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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.
| 301. |
Obtain the reduction formula for `intcos^(n)xdx` |
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Answer» Let `I_(n) = intcos^(n)xdx` `I_(n)=intcosx(cosx)^(n-1)dx` `I_(n)=(sinx)(cosx)^(n-1)-int(n-1)(cosx)^(n-2)(-sinx)sinxdx` `I_(n)=(sinx)(cosx)^9n-1+(n-1)int(cosx)^(n-2)dx-(n-1)int(cosx)^(n)dx` `I_(n) = (sinx)(cosx)^(n-1)+(-1)I_(n-2)-(n-1)I_(n-2)` `I_(n)+(n-1)I_(n)=(sinx)(cosx)^(n-1)+(n-1)I-(n-2)` `I_(n)=((sinx)cosx))^(n-1)+(n-1)I_(n-2)` `I_(n) = ((sinx)(cosx)^(n-1))/(n) + (n-1)/(n)I-(n-2), nge2` |
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| 302. |
The relation between `I_(4,2)` and `I_(4,4)` isA. `I_(4,2)=1/3(sin^(5)xcos^(3)x+8I_(4,4))`B. `I_(4,2)=1/3(-sin^(5)xcos^(3)x+8I_(4,4))`C. `I_(4,2)=1/3(sin^(5)xcos^(3)x-8I_(4,4))`D. `I_(4,2)=1/3(sini^(5)xcos^(3)x+6I_(4,4))` |
| Answer» Correct Answer - B | |
| 303. |
Deduce the reduction formula for `I_(n)=int(dx)/(1+x^(4))^(n)` and Hence evaluate `I_(2)=int(x)/(1+x^(4))^(2)`. |
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Answer» `I_(n)=x/((4(n-1)(1+x^(4))^(n-1))+(4n-5)/(4(n-1))I_(n-1)` `I_(2)=x/(4(1+x^(4)) + (1/(2sqrt(2))tan^(-1)(x-1/x)/sqrt(2) -1/(4sqrt(2))"ln"(x+1/x-sqrt(2))/(x+1/x+sqrt(2)))+C` |
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| 304. |
If `intt(x)= Psi(x^(3))dx+C`A. `1/3[x^(3)Psi(x^(3))-intx^(2)Psi(x^(3))dx]+C`B. `1/3x^(2)Psi(x^(3)-3intx^(3)dx+C)`C. `1/3x^(3)Psi(x^(3))-intx^(2)Psi(x^(2))dx+C`D. `1/3[x^(3)Psi(x^(3))-intx^(3)Psi(x^(3))dx]+C` |
| Answer» Correct Answer - 3 | |
| 305. |
If `int f(x)dx=psi(x)`, then `int x^5f(x^3)dx`A. ` (1)/(3)[x^(3)Psi(x^(3))-int x^(2) Psi(x^(3))dx]+c`B. ` (1)/(3)x^(3)Psi(x^(3))-3int x^(3) Psi(x^(3))dx+c`C. ` (1)/(3)x^(3)Psi(x^(3))-int x^(2) Psi(x^(3))dx+c`D. ` (1)/(3)[x^(3)Psi(x^(3))-int x^(3) Psi(x^(3))dx]+c` |
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Answer» Correct Answer - C ` "Given that " int f(x)dx=Psi(x). " Let " I=int x^(5)f(x^(3))dx` `"or " I=int t f(t)(dt)/(3) " " ("Put " x^(3)=t)` `=(1)/(3)[t Psi(t)-int Psi(t)dt]+c` ` :. I=(x^(3))/(3) Psi(x^(3))-(1)/(3)int3x^(2)Psi(x^(3))dx+c` `=(x^(3))/(3)Psi(x^(3))-int x^(2) Psi(x^(3))dx+c` |
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| 306. |
`int(x^(3)-1)/(x^(3)+x)` dx is equal to:A. `x-log_(e)|x|+log_(e)(x^(2)+1)-tan^(-1)x+C`B. `x-log_(e)|x|+(1)/(2)log_(e)(x^(2)+1)-tan^(-1)x+C`C. `x+log_(e)|x|+(1)/(2)log_(e)(x^(2)+1)+tan^(-1)x+C`D. none of these |
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Answer» Correct Answer - B `int(x^(3)-1)/(x^(3)+x)dx=int(1+(1)/(1+x^(2))-(1)/(x)+(x)/(x^(2)+1))dx` `=x-tan^(-1)x-log_(e)|x|+(1)/(2)log_(e)(x^(2)+1)+C` |
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| 307. |
Evaluate: `intsqrt("tan theta d theta")` |
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Answer» Let `I=int sqrt(tan theta) d theta.` Let `tan theta=x^(2). " Then, " d(tan theta)=d(x^(2)) " or " sec^(2) theta d theta =2xdx` or ` d theta=(2x*dx)/(sec^(2) theta)=(2xdx)/(1+tan^(2) theta)=(2xdx)/(1+x^(4))` `I=int sqrt(x^(2))xx(2xdx)/(1+x^(4))` `=int(2x^(2))/(x^(4)+1)dx` `=int(2)/(x^(2)+1//x^(2))dx` `=int(1+1//x^(2)+1-1//x^(2))/(x^(2)+1//x^(2))dx` `=int(1+1//x^(2))/(x^(2)+1//x^(2))dx+int(1-1//x^(2))/(x^(2)+1//x^(2))dx` `=int(1+1//x^(2))/((x-1//x)^(2)+2)dx+int(1-1//x^(2))/((x+1//x)^(2)-2)dx` Putting `x-(1)/(x)=u` in first integral and `x+(1)/(x)=v` in second integral, we get `I=int(du)/(u^(2)+(sqrt(2))^(2))+int(dv)/(v^(2)-(sqrt(2))^(2))` ` =(1)/(sqrt(2))tan^(-1)((u)/(sqrt(2)))+(1)/(2sqrt(2))log|(v-sqrt(2))/(v+sqrt(2))|+C` `=(1)/(sqrt(2))tan^(-1)((x-1//x)/(sqrt(2)))+(1)/(2sqrt(2))log|(x+1//x-sqrt(2))/(x+1//x+sqrt(2))|+C` `=(1)/(sqrt(2))tan^(-1)((x^(2)-1)/(x sqrt(2)))+(1)/(2sqrt(2))log|(x^(2)-x sqrt(2)+1)/(x^(2)+x sqrt(2)+1)|+C` `=(1)/(sqrt(2))tan^(-1)((tan theta-1)/(sqrt(2tan theta)))+(1)/(2sqrt(2))log|(tan theta-sqrt(2tan theta)+1)/(tan theta+ sqrt(2tan theta)+1)|+C` |
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| 308. |
Evaluate:`int(x^2+4)/(x^4+16)dx` |
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Answer» `I=int(x^(2)+4)/(x^(4)+16)dx` `=int(1+(4)/(x^(2)))/(x^(2)+(16)/(x^(2)))dx` `=int(1+(4)/(x^(2)))/(x^(2)+((4)/(x))^(2)-8+8)dx` `=int(1+(4)/x^(2))/((x-(4)/(x))^(2)+8)dx` Let `x-(4)/(x)=t. " Then " d(x-(4)/(x))=dt " or " (1+(4)/(x^(2)))dx=dt.` Therefore, `I=int(dt)/(t^(2)+(2sqrt(2))^(2))=(1)/(2sqrt(2))tan^(-1)((t)/(2sqrt(2)))+C` `=(1)/(2sqrt(2))tan^(-1)((x-(4)/(x))/(2sqrt(2)))+C` `=(1)/(2sqrt(2))tan^(-1)((x^(2)-4)/(2x sqrt(2)))+C` |
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| 309. |
Evaluate: `int1/(sinx(2cos^(2)x-1))dx` |
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Answer» Putting `cosx=t`, we get `I=int(1/(sinx(2cos^(2)x-1)))dx=int(1/(sinx(2t^(2)-1))xx -(dt)/(sinx))=-int1/((1-t^(2)(2t^(2)-1))`dt `therefore I=-int(1/(1-t^(2))+2/(2t^(2)-1))dt=-int1/(1-t^(2))dt-2int1/(2t^(2)-1)dt` `=-1/2ln |(1+t)/(1-t)|-sqrt(2)/2ln |(sqrt(2)t-1)/(sqrt(2)t+1)|+C=-1/2ln|(1+cosx)/(1-cosx)|-1/sqrt(2)ln|(sqrt(2)cosx-1)/(sqrt(2)cosx+1)|+C` |
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| 310. |
Evaluate:`int(e^(2x)-2e^x)/(e^(2x)+1)dx` |
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Answer» Correct Answer - `(1)/(2)log(e^(2x)+1)-2 tan^(-1)(e^(x))+C` `int(e^(2x)-2e^(x))/(e^(2x)+1)dx=(1)/(2)int(2e^(2x))/(e^(2x)+1)dx-2int(e^(x)dx)/((e^(x))^(2)+1) ` `=(1)/(2)log(e^(2x)+1)-2int(dz)/(z^(2)+1), " where " z=e^(x) ` ` =(1)/(2)log(e^(2x)+1)-2 tan^(-1)(e^(x))+C` |
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| 311. |
The value of `int(cos^(3)x)/(sin^(2)x+sinx)`dx is equal to:A. `ln|sinx|+sinx+C`B. `"ln|sinx|-sinx+C`C. `-ln|" cosec "x|-sinx+C`D. `-ln|sinx|+sinx+C` |
| Answer» Correct Answer - BC | |
| 312. |
Evaluate: `intsqrt(x^(2)+2x+5)`dx |
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Answer» We have, `intsqrt(x^(2)+2x+5) = intsqrt(x^(2)+2x+1+4)dx=intsqrt((x+1)^(2)+2^(2))` `=1/2(x+1)sqrt((x+1)^(2)+2^(2)+1/2.(2)^(2)ln|(x+1)+sqrt(x^(2)+2x+5)|)+C` |
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| 313. |
Resolve `(2x-3)/((x-1)(x^(2)+1)^(2))` into partial fractions, |
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Answer» Let `(2x-3)/((x-1)(x^(2)+1)^(2))=A/(x-1)+(Bx+C)/(x^(2)+1)+(Dx+E)/(x^(2)+1)^(2)`. Then, `2x-3=A(x^(2)+1)^(2)+(Bx+C)(x-1)(x^(2)+1)+(Dx+E)(x-1)`…………(i) Putting x=1 in eq (i) , we get `-1=A(1+1)^(2) rArr A=-` Comparing coefficients of like powers of x on both side of (i), we have `A+B=0, C-B=0, 2A+B-C+D=0, C+E-B-D=2` and `A-C-E=-3`. Putting `A=-1/4` and solving these equations, we get `B=1/4=C,D=1/4` and `E=5/2 therefore (2x-3)/((x-1)(x^(2)+1)^(2))=(-1)/(4(x-1))+(x+1)/(4(x^(2)+1))+(9x+5)/(2(x^(2)+1)^(2))` |
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| 314. |
`int (x^2 -1 )/ (x^3 sqrt(2x^4 - 2x^2 +1))dx` is equal toA. `2 sqrt(2-(2)/(x^(2))+(1)/(x^(4)))+c`B. `2 sqrt(2+(2)/(x^(2))+(1)/(x^(4)))+c`C. `(1)/(2) sqrt(2-(2)/(x^(2))+(1)/(x^(4)))+c`D. None of these |
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Answer» Correct Answer - C Let `I=int((x^(2)-1)dx)/(x^(3)sqrt(2x^(4)-2x^(2)+1))` [dividing numerator and enominator by `x^(5)`] `=int(((1)/(x^(3))-(1)/(x^(5)))dx)/(sqrt(2-(2)/(x^(2))+(1)/(x^(4))))` `Put" "2-(2)/(x^(2))+(1)/(x^(4))=t` `rArr ((4)/(x^(3))-(4)/(x^(5)))dx=dt` `therefore I = (1)/(4)int(dt)/(sqrt(t))=(1)/(4)*(t^(1//2))/(1//2)+c` `=(1)/(2)sqrt(2-(2)/(x^(2))+(1)/(x^(4)))+c` |
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| 315. |
Evaluate:`inttan^(-1)sqrt((1-x)/(1+x))dx` |
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Answer» Correct Answer - `theta cos 2 theta-(sin2 theta)/(2)+C` ` I=int tan^(-1)sqrt((1-x)/(1+x))dx` ` "Let "x=cos2 theta " or "dx=-2sin2 theta d theta ` `:. I=int tan^(-1)sqrt((1-cos2 theta)/(1+cos2 theta))(-2sin2 theta)d theta ` `=-2int tan^(-1)(tan theta)sin2 theta d theta ` `=-2int theta sin 2 theta d theta ` `=-2[-(theta cos2 theta)/(2)+int(cos2 theta)/(2) d theta]` `=theta cos 2 theta-(sin2 theta)/(2)+C` |
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| 316. |
What is `int sin^(2) x dx + int cos^(2) x dx` equal to ? Where c is an arbitary constantA. `x + c`B. `(x^(2))/(2) + c`C. `x^(2) + c`D. None of these |
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Answer» Correct Answer - A `int sin^(2)x dx + int cos^(2) x dx = int (sin^(2) x + (cos^(2)x) dx` `int dx = x + c` |
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| 317. |
The value of `int("ln(tanx))/(sinxcosx)` dx is equal toA. `1/2"ln"^(2)(cotx)+C`B. `1/2"ln"^(2)(sec x)+C`C. `1/2"ln"^(2)(sinxsecx)+C`D. `1/2"ln"^(2)(cosx" cosec "x)+C` |
| Answer» Correct Answer - ACD | |
| 318. |
Evaluate: `int(dx)/sqrt(2-6x-9x^(2))`dx |
| Answer» `int (dx)/sqrt(2-6x-9x^(2))` dx `=int1/sqrt(3-(3x+1)^(2))dx=1/3sin^(-1)(3x+1)/(sqrt(3))+C` | |
| 319. |
Evaluate `int cos^(3)x sqrt(sinx)dx`. |
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Answer» [ Here, the power of cos x is 3, which is an odd positive integer. Therefore, put `dz=cos x dx`. Now, `int cos^(3)x sqrt(sinx)dx=int cos^(2)x sqrt(sinx)cos x dx` `=int(1-sin^(2)x)sqrt(sinx)cosx dx` `=int (1-z^(2))sqrt(z) dz` `=int (sqrt(z)-z^(5//2))dz` `=(z^(3//2))/(3//2)-(z^(7//2))/(7//2)+c=(2)/(3)z^(3//2)-(2)/(7)z^(7//2)+c` `=(2)/(3)"sin"^(3//2)x-(2)/(7)"sin"^(7//2)x+c` |
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| 320. |
If `int (4e^x+6e^-x)/(9e^x-4e^-x)dx=Ax+B ln (9e^(2x)-4)+C`, then |
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Answer» Correct Answer - `A = -(3)/(2), B = (35)/(36) and C in R` Given, `int (4e^(x)+6e^(-x))/(9e^(x) - 4e^(-x))dx = Ax + B log (9e^(2x)-4)+c` `LHS=int(4e^(2x)+6)/(9e^(2x)-4)dx` `Let" "4e^(2x)+6=A(9e^(2x)-4)+B(18e^(2x))` `rArr" "9A + 18B = 4 and -4A = 6` `rArr" "A = -(3)/(2) and B = (35)/(36)` `therefore int(A(9e^(2x)-4)+B(18e^(2x)))/(9e^(2x)-4)dx = A int 1 dx + B int (1)/(t) "dt where t" = 9e^(2x)-4` `=A x + B log(9e^(2x)-4)+c` `=-(3)/(2)x+(35)/(36)log(9e^(2x)-4)+c` `therefore" "A = -(3)/(2), B = (35)/(36)` and c = any real number |
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| 321. |
If `I=int(e^x)/(e^(4x)+e^(2e)+1) dx. J=int(e^(-x))/(e^(-4x)+e^(-2x)+1) dx.` Then for an arbitrary constant c, the value of `J-I` equal toA. `(1)/(2)log|(e^(4x)-e^(2x)+1)/(e^(4x)+e^(2x)+1)|+c`B. `(1)/(2)log|(e^(2x)+e^(x)+1)/(e^(2x)-e^(x)+1)|+c`C. `(1)/(2)log|(e^(2x)-e^(x)+1)/(e^(2x)+e^(x)+1)|+c`D. `(1)/(2)log|(e^(4x)+e^(2x)+1)/(e^(4x)-e^(2x)+1)|+c` |
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Answer» Correct Answer - C Since, `I = int(e^(x))/(e^(4x)+e^(2x)+1)dx and J = int (e^(3x))/(1+e^(2x)+e^(4x))dx` `therefore J - I = int((e^(3x)-e^(x)))/(1+e^(2x) + e^(4x))dx` Put `e^(x)=u rArr e^(x)dx=du` `therefore J-I = int((u^(2)-1))/(1+u^(2)+u^(4))du = int((1-(1)/(u^(2))))/(1+(1)/(u^(2))+u^(2))du` `= int ((1-(1)/(u^(2))))/((u+(1)/(u))^(2)-1)du` `Put" "u+(1)/(u)=t` `rArr" "(1-(1)/(u^(2)))du = dt` `" "=int(dt)/(t^(2)-1)=(1)/(2)log|(t-1)/(t+1)|+c` `" "=(1)/(2)log|(u^(2)-u+1)/(u^(2)+u+1)|+c` `" "=(1)/(2)log|(e^(2x)-e^(x)+1)/(e^(2x)+e^(x)+1)|+c` |
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| 322. |
What is `int e^(e^(x)) e^(x) dx` equal to ?A. `e^(e^(x)) + c`B. `2e^(e^(x)) + c`C. `e^(e^(x)) e^(x) + c`D. `2e^(e^(x)) e^(x) + c` |
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Answer» Correct Answer - A `I = int e^(x) e^(x) dx` Let `e^(x) = y rArr e^(x) dx = dy` `dx = (dy)/(e^(x))` `I = int e^(y) e^(x) (dy)/(e^(x)) = int e^(y) dy = e^(y) + c` `I = int e^(y) e^(x) (dy)/(e^(x)) = int e^(y) dy = e^(y) + c` `I = e^(e^(x)) + c` |
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| 323. |
Evaluate: `intcos2x" ln "(1+tanx)dx` |
| Answer» `1/2[sin2x."ln"(1+tanx)-x+"ln"|sinx+cosx|]+C` | |
| 324. |
The integral `intcos(log_(e)x)dx` is equal to: (where C is a constant of integration)A. `(x)/(2)[cos(log_(e)x)+sin(log_(e)x)]+C`B. `x[cos(log_(e)x)+sin(log_(e)x)]+C`C. `x[cos(log_(e)x)-sin(log_(e)x)]+C`D. `(x)/(2)[cos(log_(e)x)-sin(log_(e)x)]+C` |
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Answer» Correct Answer - A Let `I=int cos(log_(e)x)dx` `=x cos(log_(e)x)-intx(-sin(log_(e)x))(1)/(x)*dx" "["using integration by parts"]` `= x cos(log_(e)x)+int sin(log_(e)x)dx` `=x cos(log_(e)x)+ x sin(log_(e)x)-int x(cos(log_(e)x))(1)/(x)dx" "["again, using integration by parts"]` `rArr I = x cos(log_(e)x)+x sin(log_(e)x)-I` `rArr I=(x)/(2)[cos(log_(e)x)+sin(log_(e)x)]+C`. |
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| 325. |
What is `int(dx)/(2^(x) -1)` equal to ?A. `ln (2^(x) -1) + c`B. `(ln(1 -2^(-x)))/(ln 2) + c`C. `(ln (2^(-x) -1))/(2ln 2) + c`D. `(ln(1 + 2^(-x)))/(ln 2) + c` |
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Answer» Correct Answer - B `int (dx)/(2^(x) -1) =int (dx)/((1)/(2^(-x)))` `= int (2^(-x))/(1 -2^(-x)).dx` Let `1 - 2^(-x) = t` `rArr 2^(-x) .log 2 = (dt)/(dx) rArr 2^(-x) = (1)/(log 2).(dt)/(dx) rArr 2^(-x).dx = (dt)/(log 2)` `:. int (2^(-x))/(1 -2^(-x)) dx = (1)/(log 2) int (dt)/(t) = (1)/(log 2) (log t) + c` `= (1)/(log2) (log (1 - 2^(-x))) + c` |
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| 326. |
Evaluate:`int(sin^(-1)sqrt(x)-cos^(-1)sqrt(x))/(sin^(-1)sqrt(x)+cos^(-1)sqrt(x))dx`A. `-x+2/pi(2x-1)sin^(-1)sqrt(x)+2/pisqrt(x-x^(2))+C`B. `x-(4x)/(pi)cos^(-1)-2/pisin^(-1)sqrt(x)+2/sqrt(x)sqrt(1-x)+C`C. `-x+2/pi(2x+1)cos^(-1)sqrt(x)+2/pisqrt(x)sqrt(1-x)+C`D. `x-(4x)/pisin^(-1)sqrt(x)+C` |
| Answer» Correct Answer - AB | |
| 327. |
If `int (4e^x+6e^-x)/(9e^x-4e^-x)dx=Ax+B ln (9e^(2x)-4)+C`, thenA. A+18B=16B. 18B-A=19C. A-18B=17D. A+18B=32 |
| Answer» Correct Answer - AB | |
| 328. |
What is `int (x cos x + sin x) dx` equal to ? Where c is an arbitrary constantA. `x sin x + c`B. `x cos x + c`C. `-x sin x + c`D. `-x cos x + c` |
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Answer» Correct Answer - A `int (x cos x + sin x) dx = int x cos x dx + int sin x dx` `= x sin x - int sin x dx + int sin x dx = x sin x + c` |
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| 329. |
Let `alpha in (0, pi//2)` be fixed. If the integral `int("tan x" + "tan" alpha)/("tan x" - "tan" alpha)dx = A(x)"cos 2 alpha+B(x)` `"sin" 2 alpha +C`,where C is a constant of integration, then the functions A(x) and B(x) are respectively.A. `x+alpha "and log"_(e)|"sin"(x+alpha)|`B. `x-alpha "and log"_(e)|"sin"(x-alpha)|`C. `x-alpha "and log"_(e)|"cos"(x-alpha)|`D. `x+alpha "and log"_(e)|"sin"(x-alpha)|` |
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Answer» Correct Answer - B Let `I = int(tan x + tan alpha)/(tan x - tan alpha)dx, alpha in (0,(pi)/(2))` `= int((sinx)/(cosx)+(sin alpha)/(cos alpha))/((sinx)/(cosx)-(sin alpha)/(cos alpha))dx` `=int(sin x cos alpha + sin alpha cos x)/(sin x cos alpha - sin alpha cos x)dx` `=int(sin(x+alpha))/(sin(x-alpha))dx` Now, put `x-alpha = t rArr dx = dt, so` `I=int(sin(t+2 alpha))/(sin t)dt` `=int(sin t cos 2 alpha + sin 2 alpha cos t)/(sin t)dt` `=int(cos 2 alpha + sin 2 alpha (cos t)/(sin t))dt` `= t(cos 2 alpha)+(sin 2 alpha)log_(e)|sin t|+C` `=(x-alpha) cos 2 alpha +(sin 2 alpha)log_(e)|sin(x-alpha)|+C` `= A(x) cos 2 alpha + B(x) sin 2 alpha + C` (given) Now on comparing, we get `A(x) = x - alpha and B(x) = log_(e)|sin(x-alpha)|` |
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| 330. |
What is the equation of a curve passing through (0, 1) and whose differential equation is given by `dy = y tan x dx`?A. `y = cos x`B. `y = sin x`C. `y = sec x`D. `y = "cosec" x` |
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Answer» Correct Answer - C `dy = y tan x dx` `int (dy)/(y) = int tan x dx` `log |y| = log |sec x| + log |c|` `log |y| = log |c sec x|` `y = c sec x` at `x = 0, y = 1` `y = c` Soluttion is given by `y = sec x` |
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| 331. |
The integral `int(1+x-1/x)e^(x+1/x)dx` is equal toA. `(x-1)e^(x+(1)/(x))+c`B. `xe^(x+(1)/(x))+c`C. `(x+1)e^(x+(1)/(x))+c`D. `-xe^(x+(1)/(x))+c` |
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Answer» Correct Answer - B `int(1+x-(1)/(x))e^(x+(1)/(x))dx` `=int e^(x+(1)/(x))dx+int(1-(1)/(x^(2)))e^(x+(1)/(x))dx` `=int e^(x+(1)/(x))dx + xe^(x+(1)/(x))-int(d)/(dx)(x)e^(x+(1)/(x))dx` `=int e^(x+(1)/(x))dx + xe^(x+(1)/(x))-int e^(x+(1)/(x))dx" "[therefore int(1-(1)/(x^(2)))e^(x+(1)/(x))dx = e^(x+(1)/(x))]` `=int e^(x+(1)/(x))dx + xe^(x+(1)/(x))-int ex^(x+(1)/(x))dx = xe^(x+(1)/(x))+c` |
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| 332. |
If `intx^5e^(4x^2)dx=(1)/(48)e^(4x^2)(f(x))+c`, where c is contant of intergration then `f(x)` equals to (a) `-4x^3-1` (b) `-1-2x^3` (c) `4x^3+1` (d) `1-2x^3`A. `-4x^(3)-1`B. `4x^(3)+1`C. `-2x^(3)-1`D. `-2x^(3)+1` |
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Answer» Correct Answer - A Given, `int x^(5) e^(-4x^(3))dx=(1)/(48)e^(-4x^(3))f(x)+C` In LHS, put `x^(3) = t` `rArr" "3x^(2)dx=dt` So, `int x^(5)e^(-4x^(3))dx =(1)/(3)int te^(-4t)dt` `=(1)/(3)[t(e^(-4t))/(-4)-int (e^(-4t))/(-4)dt]" "["using integration by parts"]` `=(1)/(3)[(te^(4t))/(-4)+(e^(-4t))/(-16)]+C` `=-(1)/(48)e^(-4t)[4t+1]+C` `=-(e^(-4x^(3)))/(48)[4x^(3)+1]+C" "[therefore t=x^(3)]` `therefore" "f(x) = -1 -4X^(3)` (comparing with given equation) |
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| 333. |
The integral `int sec^(2//3) "x cosec"^(4//3)"x dx"` is equal to (here C is a constant of integration)A. `3tan^(-1//3)x+C`B. `-3tan^(-1//3)x+C`C. `-3cot^(-1//3)x+C`D. `-(3)/(4)tan^(-4//3)x+C` |
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Answer» Correct Answer - B Let `I=int sec^((2)/(3))x cos ec^((4)/(3))x dx = int(dx)/(cos^((2)/(3))x sin^((4)/(3))x) int(dx)/(((sin x)/(cos x))^((4)/(3))cos^((4)/(3)) x cos^((2)/(3))x)` [dividing and multiplying by `cos^(4//3)` x in denominator] `=int(dx)/(tan^((4)/(3)) x cos^(2) x)=int(sec^(2)xdx)/((tan x)^((4)/(3)))` Now, put `tan x = t rArr sec^(2) x dx = dt` `therefore I=int(dt)/(t^(4//3))=(t^((-4)/(3)+1))/((-4)/(3)+1)+C` `=-3(1)/(t^((1)/(3)))+C =(-3)/((tan x)^((1)/(3)))+C=-3tan^(-(1)/(3))x+C` |
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| 334. |
The integral `int (dx)/(a cos x + b sin x)` is of the form `(1)/(r) " In" [tan ((x + alpha)/(2))]` What is r equal to ?A. `a^(2) + b^(2)`B. `sqrt(a^(2) + b^(2))`C. `a + b`D. `sqrt(a^(2) + b^(2))` |
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Answer» Correct Answer - B Given that, `int (dx)/(a cos x + b sin x) = (1)/(r) ln [tan ((x + alpha)/(2))]` Let `a = r sin alpha, b = r cos alpha` `int (dx)/(r sin alpha cos x + r cos alpha sin x) = (1)/(r) int (1)/(sin (x + alpha))` `= (1)/(r) int "cosec"(x + alpha) dx = (1)/(r) ln [tan((x + alpha)/(2))]` `a = r sin alpha rArr a^(2) = r^(2) sin^(2) alpha`...(i) `b = r cos alpha rArr b^(2) = r^(2) cos^(2) alpha`...(ii) Adding (i) and (ii), we get `r^(2) = a^(2) + b^(2)` `rArr r= sqrt(a^(2) + b^(2))` |
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| 335. |
`int(x+x^(2/3)+x^(1/6))/(x(1+x^(1/3)))dx` equalsA. `(3x^((2)/(3)))/(4)+6tan^(-1)(x^((1)/(6)))+C`B. `(3x^((2)/(3)))/(2)+6tan^(-1)(x^((1)/(6)))+C`C. `(3x^((2)/(3)))/(10)+6tan^(-1)(x^((1)/(6)))+C`D. `(3x^((2)/(3)))/(5)+6tan^(-1)(x^((1)/(6)))+C` |
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Answer» Correct Answer - B Put `x=u^(6)` `therefore" "I=6int(u^(5)+u^(3)+1)/(1+u^(2))du` `=6intu^(3)du+6int(du)/(1+u^(2))` `=(3u^(4))/(2)+6 tan^(-1)u+C` `=(3x^((2)/(3)))/(2)+6 tan^(-1)(x^((1)/(6)))+C` |
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| 336. |
Consider `int x tan^(-1) x dx = A (x^(2) + 1) tan^(-1) x + Bx + C`, where C is the constant of integration What is the value of A ?A. 1B. `1//2`C. `-1//2`D. `1//4` |
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Answer» Correct Answer - B Given, `int x tan ^(-1) x dx = A(x^(2) + 1) tan^(-1) x + Bx + C` where, C is the constant of integration Consider, `int underset(II)(x) underset(I)(tan)^(-1) x dx` `= tan^(-1) x.(x^(2))/(2) - int (d)/(dx) (tan^(-1)x) .(x^(2))/(2) dx` (using integration by parts) `= (x^(2).tan^(-1)x)/(2) - (1)/(2) int (x^(2))/(1 + x^(2)) dx` `= (x^(2) tan^(-1) x)/(2) - (1)/(2) (int ((1 + x^(2) -1)/(1 + x^(2)))dx)` `= (x^(2) tan^(-1) x)/(2) - (1)/(2) (int dx - int (dx)/(1 + x^(2)))` `= (x^(2) tan^(-1) x)/(2) - (1)/(2) (x - tan^(-1) x) + C` `= (x^(2) tan^(-1)x)/(2) - (x)/(2) + (tan^(-1)x)/(2) + C` `= (1)/(2) (x^(2) + 1 ) tan^(-1) x - (x)/(2) + C` `A = (1)/(2)`, hence option (b) is correct |
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| 337. |
The integral `int(2x^(3)-1)/(x^(4)+x)dx` is equal to (here C is a constant of intergration)A. `(1)/(2)"log"_(e)(|x^(3)+1|)/(x^(2))+C`B. `(1)/(2)"log"_(e)(|x^(3)+1|^(2))/(|x^(3)|)+C`C. `"log"_(e)|(x^(3)+1)/(x)|+C`D. `"log"_(e)(|x^(3)+1|)/(x^(2))+C` |
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Answer» Correct Answer - C Key Idea (i) Divide each term of numerator and denominator by `x^(2)`. (ii) Let `x^(2)+(1)/(x) = t` Let integral is `I = int(2x^(3)-1)/(x^(4)+x)dx = int(2x-1//x^(2))/(x^(2)+(1)/(x))dx` [dividing each term of numerator and denominator by `x^(2)`] Put `x^(2)+(1)/(x)=t rArr (2x+(-(1)/(x^(2))))dx = dt` `therefore I = int(dt)/(t)=log_(e)|(t)|+C` `=log_(e)|(x^(2)+(1)/(x))|+C` `=log_(e)|(x^(3)+1)/(x)|+C` |
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| 338. |
Consider `int x tan^(-1) x dx = A (x^(2) + 1) tan^(-1) x + Bx + C`, where C is the constant of integration What is the value of B ?A. 1B. `1//2`C. `-1//2`D. `1//4` |
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Answer» Correct Answer - C Given, `int x tan ^(-1) x dx = A(x^(2) + 1) tan^(-1) x + Bx + C` where, C is the constant of integration Consider, `int underset(II)(x) underset(I)(tan)^(-1) x dx` `= tan^(-1) x.(x^(2))/(2) - int (d)/(dx) (tan^(-1)x) .(x^(2))/(2) dx` (using integration by parts) `= (x^(2).tan^(-1)x)/(2) - (1)/(2) int (x^(2))/(1 + x^(2)) dx` `= (x^(2) tan^(-1) x)/(2) - (1)/(2) (int ((1 + x^(2) -1)/(1 + x^(2)))dx)` `= (x^(2) tan^(-1) x)/(2) - (1)/(2) (int dx - int (dx)/(1 + x^(2)))` `= (x^(2) tan^(-1) x)/(2) - (1)/(2) (x - tan^(-1) x) + C` `= (x^(2) tan^(-1)x)/(2) - (x)/(2) + (tan^(-1)x)/(2) + C` `= (1)/(2) (x^(2) + 1 ) tan^(-1) x - (x)/(2) + C` `B = - (1)/(2)`, hence option (c) is correct |
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| 339. |
What is `int e^("In"(tan x)) dx` equal to ?A. In `|tan x| + c`B. In `|sec x| + c`C. `tan x + c`D. `e^(tan x) + c` |
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Answer» Correct Answer - B `int e^(l n(tan x)) dx = int tan xdx` `= l n |sec x| + c` |
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| 340. |
If `int(x^2-x+1)/((x^2+1)^(3/2))e^x dx=e^xf(x)+c ,t h e n``f(x)`is an even function`f(x)`is a bounded functionthe range of `f(x)`is `(0,1)``f(x)`has two points of extremaA. f(x) is a an even functionB. f(x) is a bounded functionC. Range of f(x) is (0,1)D. f(x) has two points of extreme. |
| Answer» Correct Answer - ABC | |
| 341. |
`"If" int(dx)/((x^(2)-2x+10)^(2))=A("tan"^(-1)((x-1)/(3))+(f(x))/(x^(2)-2x+10))+C`,where, C is a constant of integration, thenA. `A=(1)/(27)andf(x)=9(x-1)`B. `A=(1)/(81)andf(x)=3(x-1)`C. `A=(1)/(54)andf(x)=3(x-1)`D. `A=(1)/(54)andf(x)=9(x-1)^(2)` |
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Answer» Correct Answer - C Let `I+int(dx)/((x^(2)-2x+10)^(2))=int(dx)/(((x-1)^(2)+3^(2))^(2))` Now, put `x-1 = 3 tan theta rArr dx = 3 sec^(2)theta d theta` So, `I = int(3sec^(2)theta d theta)/((3^(2)tan^(2)theta+3^(2))^(2))=int(3sec^(2)theta d theta )/(3^(4)sec^(4)theta)` `=(1)/(27)intcos^(2)theta d theta = (1)/(27)int(1+cos 2 theta)/(2)d theta` `" "[therefore cos^(2)theta=(1+cos2theta)/(2)]` `=(1)/(54)int(1+cos 2 theta)d theta = (1)/(54)(theta+(sin 2 theta)/(2))+C` `=(1)/(54)tan^(-1)((x-1)/(3))+(1)/(108)((2 tan theta)/(1 + tan^(2)theta))+C` `" "[therefore sin 2 theta = (2 tan theta)/(1+tan^(2)theta)]` `=(1)/(54)tan^(-1)((x-1)/(3))+(1)/(54)(((x-1)/(3)))/(1+((x-1)/(3))^(2))+C` `=(1)/(54)tan^(-1)((x-1)/(3))+(1)/(18)((x-1)/((x-1)^(2)+3^(2)))+C` `=(1)/(54)tan^(-1)((x-1)/(3))+(1)/(18)((x-1)/(x^(2)-2x+10))+C` `=(1)/(54)[tan^(-1)((x-1)/(3))+(3(x-1))/(x^(2)-2x + 10)]+C` It is given, that `I=A[tan^(-1)((x-1)/(3))+(f(x))/(x^(2)-2x + 10)]+C` On comparing, we get `A=(1)/(54)and f(x) = 3(x-1)`. |
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| 342. |
The value of the integral `int((1-costheta)^(2/7))/((1+costheta)^(9/7))dthetai s``7/(11)(t a ntheta/2)^((11)/7)+C`(b) `7/(11)(costheta/2)^((11)/7)+C``7/(11)(s intheta/2)^((11)/7)+C`(d)none of theseA. `(7)/(11)("tan"(theta)/(2))^((11)/(7))+C`B. `(7)/(11)("cos"(theta)/(2))^((11)/(7))+C`C. `(7)/(11)("sin"(theta)/(2))^((11)/(7))+C`D. none of these |
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Answer» Correct Answer - A Let `I=int((1-cos theta)^(2//7))/((1+cos theta)^(9//7))d theta` `=int((2sin^(2)theta//2)^(2//7))/((2cos^(2)theta//2)^(9//7))d theta` `=(1)/(2)int((sin theta//2)^(4//7))/((cos theta//2)^(18//7)) d theta` Put `(theta)/(2)=t" or " (d theta)/(2)=dt` ` :. I=int((sint)^(4//7))/((cost)^(18//7)) dt " "("Here, "m+n= -2)` `=int(tant)^(4//7)sec^(2)t dt` Put `tan t=u" or "sec^(2)t dt=du` ` :. I=int u^(4//7)du=(u^(11//7))/(11//7)+c=(7)/(11)(tan t)^(11//7)+C` `=(7)/(11)("tan"(theta)/(2))^(11//7)+C` |
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| 343. |
`inttanx.tan2x.tan3xdx`=A. `-ln|cosx|-1/2ln|sec2x|+1/3ln|sec3x|+C`B. `-ln|secx|-1/2ln|sec2x|+1/3ln|sec3x|+C`C. `ln|cosx|+ln|cos2x|+ln|cos3x|+C`D. `ln|secx|+1/2ln|sec2x|+1/3|sec3x|+C` |
| Answer» Correct Answer - b | |
| 344. |
What is `int sin ^(3) x cos x dx` equal to ? Where c is the constant of integrationA. `cos^(4) x + c`B. `sin^(4) x + c`C. `((1 - sin^(2)x)^(2))/(4) + c`D. `((1 - cos^(2) x)^(2))/(4) + c` |
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Answer» Correct Answer - D Let `t = sin x rArr dt = cos x dx` `int sin^(3) x cos x dx = int t^(3) dt = (t^(4))/(4) + c = (sin^(4)x)/(4) + c` `= ((1 - cos^(2)x)^(2))/(4) + c` |
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| 345. |
`int(cosx-sinx+1-x)/(e^(x)+sinx+x)dx=log_(e)(f(x))+g(x)+C` where C is the constant of integration and f(x) is positive. Then `f(x)+g(x)` has the value equal toA. `e^(x)+sinx+2x`B. `e^(x)+sinx`C. `e^(x)-sinx`D. `e^(x)+sinx+x` |
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Answer» Correct Answer - B `I=int((e^(x)+cosx+1)-(e^(x)+sinx+x))/(e^(x)+sinx+x)` `=log_(e)(e^(x)+sinx+x)-x+C` `therefore" "f(x)=e^(x)+sin x+x and g(x)=-x` `" "f(x)+g(x)=e^(x)+sinx` |
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| 346. |
If `int x^(26).(x-1)^(17).(5x-3)dx=(x^(27).(x-1)^(18))/(k)+C` where C is a constant of integration, then the value of k is equal toA. 3B. 6C. 9D. 12 |
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Answer» Correct Answer - C Differentiating both sides gives `x^(26).(x-1)^(17)(5x-3)=(1)/(k)[x^(27).18(x-1)^(17)+(x-1)^(18)27x^(26)]` `=(x^(26)(x-1)^(17))/(k)[18x+27(x-1)]` `=(x^(26)(x-1)^(17))/(k)(45x-27)` `=9(x^(26)(x-1)^(17))/(k)(5x-3)` `rArrk=9` |
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| 347. |
If `int sqrt(1-x^2)/x^4dx=A(x).(sqrt(1-x^2))^m` where `A(x)` is a function of `x` then `(A(x))^m`= (A) `-1/(27x^9)` (B) `1/(27x)^9` (C) `1/(3x^9)` (D) `-1/(3x^9)`A. `(1)/(9x^(4))`B. `(-1)/(3x^(3))`C. `(-1)/(27x^(9))`D. `(1)/(27x^(0))` |
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Answer» Correct Answer - C ....(i) We have, `int sqrt(1-x^(2))/(x^(4))dx = A(x)(sqrt(1-x^(2))^(m)+ C` Let `I = int sqrt(1-x^(2))/(x^(4))dx = int sqrt(x^(2)((1)/(x^(2))-1))/x^(4)dx` `= int (x sqrt((1)/(x^(2))-1))/(x^(4))dx = int (1)/(x^(3))sqrt((1)/(x^(2))-1)dx` `=int(x sqrt((1)/(x^(2))-1))/(x^(4))dx=int(1)/(x^(3))sqrt((1)/(x^(2))-1)dx` Put `(1)/(x^(2))-1 = t^(2) rArr (-2)/(x^(3))dx = 2t dt rArr (1)/(x^(3))dx = - t dt` `therefore I = -intt^(2)dt = -(t^(3))/(3)+C` `= -(1)/(3)((1-x^(2))/(x^(2)))^(3//2)+C [therefore t = ((1)/(x^(2))-1)^(1//2)]` ...(ii) `= -(1)/(3)(1)/(x^(3))(sqrt(1-x^(2)))^(3)+C` On comparing Eqs. (i) and (ii), we get `A(x) = -(1)/(3x^(3)) and m = 3` `therefore (A(x))^(m) = (A(x))^(3) = -(1)/(27x^(9))` |
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| 348. |
`"If"int(dx)/(x^(3)(1+x^(6))^(2/3))=xf(x)(1+x^(6))^(1/3)+C` where, C is a constant of integration, then the function f(x) is equal toA. `-(1)/(6x^(3))`B. `-(1)/(2x^(3))`C. `-(1)/(2x^(2))`D. `(3)/(x^(2))` |
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Answer» Correct Answer - B Let `I = int(dx)/(x^(3)(1+x^(6))^(2//3))` `=int(dx)/(x^(3)x^(4)((1)/(x^(6))+1)^(2//3))=int(dx)/(x^(7)((1)/(x^(6))+1)^(2//3))` Now, put `(1)/(x^(6))+1 = t^(3)` `rArr" " -(6)/(x^(7))dx = 3t^(2)dt` `rArr" "(dx)/(x^(7))=-(t^(2))/(2)dt` So, `I=int(-(1)/(2)t^(2)dt)/(t^(2))= -(1)/(2)int dt` `= -(1)/(2)t+C = -(1)/(2)((1)/(x^(6))+1)^(1//3)+C" "[therefore t^(3)=(1)/(x^(6))+1]` `= -(1)/(2)(1)/(x^(2))(1+x^(6))^(1//3)+C` `= x*f(x)*(1+x^(6))^(1//3)+C" "["given"]` On comparing both sides, we get `f(x) = -(1)/(2x^(3))` |
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| 349. |
Evaluate: `inte^x(1+n x^(n-1)-x^(2n))/((1-x^n)sqrt(1-x^(2n)))dx` |
| Answer» `e^(x)sqrt((1+x^(n))/(1-x^(n))+C` | |
| 350. |
The value of the integral `int(x^2+x)(x^(-8)+2x^(-9))^(1/(10))dx`is`5/(11)(x^2+2x)^((11)/(10))+c`(b) `5/6(x+1x)^((11)/(10))+c``6/7(x+1)^((11)/(10))+c`(d) none oftheseA. `(5)/(11)(x^(2)+2x)^(11//10)+c`B. `(5)/(6)(x+1)^(11//10)+c`C. `(6)/(7)(x+1)^(11//10)+c`D. none of these |
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Answer» Correct Answer - A Given that `I=int(x^(2)+x)(x^(-8)+2x^(-9))^(1//10) dx` `=int (x+1)(x^(2)+2x)^(1//10)dx` Now put `x^(2)+2x=t " and "(x+1)dx=(dt)/(2)` ` :. I=int t^(1//10)(dt)/(2)=(1)/(2)xx(10)/(11)t^(11//10)+C=(5)/(11)t^(11//10)+C` ` =(5)/(11)(x^(2)+2x)^(11//10)+C` |
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