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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. |
Find:Integrate ((x4 - 1)1/4 )/(x6) dx\(\int\frac{{x^4-1}^{1/4}dx}{x^6}\) |
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Answer» Let I = \(\int\frac{{x^4-1}^{1/4}dx}{x^6}\) \(\int\frac{x^4(1-\frac1{x^4})}{x^6}dx\) = \(\int\cfrac{{(x^4(1-\frac1{x^4})}^{1/4}}{x^6}dx\) = \(\int\cfrac{{x(1-\frac1{x^4})}^{1/4}}{x^6}dx\) = \(\int\cfrac{{(x^4(1-\frac1{x^4})}^{1/4}}{x^5}dx\) Let 1 - \(\frac1{x^4}=t^4\) ⇒ \(\frac4{x^5}dx\) = 4t3dt ⇒ \(\frac1{x^5}dx\) = t3dt \(\therefore\) I = \(\int\)(t4)1/4.t3dt = \(\int\)t4dt = \(\frac{t^5}5+c\) = \(\frac15(1-\frac1{x64})^{5/4}+c\) (\(\because\) t = (1 - \(\frac1{x^4}\))1/4) Hence, \(\int\frac{(x^4-1)^{1/4}}{x^6}dx\) = \(\frac15(1-\frac1{x^4})^{5/4}\) + c |
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| 2. |
`int((2+secx)secx)/((1+2secx)^2)dx=`A. `(1)/("2 cosec x"+cotx)+C`B. `"2 cosec x"+cotx+C`C. `(1)/("2 cosec x"-cotx)+C`D. `"2 cosec x"-cotx+C` |
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Answer» Correct Answer - A `I=int((2cosx+1))/((2+cosx)^(2))dx` `=int((2+cosx)cosx+sin^(2)x)/((2+cosx)^(2))dx` `=int(cosx)/(2+cosx)dx-int(-sin^(2)x)/((2+cosx)^(2))dx` `=(sinx)/(2+cosx)+C` |
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| 3. |
If `int(e^(4x)-1)/(e^(2x))log((e^(2x)+1)/(e^(2x-1)))dx=(t^(2))/(2)logt-(t^(2))/(4)-(u^(2))/(2)logu+(u^(2))/(4)+C,` thenA. `u=e^(x)+e^(-x)`B. `u=e^(x)-e^(-x)`C. `t=e^(x)+e^(-x)`D. `t=e^(x)-e^(-x)` |
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Answer» Correct Answer - B::C `I=int{(e^(2x)-e^(-2x))ln(e^(x)+e^(-x))-(e^(2x)-e^(-2x))ln(e^(x)-e^(-x))}dx` `=int tln t dt- int u ln u du ("where t"=e^(x)+e^(-x) and u=e^(x)-e^(-x))` `=(t^(2))/(2)ln t-(t^(2))/(4)-(u^(2))/(2)ln u+(u^(2))/(4)+C` |
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| 4. |
Let F(x) be an indefinite integral of `sin^(2)x` Statement-1: The function F(x) satisfies `F(x+pi)=F(x)` for all real x. because Statement-2: `sin^(3)(x+pi)=sin^(2)x` for all real x. A) Statement-1: True , statement-2 is true, Statement -2 is not a correct explanation for statement -1 c) Statement-1 is True, Statement -2 is False. D) Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - d | |
| 5. |
Let F(x) be an indefinite integral of `sin^(2)x` Statement-1: The function F(x) satisfies `F(x+pi)=F(x)` for all real x. because Statement-2: `sin^(3)(x+pi)=sin^(2)x` for all real x. A) Statement-1: True , statement-2 is true, Statement -2 is not a correct explanation for statement -1 c) Statement-1 is True, Statement -2 is False. D) Statement-1 is False, Statement-2 is True.A. Both the Statement are true and Statement 2 is the correct explanation of Statement 1B. Both the Statement are true but Statement 2 is not the correct explanation of Statement 1C. Statement 1 is true but Statement 2 is falseD. Statement 1 is false but Statement 2 is true |
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Answer» Correct Answer - B `f(x) = int sin^(2) x.dx = int (1 - cos 2x)/(2) .dx` `= (1)/(2) x - (sin 2x)/(2) (1)/(2) + c = (1)/(2)x - (1)/(4) sin 2x + c` (i) `f(pi + x) = (1)/(2) (pi + x) - (1)/(4) sin 2 (pi + x)` `= (1)/(2) pi + (1)/(2) x - (1)/(2) sin (2pi + 2x) + c` `= (x)/(2) - (1)/(2) sin 2x + c = f (x)` So, statement 1 is true (ii) `sin^(2) (pi + x) = sin^(2)x` `(-sin x)^(2) = sin^(2)x` `rArr sin^(2) x = sin^(2) x` So, statement 2 is true |
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| 6. |
`int(x^3-x)/(1+x^6)dx` is equal toA. `(1)/(6)log.(x^(4)-x^(2)+1)/(x(x^(2)+1))+C`B. `(1)/(6)tan^(-1).((x^(2)+1)^(2))/(2)+C`C. `log.(x^(4)-x^(2)+1)/((1+x^(2))^(2))+C`D. `tan^(-1).((x^(2)+1)^(2))/(2)+C` |
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Answer» Correct Answer - A `I=(1)/(2)int(t-1)/(1+t^(3))dt (" where "t=x^(2))` `=(1)/(2)int((-2)/(3(1+t))+(1)/(3)((2t-1))/(t^(2)-t+1))dt " (After partial fractions)"` `=(1)/(6).ln(x^(4)-x^(2)+1)/((x^(2)+1)^(2))+C` |
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| 7. |
The integral `int(dx)/(x^(2)(x^(4)+1)^(3//4))` equalA. `(x^(4)+1)/(x^(4))^(1//4)+C`B. `(x^(4)+1)^(1//4)+C`C. `-(x^(4)+1)^(1//4)+C`D. `-(x^(4)+1)/(x^(4))+C` |
| Answer» Correct Answer - 4 | |
| 8. |
`"The integral " int(dx)/(x^(2)(x^(4)+1)^(3//4))" equals"`A. `((x^(4)+1)/(x^(4)))^(1//4)+c`B. `(x^(4)+1)^(1//4)+c`C. `-(x^(4)+1)^(1//4)+c`D. `-((x^(4)+1)/(x^(4)))^(1//4)+c` |
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Answer» Correct Answer - D ` I=int(dx)/(x^(2)(x^(4)+1)^(3//4))` ` int(dx)/(x^(5)(1+(1)/(x^(4)))^(3//4))` `"Let "1+(1)/(x^(4))=t^(4)` `implies (-4)/(x^(5))dx=4t^(3)dt " or " (dx)/(x^(5))=-t^(3)dt` ` :. I=int(-t^(3)dt)/(t^(3))=-t+c=-(1+(1)/(x^(4)))^(1//4)+c` |
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| 9. |
The integral `int[2x^[12]+5x^9]/[x^5+x^3+1]^3.dx` is equal to- (A) `x^10 / (2(x^5 + x^3 +1)^2) ` (B) `x^5/ (2(x^5 + x^3 +1)^2) ` (C) `-x^10 / (2(x^5 + x^3 +1)^2) ` (D) `- x^5 / (2(x^5 + x^3 +1)^2) `A. `(x^(10))/(2(x^(5)+x^(3)+1)^(2))+C`B. `(x^(5))/(2(x^(5)+x^(3)+1)^(2))+C`C. `(-x^(10))/(2(x^(5)+x^(3)+1)^(2))`D. `(-x^(5))/((x^(5)+x^(3)+1)^(2))+C` |
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Answer» Correct Answer - A ` I=int(2x^(12)+5x^(9))/((x^(5)+x^(3)+1)^(3))dx` `=int(((2)/(x^(3))+(5)/(x^(6))))/((1+(1)/(x^(2))+(1)/(x^(5)))^(3))dx` ` "Let " 1+(1)/(x^(2))+(1)/(x^(5))=t` ` :. (dt)/(dx)=(-2)/(x^(3))-(5)/(x^(6))` ` :. int(-dt)/(t^(3))=(1)/(2t^(2))+C` `=(1)/(2(1+(1)/(x^(2))+(1)/(x^(5)))^(2))+C` `=(x^(10))/(2(x^(5)+x^(3)+1)^(2))+C` |
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| 10. |
The integral `int[2x^[12]+5x^9]/[x^5+x^3+1]^3.dx` is equal to- (A) `x^10 / (2(x^5 + x^3 +1)^2) ` (B) `x^5/ (2(x^5 + x^3 +1)^2) ` (C) `-x^10 / (2(x^5 + x^3 +1)^2) ` (D) `- x^5 / (2(x^5 + x^3 +1)^2) `A. `(x^(10))/(2(x^(5)+x^(3)+1)^(2))+C`B. `x^(5)/(2(x^(5)+x^(3)+1)^(2))+C`C. `-x^(10)/(2(x^(5)+x^(3)+1)^(2))+C`D. `-x^(5)/(x^(5)+x^(3)+1)^(2)+C` |
| Answer» Correct Answer - 1 | |
| 11. |
The integral `int(2x^(12)+5x^9)/((x^5+x^3+1)^3)dx`is equal to:A. `(x^(10))/(2(1+x^(3)+x^(5))^(4))+c`B. `(x^(2)+2x)/((x^(5)+x^(3)+1)^(4))+c`C. `(x^(10))/(2(x^(5)+x^(3)+1)^(2))+c`D. `(2x^(10))/((x^(5)+x^(3)+1)^(3))+c` |
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Answer» Correct Answer - C `I=int(2x^(12)+5x^(9))/((x^(5)+x^(3)+1)^(3))dx` `=int (((2)/(x^(3))+(5)/(x^(6)))dx)/((1+(1)/(x^(2))+(1)/(x^(5)))^(3))` `=int (-dt)/(t^(3)), " where " t=1+(1)/(x^(2))+(1)/(x^(5))` `=(1)/(2t^(2))+C` ` =(1)/(2(1+(1)/(x^(2))+(1)/(x^(5)))^(2))+c` `=(x^(10))/(2(x^(5)+x^(3)+1)^(2))+c` |
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| 12. |
The value of `int(cos^(3)x)/(sin^(2)x+sinx)dx` is equal toA. `log_(e)|sinx|+sinx+C`B. `log_(e)|sinx|-sinx+C`C. `-log_(e)|sinx|-sinx+C`D. `-log_(e)|sinx|+sinx+C` |
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Answer» Correct Answer - B `I=int(cos^(3)x)/(sin^(2)x+sinx)dx` `=int(cosx.(1-sin^(2)x))/(sinx(1+sinx))dx` Put `sinx=t,` then `cos xdx =dt` `rArr" "I=int((1-t)(1+t)dt)/(t(1+t))` `=int((1-t)dt)/(t)` `=int((1)/(t)-1)dt` `=log_(e)|t|-t+C` `=log_(e)|sinx|-sinx+C` |
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| 13. |
`int (x^3-1)/((x^4+1)(x+1)) dx` isA. `(1)/(4)ln(1+x^(4))+(1)/(3)ln(1+x^(3))+c`B. `sinx|-sinx+C`C. `(1)/(4)ln(1+x^(4))-ln(1+x)+c`D. `(1)/(4)ln(1+x^(4))+ln(1+x)+c` |
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Answer» Correct Answer - C `int(x^(3)-1)/((x^(4)+1)(x+1))dx=int((x^(4)+x^(3))-(x^(4)+1))/((x^(4)+1)(x+1))dx` `=int(x^(3))/(x^(4)+1)dx-int(1)/(x+1)dx` `=(1)/(4)ln(x^(4)+1)-ln(x+1)+c` |
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| 14. |
`int (sin^2 x cos^2x)/(sin^5x+cos^3xsin^2x+sin^3xcos^2x+cos^5x)^2 dx`A. `1/(1+cot^(3)x)`+CB. `-1/(1+cot^(3)x)+C`C. `1/(3(1+tan^(3)x))+C`D. `-1/(3(1+tan^(3)x))+C` |
| Answer» Correct Answer - 4 | |
| 15. |
The value of the integral`int(cos^3x+cos^5x)/(sin^2x+sin^4x)dxi s``sinx-6tan^(-1)(sinx)+C``sinx-2(sinx)^(-1)+C``sinx-2(sinx)^(-1)-6tan^(-1)(sinx)+C``sinx-2(sinx)^(-1)+5tan^(-1)(sinx)+C`A. `sin x - 6 tan^(-1)(sin x) + c`B. `sin x - 2 (sin x)^(-1) + c`C. `sin x - 2 (sin x)^(-1) - 6 tan^(-1)(sin x) + c`D. `sin x -2 (sin x)^(-1) + 5 tan^(-1) (sin x) + c` |
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Answer» Correct Answer - C Let `I=int(cos^(3)x+cos^(5)x)/(sin^(2)x+sin^(4)x)dx` `=int((cos^(2)+xos^(4)x)*cos x dx)/((sin^(2)x+sin^(4)x))` Put `sin x = t rArr cos x dx = dt` `therefore" "I=int([(1-t^(2))+(1-t^(2))^(2)])/(t^(2)+t^(4))dt` `rArr" "I=int(1-t^(2)+1-2t^(2)+t^(4))/(t^(2)+t^(4))dt` `rArr" "I=int(2-3t^(2)+t^(4))/(t^(2)(t^(2)+1))dt" "....(i)` Using partial fraction for `(y^(2)-3y+2)/(y(y+1))=1+(A)/(y)+(B)/(y=1)" "["where", y = t^(2)]` `rArr" "A=2, B=-6` `therefore" "(y^(2)-3y+2)/(y(y+1))=1+(2)/(y)-(6)/(y+1)` Now, Eq. (i) reduces to, `I=int(1+(2)/(t^(2))-(6)/(1+t^(2)))dt` `=t-(2)/(t)-6 tan^(-1)(t)+c` `=sin x -(2)/(sin x)-6 tan^(-1)(sin x)+c` |
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| 16. |
`int (cos 5x + cos 4x)/(1-2cos3x) dx`A. `(sin2x)/(2)+cosx+c`B. `(sin2x)/(2)-cosx+c`C. `-(sin2x)/(2)-sinx+c`D. `(sin2x)/(2)-cosx+c` |
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Answer» Correct Answer - C `I=int(cos5x+cos4x)/(1-2cos3x)dx=int(2"cos"(9x)/(2)"cos"(x)/(2))/(1-2(2"cos"^(2)(3x)/(2)-1))dx` `=int(2"cos"(9x)/(2)"cos"(x)/(2))/(3-4"cos"^(2)(3x)/(2))dx` `=int(2"cos"(9x)/(2)"cos"(x)/(2)"cos"(3x)/(2))/(3"cos"(3x)/(2)-4"cos"^(3)(3x)/(2))dx` `=int(2"cos"(9x)/(2)"cos"(3x)/(2)"cos"(x)/(2))/(-"cos"(9x)/(2))dx` `=-int 2"cos"(3x)/(2)"cos"(x)/(2)dx` `=-int(cos2x+cosx)dx` `=-(sin2x)/(2)-sinx+c` |
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| 17. |
Evaluate: `int(cos5x+cos4x)/(1-2cos3x) dx`A. `-((sin2x)/(2)+cosx)+C`B. `-((sin2x)/(2)+cosx)+C`C. `-((cos2x)/(2)+cosx)+C`D. `-((sin2x)/(2)+sinx)+C` |
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Answer» Correct Answer - D `int(cos5x+cos4x)/(1-2cos3x)dx` `=int(2cos.(9x)/(2)cos.(x)/(2)dx)/(1-2(2cos^(2)((3x)/(2))-1))` `=int(2cos.(x)/(2)(4cos^(3),(3x)/(2)-3cos.(3x)/(2)))/(3-4cos^(2).(3x)/(2))dx` `=-int2cos.(3x)/(2)cos.(x)/(2)dx=-intcos2xdx-int cosxdx` `=-((sin2x)/(2)+sinx)+C` |
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| 18. |
`int(secx."cosec"x)/(2cotx-secx"cosec x")dx` di equal toA. `(1)/(2)ln|sec2x+tan2x|+C`B. `ln|secx+"cosec x"|+C`C. `ln |secx+Tanx|+C`D. `(1)/(2)ln|secx+" cosec x"|+C` |
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Answer» Correct Answer - A `I=int(sec x "cosec x")/(2cot x -sec x" cosec "x)dx` `=int(dx)/(2cos^(2)x-1)` `=intsec2xdx` `=(1)/(2)ln|sec 2x+tan2x_+C` |
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| 19. |
`int(3x^(2)+2x)/(x^(6)+2x^(5)+x^(4)+2x^(3)+2x^(2)+5)dx=`A. `(1)/(4)tan^(-1)((x^(3)+x^(2)+1)/(2))+c`B. `(1)/(2)tan^(-1)((x^(3)+x^(2)+1)/(2))+c`C. `sin^(-1)((x^(3)+x^(2)+1)/(2))+c`D. `(1)/(2)tan^(-1)((x^(3)+x^(2))/(2))+c` |
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Answer» Correct Answer - B `I=int(3x^(2)+2x)/((x^(3)+x^(2)+1)^(2))dx` Put `x^(3)+x^(2)+1=t` `therefore" "I=int(1)/(t^(2)+4)dt` `=(1)/(2)tan^(-1)((t)/(2))+c` `=(1)/(2)tan^(-1)((x^(3)+x^(2)+1)/(2))+c` |
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| 20. |
Find `intsin^(3)x cos^(5)x dx`. |
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Answer» [Here, powers of both cos x and sin x are odd positive integers, therefore, put `z=cosx or z=sin x`, but the power of `cosx` is greater. Therefore, it is convenient to put `z=cosx`.] `intsin^(3)x cos^(5)x dx` `=intsin^(2)x cos^(5)xsin x dx` `=int(1-cos^(2)x)cos^(5)x sinx dx` `=int(1-z^(2))z^(5)(-dz)` `= -int(z^(5)-z^(7))dz` `= -((z^(6))/(6)-(z^(8))/(8))+c= -(cos^(6)x)/(6)+(cos^(8)x)/(8)+c` |
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| 21. |
`int(sqrt(1-x^(2))-x)/(sqrt(1-x^(2))(1+xsqrt(1-x^(2))))dx` isA. `2tan^(-1)(x+sqrt(1-x^(2)))+c`B. `tan^(-1)(x+sqrt(1-x^(2)))+c`C. `2tan^(-1)(x-sqrt(1-x^(2)))+c`D. `2cot^(-1)(x+sqrt(1-x^(2)))+c` |
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Answer» Correct Answer - A Let `x=sin theta, dx = cos theta d theta` `therefore" "i=int((cos theta-sin theta)cos theta d theta)/(cos theta(1+sin thetacostheta))` `=2int((cos theta-sin theta)d theta)/(1+(cos theta+sin theta))` `=2tan^(-1)(cos theta+sin theta)+c` `=2 tan^(-1)(x+sqrt(1-x^(2)))+c` |
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| 22. |
`intsqrt((x-1)/(x+1))dx` is equal toA. `sin^(-1)1/x+sqrt(x^(2)+1)/(x)+C`B. `sqrt(x^(2)-1)/(x)+cos^(-1)1/x+C`C. `sec^(-1)x-sqrt(x^(2)-1)/(x)+C`D. `tan^(-1)sqrt(x^(2)-1)/(x)+C` |
| Answer» Correct Answer - c | |
| 23. |
`intsin^-1sqrt(x/(a+x))dx`A. `(a+x)arc tansqrt(x/2)- sqrt(ax)+C`B. `(a+x)arc tansqrt(a/x)+sqrt(ax)+C`C. `(a-x)arc tansqrt(x/a)tansqrt(x/a)-sqrt(ax)+C`D. `(a+x)arc cotsqrt(x/a)-sqrt(ax)+C` |
| Answer» Correct Answer - a | |
| 24. |
Evaluate: `int(dx)/((x-1)xsqrt(x^(2)-1)` |
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Answer» `int(dx)/((x-1)xsqrt(x^(2)-1)) ("put" x-1=1/t rArr dx=-1/t^(2)dt)` `rArr I=int(-1/t^(2)dt)/(1/2sqrt(1/t+1)^(2)-(1/t+1)-1) = int(-dt)/sqrt(-t^(2)+t+1) = int(-dt)/sqrt(sqrt(5)/(2))^(2)-(t-1/2)^(2)` `=-sin^(-1)(t-1/2)/(sqrt(5)/2) + C=-sin^(-1)(2t-1)/(sqrt(5))+C`, where `t=1/(x-1)` |
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| 25. |
What is `int e^(x) (sqrtx + (1)/(2 sqrtx)) dx` equal to (where C is a constant of integration)A. `xe^(x) +C`B. `e^(x) (sqrtx) + C`C. `2e^(x) (sqrtx) + C`D. `2xe^(x) + C` |
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Answer» Correct Answer - B Let `I = int e^(x) (sqrtx + (1)/(2 sqrtx)) dx` `= int e^(x) sqrtx dx + int e^(x).(1)/(2 sqrtx) dx` `= e^(x).sqrtx - int e^(x).(1)/(2 sqrtx) dx + int e^(x) .(1)/(2 sqrtx) dx` `= e^(x).sqrtx + C` where C is constant of integration |
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| 26. |
The value of `int(dx)/(xsqrt(1-x^(3)))` is equal toA. `1/3ln |(sqrt(1-x^(2))-1)/(sqrt(1-x^(2))+1)|+C`B. `1/2"ln"|(sqrt(1-x^(2))+1)/(sqrt(1-x^(2))-1)|+C`C. `1/3ln|1/sqrt(1-x^(3))|+C`D. `1/3ln|1-x^(3)|+C` |
| Answer» Correct Answer - A | |
| 27. |
The value of `int((ax^2-b)dx)/(xsqrt(c^2x^2-(ax^2+b)^2))` is equal toA. `(1)/(c)sin^(-1)(ax+(b)/(x))+k`B. `c sin^(-1)(a+(b)/(x))+k`C. `sin^(-1)((ax+(b)/(x))/(c))+k`D. none of these |
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Answer» Correct Answer - C Let `I=int ((ax^(2)-b)dx)/(xsqrt(c^(2)x^(2)-(ax^(2)+b)^(2)))` `=int((a-(b)/(x^(2)))dx)/(sqrt(c^(2)-(ax+(b)/(x))^(2))), " "{("Put "ax+(b)/(x)=t),( :.(a-(b)/(x^(2)))dx=dt):}` `=int(dt)/(sqrt(c^(2)-t^(2)))` ` =sin^(-1)((t)/(c))+k` `=sin^(-1)((ax+(b)/(x))/(c))+C` |
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| 28. |
Evaluate `int(dx)/(cos xsqrt(cos2x))`. |
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Answer» `I=int(dx)/(cos xsqrt(cos2x))` `=int(cosx dx)/(cos^(2) xsqrt(1-2sin^(2)x))` `=int(cosx dx)/((1-sin^(2) x)sqrt(1-2sin^(2)x))` `=int(dt)/((1-t^(2)) sqrt(1-2t^(2))) " "("Putting " sinx=t)` Put `t=(1)/(y)` ` :. dt= -(1)/(y^(2))dy` ` :. I= -int(dy)/(y^(2)(1-(1)/(y^(2)))sqrt(1-(2)/(y^(2))))` `= -int(ydy)/((y^(2)-1)sqrt(y^(2)-2))` Now put `y^(2)-2=z^(2)` ` :. ydy=zdz` ` :.I= -int(dz)/(z^(2)+1)` `= -tan^(-1)z+c` `= -tan^(-1)sqrt(y^(2)-2)+c` `= -"tan"^(-1)sqrt((1)/(t^(2))-2)+c` `= -"tan"^(-1)sqrt((1)/(sin^(2)x)-2)+c` |
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| 29. |
The value of `int("lnt"|x|)/(xsqrt(1+ln|x|)`dx equals:A. `2/3sqrt((1+ln|x|)("ln|x|-2))+C`B. `2/3sqrt((1+"ln"|x|)(ln|x|+2))+C`C. `1/2sqrt((1+ln|x|)(ln|x|-2))+C`D. `2sqrt(1+ln|x|(3ln|x|=2)+C` |
| Answer» Correct Answer - a | |
| 30. |
The value of `int e^sqrtx/sqrtx(sqrtx + x) dx` is equal toA. `2e^(sqrt(x))-x+1]+C`B. `2e^(sqrt(x))[x-2sqrt(x)+1)]+C`C. `2e^sqrt(x)[x-sqrt(x)+1]+C`D. `2e^sqrt(x)(x+sqrt(x)+1)+C` |
| Answer» Correct Answer - c | |
| 31. |
The value of `int (dx)/((1+sqrtx)(sqrt(x-x^2)))` is equal toA. `(1+sqrt(x))/((1-x)^(2))+c`B. `(1+sqrt(x))/((1+x)^(2))+c`C. `(1-sqrt(x))/((1-x)^(2))+c`D. `(2(sqrt(x)-1))/(sqrt((1-x)))+c` |
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Answer» Correct Answer - D Let `I=int(dx)/((1+sqrt(x))sqrt((x-x^(2))))` If `sqrt(x)=sinp, " then " (1)/(2sqrt(x))dx=cos p dp` ` :. I=int (2sin p cos p dp)/((1+sinp)sinp cosp)` `=2int (dp)/((1+sin p))` `=2int((1-sinp)dp)/(cos^(2)p)` `=2{int sec^(2)p dp-int(tanp sec p)dp}` `=2(tanp-secp)+C` `=2(sqrt((x)/((1-x)))-(1)/(sqrt((1-x))))+C=(2(sqrt(x)-1))/(sqrt((1-x)))+C` |
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| 32. |
`int(1)/((1+sqrtx)sqrt(x-x^(2)))dx` is equal toA. `2(sqrt((x)/sqrt(1-x))-(1)/(sqrt(1-x)))+c`B. `2(sqrt((x)/sqrt(1-x))-(1)/(1-x))+c`C. `2(sqrt((x)/(1-x))-(1)/(sqrt(1-x)))+c`D. `2(sqrt((x)/(1-x))-(1)/(1-x))+c` |
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Answer» Correct Answer - C `Let int(1)/((1+sqrtx)sqrt(x-1))dx` Putting `x=sin^(2)t and dx=2 sin t cos t dt,` we get `I=int(2sint cos tdt)/((1+sint)sqrt(sin^(2)t-sin^(4)t))` `rArr" "I=2int(1)/(1+sin t)dt=2int(1-sint)/(cos^(2)t)dt` `=2int(sec^(2)t-sect tan t)dt` `=2(tan t-sec t)+c` `rArr" "2(sqrt((x)/(1-x))-(1)/(sqrt(1-x)))+c` |
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| 33. |
`int(dx)/(xsqrt(x^(6)+1))` equalsA. `sec^(-1)x^(3)+C`B. `(1)/(6)log((sqrt(x^(6)+1)-1)/(sqrt(x^(6)+1)+1))+C`C. `(1)/(3)log((sqrt(x^(3)+1)-1)/(sqrt(x^(3)+1)+1))+C`D. `(1)/(3)log((sqrt(x^(3)+1)+1)/(sqrt(x^(3)+1)-1))+C` |
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Answer» Correct Answer - B `I=int(dx)/(xsqrt(x^(6)+1))."Put "x^(6)+1=v^(2)` `rArr" "6x^(5)dx=2vdv` `therefore" "I=(1)/(2)int(v)/((v^(2)-1))dv` `=(1)/(3)int(dv)/(v^(2)-1)=(1)/(6)log((sqrt(x^(6)+1)-1)/(sqrt(x^(6)+1)+1))+C` |
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| 34. |
`int(dx)/(x^(2)sqrt(16-x^(2)))` has the value equal toA. `C-(1)/(4)tan^(-1)sec((x)/(4))`B. `(1)/(4)tan^(-1)sec((x)/(4))+C`C. `C-(sqrt(16-x^(2)))/(16x)`D. `(sqrt(16-x^(2)))/(16x)+C` |
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Answer» Correct Answer - C `I=int(1)/(x^(2)sqrt(16-x^(2)))dx` Put `x=(1)/(t),` `dx=-(1)/(t^(2))dt therefore I=int(-(1)/(t^(2))dt)/((1)/(t)xx(1)/(t^(2))sqrt(16t^(2)-1))=int(-tdt)/(sqrt(16t^(2)-1))` Let `16t^(2)-1=u^(2), 32tdt = 2u du,` `tdt=(u)/(16)du thereforeI=-(1)/(16)int(udu)/(u)=-(u)/(16)+C=-(sqrt(16-x^(2)))/(16x)+C` |
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| 35. |
`int(dx)/((1+sqrtx)^(2010))=2[(1)/(alpha(1+sqrtx)^(alpha))-(1)/(beta(1+sqrtx))^(beta)]+c` where `alpha, betagt0" then "alpha-beta` isA. 1B. 2C. `-1`D. `-2` |
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Answer» Correct Answer - A `int(dx)/((1+sqrtx)^(2010))` `=int(sqrtx)/(sqrtx(1+sqrtx)^(2010))dx` `=2int(t-1)/(t^(2010))dt(t=1+sqrtx)` `=2[(1)/(2009t^(2006))-(1)/(2008t^(2008))]+c` `rArr" "alpha=2009, beta=2008` |
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| 36. |
If `int(tan^(9)x)dx=f(x)+log|cosx|,` where f(x) is a polynomial of degree n in tan x, then the value of n isA. 6B. 7C. 8D. none of these |
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Answer» Correct Answer - C Let `I_(9)=int tan^(9)xdx=int tan^(7)x(sec^(2)x-1)dx` `"i.e., "I_(9)=(tan^(8)x)/(8)-I_(7)` `"Similarly, "I_(7)=(tan^(6)x)/(6)-I_(5)` ………………………………….. ……………………………….. ………………………………. `I_(3)=(tan^(2)x)/(2)-I_(1)=(tan^(2)x)/(2)-int tanxdx` `=(tan^(2)x)/(2)+log|cos x|` `therefore I_(9)=" (a polynomial of degree 8 in tan x)"+ log |cos x|+C` |
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| 37. |
`int (px^(p+2q-1)-qx^(q-1))/(x^(2p+2q)+2x^(p+q)+1) dx` is equal to(1) `-x^p/(x^(p+q)+1)+C` (2) `x^q/(x^(p+q)+1)+C` (3) `-x^q/(x^(p+q)+1)+C` (4) `x^p/(x^(p+q)+1)+C` A. ` -(x^(p))/(x^(p+q)+1)+C`B. ` (x^(q))/(x^(p+q)+1)+C`C. ` -(x^(q))/(x^(p+q)+1)+C`D. ` (x^(p))/(x^(p+q)+1)+C` |
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Answer» Correct Answer - C `int(px^(p+2q-1)-qx^(q-1))/((x^(p+q)+1)^(2))dx=int(px^(p-1)-qx^(q-1))/((x^(p)+x^(-q))^(2))dx " " ("Dividing "N^(r)" and "D^(r) " by "x^(2q))` `=int(dt)/(t^(2))= -(1)/(t)+C, " where " t=x^(p)+x^(-q)` `= - (1)/(x^(p)+x^(-q))+C` `= -(x^(q))/(x^(p+q)+1)+C` |
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| 38. |
Evaluate:`int"t"a n^(-1)sqrt(x)dx` |
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Answer» Correct Answer - `(x+1)tan^(-1)sqrt(x)-sqrt(x)+C` `int tan^(-1)sqrt(x)*1dx=(tan^(-1)sqrt(x))-int(1)/(1+x)(1)/(2sqrt(x))x dx " (Integrating by parts)" ` `=x tan^(-1)sqrt(x)-(1)/(2)int(sqrt(x)dx)/(1+x)` `=x tan^(-1)sqrt(x)-(1)/(2)[2(sqrt(x)-tan^(-1)sqrt(x))]+C` `=x tan^(-1)sqrt(x)-sqrt(x)+tan^(-1)sqrt(x)+C` `=(x+1)tan^(-1)sqrt(x)-sqrt(x)+C` |
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| 39. |
Evaluate:`int(e^x(2-x^2)dx)/((1-x)sqrt(1-x^2))` |
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Answer» Correct Answer - `e^(x)sqrt((1+x)/(1-x))+C` `int (e^(x)(2-x^(2)))/((1-x)sqrt(1-x^(2)))dx=int(e^(x)(1-x^(2)+1))/((1-x)sqrt(1-x^(2)))dx` `=int e^(x)[((1-x^(2)))/((1-x)sqrt(1-x^(2)))+(1)/((1-x)sqrt(1-x^(2)))]dx` `=int e^(x)[(1+x)/(sqrt(1-x^(2)))+(1)/((1-x)sqrt(1-x^(2)))]dx` `=int e^(x)[sqrt((1+x)/(1-x))+(d)/(dx)(sqrt((1+x)/(1-x)))]dx` `=e^(x)sqrt((1+x)/(1-x))+C` |
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| 40. |
Evaluate `int(log_(e)x)^(2)dx` |
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Answer» Correct Answer - `x*(log_(e)x)^(2)-2(xlog_(e)x-x)+c ` `int 1*(log_(e)x)^(2)dx` `=x*(log_(e)x)^(2)-int 2(log_(e)x)/(x)*xdx ` `=x*(log_(e)x)^(2)-2int log_(e)xdx ` `=x*(log_(e)x)^(2)-2(xlog_(e)x-x)+c ` |
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| 41. |
Evaluate `int sqrt(x^(2)+2x+5)dx` |
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Answer» Correct Answer - `(1)/(2)(x+1)sqrt(x^(2)+2x+5)+2log|(x+1)+sqrt(x^(2)+2x+5)|+C` `int sqrt(x^(2)+2x+5)dx` `=int sqrt((x+1)^(2)+4)dx` `=(1)/(2)(x+1)sqrt((x+1)^(2)+2^(2)) +(1)/(2)*(2)^(2)log|(x+1)+sqrt((x+1)^(2)+2^(2))|+C` `=(1)/(2)(x+1)sqrt(x^(2)+2x+5)+2log|(x+1)+sqrt(x^(2)+2x+5)|+C` |
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| 42. |
Evaluate:`int(2x)/((x^2+1)(x^2+2))dx` |
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Answer» Let `I=int (2x)/((x^(2)+1)(x^(2)+2))dx` Putting `x^(2)=t` and `2x dx=dt`, we get `I=int (dt)/((t+1)(t+2))` `=int((1)/(t+1)-(1)/(t+2))dt` `=log|t+1|-log|t+2|+C` `=log|x^(2)+1|-log|x^(2)+2|+C` |
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| 43. |
Integrate `(x^(2))/((a + bx)^(2))`. |
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Answer» Correct Answer - `(1)/(b^(3))(a+bx-2a "log"(a+bx)-(a^(2))/(a+bx)+c)` Let `I=(x^(2))/((a+bx)^(2))` Put `a+bx=t rArr "b dx" =dt` `therefore I=int((t-a)/(b))^(2)/t^(2)*(dt)/(b)=(1)/(b^(3))int((t^(2)-2 at + a^(2))/(t^(2)))dt` `=(1)/(b^(3))int(1-(2a)/(t)+(a^(2))/(t^(2)))dt` `(1)/(b^(3))(t-2 "a log t"-(a^(2))/(t))+c` `=(1)/(b^(3))(a+bx-2 "a log"(a+bx)-(a^(2))/(a+bx)+c)` |
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| 44. |
Integrate the curve `(x)/(1+x^(4))` |
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Answer» Correct Answer - `(1)/(2)"tan"^(-1)(x^(2))+c` Let `I=int (xdx)/(1+x^(4))=(1)/(2)int(2x)/(1+(x^(2))^(2))dx` Put `x^(2)=u rArr 2 xdx = du` `therefore" "I=(1)/(2) int(du)/(1+u^(2))=-(1)/(2)tan^(-1)(u)+x=(1)/(2)tan^(-1)(x^(2))+c` |
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| 45. |
Evaluate:`int(2x-1)/((x-1)(x+2)(x-3))dx` |
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Answer» Since all the factors in the denominator are linear, we have `int(2x-1)/((x-1)(x+2)(x-3))dx` `=int[(1)/((x-1)(3)(-2))+(-5)/((-3)(x+2)(-5))+(5)/((2)(5)(x-3 ))]dx` ` = -(1)/(6) log |x-1|-(1)/(3)log|x+2|+(1)/(2)log|x-3|+C` |
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| 46. |
Evaluate: `int(dx)/(sinx+sqrt(3)cosx)` |
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Answer» Let 1=`rcostheta` and `sqrt(3) = rsintheta rArr r=sqrt((1)^(2)+sqrt(3)^(2))=2` `tantheta=sqrt(3) rArr theta=pi/3` `therefore int(dx)(sinx+sqrt(3)cosx) = 1/rint(dx)/(sinxcostheta+cosxsintheta) = 1/rint(dx)/(sin(x+theta))` `=1/rint"cosec"(x+theta)dx = 1/r ln|tan(x/2+theta/2)|=1/2ln|tan(x/2+pi/6)|+C` |
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| 47. |
Integrate `sinx.sin2x.sin3x+sec^2x*cos^2 2x+sin^4x*cos^4x` |
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Answer» Correct Answer - `("cos"4x)/(16)-("cos"2x)/(8)+("cos"6x)/(24)+"sin"2x + "tan" x - 2x + (3x)/(128)-("sin"4x)/(128)+("sin"8x)/(1024)` Let `I_(1)=int sin x sin 2x sin 3"x dx"` `(1)/(4)int(sin 4x+sin 2x - sin 6x)dx` `=-(cos 4x)/(16)-(cos 2x)/(8)+(cos 6x)/(24)` `I_(2)=int sec^(2)x*cos^(2)2xdx` `=int sec^(2)x(2 cos^(2)x-1)^(2)dx` `=int(4 cos^(2)x+sec^(2)x-4)dx` `=int(2 cos 2x + sec^(2)x x-2)dx` `=sin 2x + tan x -2x and I_(3)=int sin^(4)x cos^(4)xdx` `=(1)/(128)int(3-4 cos 4x + cos 8x)dx` `=(3x)/(128)-(sin 4x)/(128)+(sin 8x)/(1024)` `therefore" "I=I_(1)+I_(2)+I_(3)` `=-(cos 4x)/(16)-(cos 2x)/(8)+(cos 6x)/(24)+sin 2x + tan x-2x + (3x)/(128)-(sin 4x)/(128)+(sin 8x)/(1024)` |
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| 48. |
Evaluate:`int(sinx)/(2+sin2x)dx` |
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Answer» `int(sinx)/(2+sin2x)dx` `=(1)/(2)int(sinx+cosx-(cosx-sinx))/(2+sin2x)dx` `=(1)/(2)int(sinx+cosx)/(2+sin2x)dx-(1)/(2)int(cosx-sinx)/(2+sin2x)dx` `=(1)/(2)int(sinx+cosx)/(3-(sinx-cosx)^(2))dx-(1)/(2)int(cosx-sinx)/(1+(sinx+cosx)^(2))dx` `=(1)/(2)int(dt)/(3-t^(2))-(1)/(2)int(du)/(1+u^(2)) " "{:((" where " t=sinx-cosx),(" and " u=sinx+cosx)):}` `=(1)/(2)(1)/(2sqrt(3))log|(sqrt(3)-t)/(sqrt(3)+t)|-(1)/(2)"tan"^(-1)u+c` `=(1)/(4sqrt(3))log|(sqrt(3)-(sinx-cosx))/(sqrt(3)+(sinx-cosx))|-(1)/(2)"tan"^(-1)(sinx+cosx)+c` |
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| 49. |
Evaluate: `int(sin2x)/(a^2sin^2x+b^2cos^2x) dx` |
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Answer» `(d)/(dx)(a^(2)sin^(2)x+b^(2)cos^(2)x)=(a^(2)-b^(2))sin2x` Now, `int(sin2x)/(a^(2)sin^(2)x+b^(2)cos^(2)x)dx` `=(1)/((a^(2)-b^(2)))int((a^(2)-b^(2))sin2x)/(a^(2)sin^(2)x+b^(2)cos^(2)x)dx` `=(1)/((a^(2)-b^(2)))int((d)/(dx)(a^(2)sin^(2)x+b^(2)cos^(2)x))/(a^(2)sin^(2)x+b^(2)cos^(2)x)dx` `=(1)/((a^(2)-b^(2)))log|a^(2)sin^(2)x+b^(2)cos^(2)x|+C` |
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| 50. |
Evaluate `int(1)/(sqrt(3)sinx+cosx)dx`. |
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Answer» `int(1)/(sqrt(3)sinx+cosx)dx` `=(1)/(2)int(1)/((sqrt(3))/(2)sinx+(1)/(2)cos x)dx` `=(1)/(2)int(1)/("sin"(x+(pi)/(6)))dx` `=(1)/(2)int "cosec"(x+(pi)/(6))dx` ` =(1)/(2)log_(e) |tan((x)/(2)+(pi)/(12))|+c` |
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