(2)*sqrt(u^(2)+(3)/(4))+(3)/(8)log|u+sqrt(u^(2)+(3)/(4))|}+C` `{becauseintsqrt(u^(2)+a^(2))du=(u)/(2)sqrt(u^(2)+a^(2))+(a^(2))/(2)log|u+sqrt(u^(2)+a^(2))|+C}` `=(x^(2)+x+1)^(3//2)-(7u)/(8)sqrt(4u^(2)+3)-(21)/(16)log|u+sqrt(u^(2)+(3)/(4))|+C` `=(x^(2)+x+1)^(3//2)-(7)/(8)(x+(1)/(2))sqrt(4(x+(1)/(2))^(2)+3)-(21)/(16)log|(x+(1)/(2))+sqrt((x+(1)/(2))^(2)+(3)/(4))|+C` `=(x^(2)+x+1)^(3//2)-(7(2x+1))/(8)sqrt(x^(2)+x+1)-(21)/(16)log|((2x+1))/(2)+sqrt(x^(2)+x+1)|+C`. |
|
| 36. |
`int(dx)/((1-9x^(2)))=?` |
|
Answer» Correct Answer - `(1)/(6)log|(1+3x)/(1-3x)|+C` On dividing , we get `((x^(2)-1))/((x^(2)+4))={1-(5)/((x^(2)+4))}`. |
|
| 37. |
Evaluate `intxsqrt(x^(4)+1)dx`. |
|
Answer» Putting `x^(2)=t and x dx =(1)/(2)dt` we get `I=(1)/(2)intsqrt(t^(2)+1)dt` `=(1)/(2)*[(t)/(2)sqrt(t^(2)+1)+(1)/(2)log|t+sqrt(t^(2)+1)|]+C` `=(x^(2))/(4)sqrt(x^(4)+1)+(1)/(4)log|x^(2)+sqrt(x^(4)+1)|+C`. |
|
| 38. |
Evaluate `intxsqrt(x+x^(2))dx`. |
|
Answer» Let `x=A*(d)/(dx)(x+x^(2))+B`. Then, x=A(1+2x)+B . . . (i) Comparing the coefficients of various powers of x , we get `(2A=1andA+B=0)rArr(A=(1)/(2)andB=(-1)/(2))`. `thereforex=(1)/(2)(1+2x)-(1)/(2)` `rArrintxsqrt(x+x^(2))dx` `=int{(1)/(2)(1+2x)-(1)/(2)}sqrt(x+x^(2))dx` `=(1)/(2)int(1+2x)sqrt(x+x^(2))dx-(1)/(2)intsqrt(x+x^(2))dx` `=(1)/(2)intsqrt(t)dt-(1)/(2)intsqrt((x^(2)+x+(1)/(4))-(1)/(4))}dx`, where ` (x+x^(2))` = t in the first integral `=(1)/(2)*(t^(3//2))/((3//2))-(1)/(2)intsqrt({(x+(1)/(2))^(2)-((1)/(2))^(2)})dx` `=(1)/(3)(x+x^(2))^(3//2)-(1)/(2)*intsqrt(u^(2)-((1)/(2))^(2))du` , where `(x+(1)/(2))=u` `=(1)/(3)(x+x^(2))^(3//2)-(1)/(2){(u)/(2)*sqrt(u^(2)(-1)/(4))-(1)/(8)log|u+sqrt(u^(2)-(1)/(4))|}+C` `{becauseintsqrt(x^(2)-a^(2))dx=(x)/(2)sqrt(x^(2)-a^(2))-(a^(2))/(2)log|x+sqrt(x^(2)-a^(2))|+C}` `=(1)/(3)(x+x^(2))^(3//2)-(1)/(4)(x+(1)/(2))sqrt((x+(1)/(2))^(2)-(1)/(4))+(1)/(16)log|(x+(1)/(2))sqrt((x+(1)/(2))^(2)-(1)/(4))|+C` `=(1)/(3)(x+x^(2))^(3//2)-(1)/(8)(2x+1)sqrt(x+x^(2))+(1)/(16)log|((2x+1))/(2)*sqrt(x+x^(2))|+C`. |
|
| 39. |
Evaluate `int(3x)/((1+2x^(4)))dx`. |
|
Answer» Putting `x^(2)=t " and "2xdx=dt`, we get `int(3x)/((1+2x^(4)))dx=(3)/(2)*int(dt)/((1+2t^(2)))` `=(3)/(4)*int(dt)/(((1)/(2)+t^(2)))=(3)/(4)*int(dt)/({((1)/(sqrt(2)))^(2)+t^(2)})` `=(3)/(4)*(1)/((1//sqrt(2)))"tan"^(-1)(t)/((1//sqrt(2))^(2))+C` `=(3)/(2sqrt(2))tan^(-1)(sqrt(2)t)+C=(3)/(2sqrt(2))tan^(-1)(sqrt(2x)^(2))+C`. |
|
| 40. |
Evaluate `int(dx)/((1+5sin^(2)x))`. |
|
Answer» On dividing num . And denom by `cos^(2)x` , we get `int(dx)/((1+5sin^(2)x))=int((1//cos^(2)x))/(((1)/(cos^(2)x)+5*(sin^(2)x)/(cos^(2)x)))dx` `=int(sec^(2)x)/((sec^(2)x5tan^(2)x))dx=int(sec^(2)x)/({(1+tan^(2)x)+5tan^(2)x})dx` `=int(sec^(2)x)/((1+6tan^(2)x))dx=int(dt)/((1+6t^(2)))`, where tan =t `=(1)/(6)int(dt)/(((1)/(6)+t^(2)))=(1)/(6)*int(dt)/({((1)/(sqrt(6)))^(2)+t^(2)})` `=(1)/(6)*(1)/((1//sqrt(6)))"tan"^(-1)(t)/((1//sqrt(6)))+C=(1)/(sqrt(6))tan^(-1)(sqrt(6t))+C` `=(1)/(sqrt(6))tan^(-1)(sqrt(6)tanx)+C`. |
|
| 41. |
Evaluate `int(dx)/((x^(2)+6x+13))`. |
|
Answer» We have `int(dx)/((x^(2)+6x+13))=int(dx)/({(x^(2)+6x+9)+4})` `=int(dx)/({(x+3)^(2)+2^(2)})=int(dt)/((t^(2)+2^(2)))`, where `(x+3)=t` `=(1)/(2)"tan"^(-1)(t)/(2)+C` `=(1)/(2)"tan"^(-1)(1)/(2)(x+3)+C`. |
|
| 42. |
Evaluate `int(dx)/(sqrt(3x^(2)+6x+12))dx`. |
|
Answer» We have `int(dx)/(sqrt(3x^(2)+6x+12))=(1)/(sqrt(3))*int(dx)/(sqrt(x^(2)+2x+4))` `=(1)/(sqrt(3))*int(dt)/(sqrt(t^(2)+(sqrt(3))^(2))), "where " (x+1)=t` `=(1)/(sqrt(3))log|t+sqrt(t^(2)+3)|+C` `=(1)/(sqrt(3))log|(x+1)+sqrt(x^(2)+2x+4)|+C`. |
|
| 43. |
Evaluate `int(2^(x))/(sqrt(1-4^(x)))dx` |
|
Answer» Putting `2^(x)=t and(2^(x)log2)dx=dt` , we get `int(2^(x))/(sqrt(1-4^(x)))dx=(1)/((log2))*int(dt)/(sqrt(1-t^(2)))` `=(1)/((log2))*sin^(-1)t+C=(1)/((log2))*sin^(-1)(2^(x))+C`. |
|
| 44. |
Evaluate: `int(2x-5) sqrt(x^2-4x+3) dx` |
|
Answer» Correct Answer - `(2)/(3)(x^(2)-4x+3)^(3//2)-(1)/(2)(x-2)sqrt(x^(2)-4x+3)+(1)/(2)log|x-2+sqrt(x^(2)-4x+3)|+C` Let `(2x-5)=A*(d)/(dx)(x^(2)-4x+3)+B=A(2x-4)+B`. |
|
| 45. |
Evaluate: `int(x-3)sqrt(x^2+3x-18) dx` |
|
Answer» Correct Answer - `(1)/(3)(x^(2)+3x-18)^(3//2)-(9)/(8){(2x+3)sqrt(x^(2)+3x-18)}-(81)/(2)log|x+(3)/(2)+sqrt(x^(2)+3x-18)|+C` Let `(x-3)=A*(d)/(dx)(x^(2)+3x-18)+B`. Then `thereforeI=(1)/(2)int(2x+3)sqrt(x^(2)+3x-18)dx-(9)/(2)intsqrt(x^(2)+3x-18)dx` `=(1)/(2)intsqrt(t)dt-(9)/(2)intsqrt((x+(3)/(2))^(2)-((9)/(2))^(2))dx`. |
|
| 46. |
`int(dx)/(sqrt(2+2x-x^(2)))=?` |
| Answer» Correct Answer - `sin^(-1)((x-1)/(sqrt(3)))+C` | |
| 47. |
`int(dx)/(sqrt(x^(2)-6x+10))` |
| Answer» Correct Answer - `log|(x-3)+sqrt(x^(2)-6x+10)|+C` | |
| 48. |
`int(dx)/(sqrt(8+2x-x^(2)))` |
| Answer» Correct Answer - `sin^(-1)((x-1)/(3))+C` | |
| 49. |
`int(dx)/(sqrt(x^2-3x+2))` |
| Answer» Correct Answer - `log|(x-(3)/(2))+sqrt(x^(2)-3x+2)|+C` | |
| 50. |
`int (dx)/(sqrt(x^(2)+6x+5))=?` |
| Answer» Correct Answer - `log|(x+3)+sqrt(x^(2)+6x+5)|+C` | |