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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.
| 251. |
What is `int tan^(2) x sec^(4) x dx` equal to ?A. `(sec^(5)x)/(5) + (sec^(3) x)/(3) + c`B. `(tan^(5)x)/(5) + (tan^(3)x)/(3) + c`C. `(tan^(5) x)/(5) + (sec^(3)x)/(3) + c`D. `(sec^(5)x)/(5) + (tan^(3)x)/(3) + c` |
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Answer» Correct Answer - B Let `I = int tan^(2) x sec^(4) x dx` Let `tan x = t` `rArr sec^(2) x dx = dt` `:. I = int tan^(2) x.sec^(2) x.sec^(2) x.dx` `= int tan^(2) x (1 + tan^(2)x) sec^(2) x.dx` `:. I = int t^(2) (1 + t^(2)) dt = int (t^(2) + t^(4)) dt` `= (t^(5))/(5) + (t^(3))/(3) + c = (tan^(5) x)/(5) + (tan^(3)x)/(3) + c` |
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| 252. |
What is `int sin x log (tan x) dx` equal to ?A. `cos x log tan x + log tan (x//2) + c`B. `-cos x log tan x + log tan (x//2) + c`C. `cos x log tan x + log cot (x//2) + c`D. `-cos x log tan x + log cot (x//2) + c` |
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Answer» Correct Answer - B `int sin x log (tan x) dx` `= - cos x log tan x - int (-cos x) (1)/(tan x). Sec^(2) x dx` `= - cos x log tan x + int (1)/(sin x) dx` `= - cos x log (tan x) + int (1 + "tan"^(2) (x)/(2))/(2 "tan "(x)/(2)) dx` Let `t = "tan"(x)/(2)` `rArr (dx)/(dt) = (2)/(1 + t^(2)) rArr dx = (2)/(1 + t^(2)).dt` So, `- cos x. log (tan x) + int (1 + "tan"^(2) (x)/(2))/(2 "tan"(x)/(2)).dx` `= - cos x. log (tan x) + int (1 + t^(2))/(2t).(2)/(1 + t^(2))dt` `= - cos x log tan x + int (1)/(t) .dt` `= - cos x log tan x + log (t) + c` `= - cos x log tan x + log tan ((x)/(2)) + c` |
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| 253. |
What is `int (d theta)/(sin^(2) theta + 2 cos^(2) theta -1)` equal to ?A. `tan theta + c`B. `cot theta + c`C. `(1)/(2) tan theta + c`D. `(1)/(2) cot theta + c` |
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Answer» Correct Answer - A Let `I = int (d theta)/(sin^(2) theta + 2 cos^(2) theta - 1)` `= int (d theta)/(1- cos^(2) theta + 2 cos^(2) theta - 1) = int (d theta)/(cos^(2) theta)` `= int sec^(2) theta d theta = tan theta + c` |
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| 254. |
`1f I=int(dx)/((2a x+x^2)^(3/2))` |
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Answer» Correct Answer - `-(x+a)/(a^(2)sqrt(2ax+x^(2)))+C` `2ax+x^(2)=(x+a)^(2)-a^(2).` Now, put `x+a=a sec theta` ` :. dx= a sec theta tan theta d theta` ` :. int (dx)/((2ax+x^(2))^(3//2))=int(a sec theta tan theta)/(a^(3) tan^(3) theta)d theta` `=(1)/(a^(2))int(cos theta)/(sin^(2) theta) d theta` `= -(1)/(a^(2)sin theta)+C` `= -(1)/(a^(2)sqrt(1-(a^(2))/((x+a)^(2))))+C` `=-(x+a)/(a^(2)sqrt(2ax+x^(2)))+C` |
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| 255. |
Evaluate:`int(sqrt(x)dx)/(1+x)` |
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Answer» Correct Answer - `2sqrt(x)-2tan^(-1)sqrt(x)+C` ` " Put " x=t^(2) or dx=2tdt.` Then, `int(sqrt(x)dx)/(1+x)=2int(t^(2)dt)/(1+t^(2)) ` ` =2int(1-(1)/(1+t^(2)))dt ` `=2(t-tan^(-1)t)+C` ` =2sqrt(x)-2tan^(-1)sqrt(x)+C` |
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| 256. |
Evaluate:`int(x^2tan^(-1)x^3)/(1+x^6)dx` |
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Answer» Correct Answer - `((tan^(-1)x^(3))^(2))/(6)+C ` `I= int tan^(-1)x^(3)*x^(2)/(1+x^(6))dx` ` " Let " tan^(-1)x^(3)=t` ` :. dt=(3x^(2))/(1+x^(6))dx ` ` :. I=(1)/(3)int tdt=(t^(2))/(6)+C=((tan^(-1)x^(3))^(2))/(6)+C ` |
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| 257. |
Evaluate: `int(x^2tan^(-1)x)/(1+x^2) dx` |
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Answer» Let `x=tan theta.` ` :. dx=sec^(2) theta d theta` ` :. I=int (x^(2)tan^(-1)x)/(1+x^(2))dx` `=int (theta tan^(2) theta)/(1+tan^(2)theta)*sec^(2) theta d theta` `= int theta tan^(2) theta d theta` `=int theta (sec^(2) theta-1) d theta` `=int theta sec^(2) theta d theta - int theta d theta` `=theta tan theta -int 1*tan theta d theta-(theta^(2))/(2) +c` `=theta tan theta -log_(e)|cos theta|-(theta^(2))/(2)+c, " where " theta = tan^(-1)x`. |
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| 258. |
Evaluate ` int 2x^(3)e^(x^(2))dx`. |
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Answer» `I= int x^(2)e^(x^(2))(2x)dx` Let `x^(2)=t` ` :. I=int t e^(t)dt` `=t*e^(t)-int 1*e^(t)dt` `=t*e^(t)-e^(t)+c` `=x^(2)*e^(x^(2))-e^(x^(2))+c` |
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| 259. |
Evaluate ` int x log x dx`. |
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Answer» `int underset (II)(x)underset(I)(log)x dx` `=logx {int x dx}-int {(d)/(dx)(logx)intxdx}dx` `=(logx)(x^(2))/(2)int(1)/(x)(x^(2))/(2)dx` `=(x^(2))/(2)log x-(1)/(2)int x dx` `=(x^(2))/(2)logx-(1)/(2)((x^(2))/(2))+C=(x^(2))/(2)logx-(1)/(4)x^(2)+C` |
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| 260. |
Evaluate:`int(a x^3+b x)/(x^4+c^2)dx` |
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Answer» Correct Answer - `(a)/(4)log(x^(4)+c^(2))+(b)/(2)(1)/(c)"tan"^(-1)(x^(2))/(c)+k` `I=int(ax^(3)+bx)/(x^(4)+c^(2))dx ` `=int[(ax^(3))/(x^(4)+c^(2))+(bx)/(x^(4)+c^(2))]dx` `=(a)/(4)int(4x^(3))/(x^(4)+c^(2))dx+(b)/(2)int(2x)/(x^(4)+c^(2))dx ` ` " " I_(1) " "+ " "I_(2)` `=(a)/(4)log(x^(4)+c^(2))+(b)/(2)int(dt)/(t^(2)+c^(2)) " " (" in " I_(2), " put " x^(2)=t)` ` =(a)/(4)log(x^(4)+c^(2))+(b)/(2)(1)/(c)"tan"^(-1)(x^(2))/(c)+k` |
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| 261. |
Evaluate:`int(log(1+1/x))/(x(1+x))dx` |
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Answer» Correct Answer - ` -((log(1+(1)/(x)))^(2))/(2) + C` Let `log(1+(1)/(x))=t` ` :. (0-(1)/(x^(2)))/(1+(1)/(x)) dx=dt ` ` implies (-dx)/(x(x+1))=dt ` ` implies int(log(1+(1)/(x)))/(x(1+x))dx=-int t dt ` `=- (t^(2))/(2) +c ` ` = -((log(1+(1)/(x)))^(2))/(2) + c` |
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| 262. |
What is `int (e^(x) + 1)^(-1) dx` equal to?A. `ln (e^(x) + 1) + c`B. `ln (e^(-x) + 1) + c`C. `-l (e^(-x) + 1) + c`D. `-(e^(x) + 1) + c` |
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Answer» Correct Answer - C Let I `= int (e^(x) + 1)^(-1) dx = int (1)/(e^(x) + 1) dx = int (e^(-x))/(1 + e^(-x)) dx` Let `1 + ^(-x) =t rArr -e^(-x) dx = dt` `:. I = - int (1)/(t) dt = - log t + c = - log (1 + e^(-x)) + c` |
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| 263. |
If `int(dx)/(f(x)) = log {f(x)}^(2) + c`, then what is f(x) equal to ?A. `2x + alpha`B. `x + alpha`C. `(x)/(2) + alpha`D. `x^(2) + alpha` |
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Answer» Correct Answer - C We check from the given option one by one. Options (a) and (b) do not satisfy. We check option (c) Let `f(x) = (x)/(2) + alpha` `:. int (dx)/((x)/(2) + alpha) = int (2dx)/((x + 2 alpha))` `=2 log (x + 2 alpha) + c_(1) = log (x + 2alpha)^(2) + c_(1)` `= log ((x)/(2) + alpha)^(2) + log 2^(2) + c_(1) = log ((x)/(2) + alpha)^(2) + c` |
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| 264. |
Evaluate: (i) `int(sinsqrt(x))/(sqrt(x)) dx`(ii) `int((x+1)e^x)/(sin^2(x e^x)) dx`A. `-e^(x) cot x + c`B. `cos^(2) (x e^(x)) + c`C. `log sin (x e^(x)) + c`D. `- cot (x e^(x)) + c` |
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Answer» Correct Answer - D Let the given integral be `I = int (e^(x) (1 + x))/(sin^(2) (xe^(x))) dx` Put `x e^(x) = t and e^(x) (1 + x) dx = dt` `rArr I int (dt)/(sin^(2)t) = int "cosec"^(2) t dt` `= - cot t + c = - cot (x e^(x)) + c` |
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| 265. |
What is the value of `int(sqrtx + x)^(-1) dx`?A. `ln (x + sqrtx) + c`B. `2 ln (1 + sqrtx) + c`C. `2 ln (x + sqrtx) + c`D. `2 ln (x - sqrtx) + c` |
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Answer» Correct Answer - B `int (x + sqrtx)^(-1) dx = int (1)/((x + sqrtx)) dx` `= int (1)/((sqrtx.sqrtx + sqrtx)).dx` Let `sqrtx + 1 = t` then, `(1)/(2sqrtx) dx = dt rArr (1)/(sqrtx) dx = 2dt` `:. Int (1)/(sqrtx (sqrtx + 1))dx = int (2dt)/(t) = 2 log t + c` `= 2 log (1 + sqrtx) + c` |
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| 266. |
What is the value of `int(dx)/((x^(2) + a^(2)) (x^(2) + b^(2)))` ?A. `int([{tan^(-1) (x//a)}//a - {tan^(-1) (x//b)}//b])/((a^(2) + b^(2))) + c`B. `int([{tan^(-1) (x//a)}//a + {tan^(-1) (x//b)}//b])/((a^(2) +b^(2))) + c`C. `int([{tan^(-1) (x//a)}//a + {tan^(-1)(x//b)}//b])/((b^(2) -a^(2))) + c`D. `int([{tan^(-1) (x//a)} //a+ {tan^(-1) (x //b)}//b])/((b^(2) - a^(2))) + c` |
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Answer» Correct Answer - D The given integral is `int (dx)/((x^(2) + a^(2)) (x^(2) + b^(2)))` Breaking the expression under integral into partial fraction `(1)/((x^(2) + a^(2)) (x^(2) + b^(2)))` `= ((1)/((x^(2) + a^(2))) - (1)/((x^(2) + b^(2)))) xx (1)/(b^(2) -a^(2))` The given integral is `(1)/((b^(2) -a^(2))) int ((1)/((x^(2) + a^(2))) - (1)/((x^(2) + b^(2))))dx` `= (1)/((b^(2) -a^(2))) int [(1)/(x^(2) + a^(2)) dx - int (1)/(x^(2) + b^(2)) dx]` `= (1)/((b^(2) -a^(2))) {(tan^(-1) ((x)/(a)))/(a) - (tan^(-1) ((x)/(b)))/(b)} + c` |
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| 267. |
`Ifint(x^4+1)/(x^6+1)dx=tan^(-1)f(x)-2/3tan^(-1)g(x)+C ,t h e n`both `f(x)a n dg(x)`are odd functions`f(x)`is monotonic function`f(x)=g(x)`has no real roots`int(f(x))/(g(x))dx=-1/x+3/(x^3)+c`A. both `f(x)` and `g(x)` are odd functionsB. `f(x)` is one-one functionC. `f(x)=g(x)` has no real rootsD. `int (f(x))/(g(x))dx=(1)/(x)+(3)/(x^(3))+c` |
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Answer» Correct Answer - A::C::D Let `I=int((x^(4)+1))/((x^(6)+1))dt=int((x^(2)+1)^(2)-2x^(2))/((x^(2)+1)(x^(4)-x^(2)+1))dx` `=int((x^(2)+1)dx)/((x^(4)-x^(2)+1))-2 int (x^(2)dx)/((x^(6)+1))` `=int((1+(1)/(x^(2)))dx)/((x^(2)-1+(1)/(x^(2))))-2 int (x^(2)dx)/((x^(3))^(2)+1)` In the first integral, put `x-(1)/(x)=t," i.e., " (1+(1)/(x^(2)))dx=dt` and in the second integral put `x^(3) =u, " i.e., " x^(2)dx=(du)/(3)` Then ` I=int (dt)/(1+t^(2))-(2)/(3)int(du)/(1+u^(2))=tan^(-1) t-(2)/(3)tan^(-1)u +C` `=tan^(-1)(x-(1)/(x))-(2)/(3)tan^(-1)(x^(3))+C` Here, `f(x)=x-(1)/(x) and g(x)=x^(3)` Draw the graphs of `f(x) and g(x)` . We find that `f(x)` is many-one and `f(x)=g(x)` has no real roots. `int(f(x))/(g(x))dx=int(x-(1)/(x))/(x^(3))dx=int((1)/(x^(2))-(1)/(x^(4)))dx= -(1)/(x)+(3)/(x^(3))+C` |
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| 268. |
Evaluate: `int(sqrt(tanx))dx` |
| Answer» `-1/2log(1+tan^(2//3)x)+1/4log(tan^(4//3)x-tan^(2//3)x+1)+sqrt(3)/2tan^(-1)+(2tan^(2//3)x-1)/sqrt(3)+c` | |
| 269. |
Evaluate: `intsqrt(secx-1) dx` |
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Answer» ` I=int sqrt(secx-1)dx=int sqrt((1-cosx)/(cosx))dx` `=int sqrt(((1-cosx))/(cosx)xx((1+cosx))/((1+cosx)))dx` `=int sqrt((1-cos^(2)x)/(cosx+cos^(2)x))dx` `=int(sinx)/(sqrt(cos^(2)x+cosx))dx` Let `cosx=t.` Then `d(cosx)=dt` or `-sinx dx=dt.` Therefore, `I=int(-dt)/(sqrt(t^(2)+t))` `=-int(dt)/(sqrt((t+(1)/(2))^(2)-((1)/(2))^(2)))` `=-log|(t+(1)/(2))+sqrt((t+(1)/(2))^(2)-((1)/(2))^(2))|+C` ` =-log|(t+(1)/(2))+sqrt(t^(2)+t)|+C` `=-log|(cosx +(1)/(2))+sqrt(cos^(2)x+cosx)|+C` |
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| 270. |
What is `int (a + b sin x)/(cos^(2) x) dx` equal to ?A. `a sec x + b tan x + c`B. `a tan x + b sec x + c`C. `a cot x + b "cosec" x + c`D. `a "cosec" x + b cot x + c` |
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Answer» Correct Answer - B Consider `int (a + b sin x)/(cos^(2)x) dx = int ((a)/(cos^(2)x) + (b sin x)/(cos^(2)x)) dx` `= int (a sec^(2) x + b tan x sec x) dx = a tan x + b sec x + c` |
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| 271. |
Evaluate `int(tanx dx)/(sqrt(2+3tan^(2)x)).` |
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Answer» `int(tanx dx)/(sqrt(2+3tan^(2)x))=int(sinx dx)/(sqrt(2cos^(2)x+3sin^(2)x))` `=int(sinx dx)/(sqrt(2cos^(2)x+3(1-cos^(2)x)))` `=int(sinx dx)/(sqrt(3-cos^(2)x))` `=-int(dt)/(sqrt(3-t^(2)))` `=-"sin"^(-1)(t)/(sqrt(3))+c` `=-"sin"^(-1)(cosx)/(sqrt(3))+c` |
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| 272. |
Evaluate: `∫(dx)/[(x+2)^(8)(x-1)^(6)]^(1//7)` |
| Answer» `7/3((x-1)/(x+2))^(1//7)+C` | |
| 273. |
Evaluate: `∫(dx)/[(x-1)(2-x)]^(3//2)` |
| Answer» `2(sqrt((x-1)/(2-x))-sqrt((2-x)/(x-1)))+C` | |
| 274. |
`intsqrt((x-3)/(x-4))dx` |
| Answer» `sqrt((x-3)(x-4))+"ln"(sqrt(x-3)+sqrt(x-4))+C` | |
| 275. |
Evaluate: `int(4^(x)+5^(x))/(7^(x))dx` |
| Answer» `int(4^(x)+5^(x))/(7^(x))dx = int[(4/7)^(x)+(5/7)^(x)]dx=((4//7)^(x))/(ln(4/7)) + (5//7)^(x)/(ln(5/7))+C` | |
| 276. |
Evaluate `int(dx)/((x^2+2x+2)sqrt(x^2+2x-4))` |
| Answer» `-1/(2sqrt(6))"ln"(sqrt(x^(2)+x-4)/sqrt(x^(2)+2x-4))-(sqrt(6)(x+1))/(sqrt(6)(x+1))+C` | |
| 277. |
Evaluate: `int2^(xlog_(2)^(3))`dx |
| Answer» We have `int2^(xlog_(2)^(3))dx=int3^(x)dx=3^(x)/(ln3)+C` | |
| 278. |
Evaluate: `int3x^(6)dx` |
| Answer» `int3x^(6)dx = 3/7x^(7)+C` | |
| 279. |
Evaluate: `int(dx)/((2x^(2)+1)sqrt(1-x^(2))` |
| Answer» `-1/sqrt(3)tan^(-1)(sqrt(x+1))/(sqrt(3)x)+C` | |
| 280. |
Evaluate: `int(x^(3)+5x^(2)-4+7/x+2/sqrt(x))` dx |
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Answer» `int(x^(3)+5x^(2)-4+7/x+2/sqrt(x))`dx `=intx^(3)dx + int5x^(2)dx-int4dx + int7/xdx+int2/sqrt(x)dx` `=intx^(3)dx+int5x^(2)dx-int4dx+int7/x dx+int2/sqrt(x)dx` `=ints^(2)dx+5intx^(2)dx-4.int1.dx+7.int1/xdx+2.intx^(1/2)dx` `=x^(4)/4+5.x^(3)/3-4x+7ln|x|+2(x^(1//2)/(1//2))+C=x^(4)/4+5/3x^(3)-4x+7ln |x|+4sqrt(x)+C` |
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| 281. |
Evaluate: `int(dx)/((x+1)sqrt(1+x-x^(2))` |
| Answer» `sin^(-1)(3/2-1/(x+1))/sqrt(5)/2+C` | |
| 282. |
Evaluate: `int(dx)/((x^(2)+5x+6)sqrt(x+1))` |
| Answer» `2tan^(-1)(sqrt(x+1))-sqrt(2)tan^(-1)(sqrt(x+1))/sqrt(2)+C` | |
| 283. |
Evaluate: `int(dx)/((x+2)sqrt(x+1))` |
| Answer» `2tan^(-1)(sqrt(x+1))+C` | |
| 284. |
Evaluate: `int1/((x+2)(x+3))dx` |
| Answer» `"ln"|(x+2)/(x+3)|+C` | |
| 285. |
Evaluate: `int(dx)/((x+1)(x^(2)+1))` |
| Answer» `1/2"ln"|x+1|-"ln"(x^(2)+1)+1/2tan^(-1)(x)+C` | |
| 286. |
Evaluate: `intsqrt(tanx)`dx |
| Answer» `1/sqrt(2)(y/sqrt(2))+1/(2sqrt(2))"ln"|(y-sqrt(2))/(y+sqrt(2))|+C`, where `y=sqrt(tanx)=1/sqrt(tanx)` | |
| 287. |
If `f(x) = int(5x^(8)+7x^(6))/((x^(2)+1+2x^(7))^(2))dx, (x ge 0)`, and f(0) = 0, then the value of f(1) isA. `-1//2`B. `1//4`C. `1//2`D. `-1//4` |
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Answer» Correct Answer - B `f(x)=int(5x^(8)+7x^(6))/(x^(14)(2+(1)/(x^(7))+(1)/(x^(5)))^(2))dx` `=int((5)/(x^(6))+(7)/(x^(8)))/((2+(1)/(x^(7))+(1)/(x^(5)))^(2))dx` `"Put, "2+(1)/(x^(7))+(1)/(x^(5))=t` `rArr" "f(x)=-int(dt)/(t^(2))` `=(1)/(t)+c` `=((x^(7))/(2x^(7)+x^(2)+1))+c` `" "f(0)=0` `rArr" "c=0` `rArr" "f(x)=((x^(7))/(2x^(7)+x^(2)+1))` `rArr" "f(1)=(1)/(4)` |
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| 288. |
Evaluate: `int(x^(2)-1)/(x^(4)-7x^(2)+1)`dx |
| Answer» `1/6"ln"|(x+1/x-3)/(x+1/x+3)|+C` | |
| 289. |
If `f(x) = int(5x^(8)+7x^(6))/((x^(2)+1+2x^(7))^(2))dx, (x ge 0)`, and f(0) = 0, then the value of f(1) isA. `-(1)/(2)`B. `-(1)/(4)`C. `(1)/(4)`D. `(1)/(2)` |
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Answer» Correct Answer - C We have, `f(x) = int(5x^(8)+7x^(6))/((x^(2)+1+2x^(7))^(2))dx` `=int(5((x^(8))/(x^(14)))+7((x^(6))/(x^(14))))/((x^(2)/(x^(7))+(1)/(x^(7))+(2x^(7))/(x^(7)))^(2))dx` (dividing both numerator and denominator by `X^(14)`) `=int(5x^(-6)+7x^(-8))/((x^(-5)+x^(-7)+2)^(2))dx` Let `x^(-5) + x^(-7)+2 = t` `rArr (-5x^(-6)-7x^(-8))dx = dt` `rArr (5x^(-6)+7x^(-8))dx = - dt` `therefore f(x) = int - (dt)/(t^(2))= - intt^(-2)dt` `= -(t^(-2+1))/(-2+1)+C=-(t^(-1))/(-1)+C=(1)/(t)+C` `=(1)/(x^(-5)+x^(-7)+2)+C=(x^(7))/(2x^(7)+x^(2)+1)+C` `therefore f(0)=0` `therefore 0 = (0)/(0+0+1)+C rArr C = 0` `therefore f(x)=(x^(7))/(2x^(7)+x^(2)+1)` `rArr" "f(1)=(1)/(2(1)^(7)+1^(2)+1)=(1)/(4)` |
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| 290. |
`int (sin^2x cos^2x)/(sin^5x+cos^3x sin^2x + sin^3x cos^2x + cos^5x)^2 dx`A. `(1)/(3(1+tan^(3)x))+C`B. `(-1)/(3(1+tan^(3)x))+C`C. `(1)/(1+cot^(3)x)+C`D. `(-1)/(1+cot^(3)x)+C` |
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Answer» Correct Answer - B We have, `I = int(sin^(2)x*cos^(2)x)/((sin^(6)x + cos^(3)x*sin^(2)x+sin^(3)x*cos^(2)x+cos^(5)x)^(2))dx` ` = int(sin^(2)x cos^(2)x)/{{sin^(3)x(sin^(2) x + cos^(2)x)+cos^(3)x(sin^(2)x +cos^(2)x)}^(2))dx` `=int(sin^(2)x cos^(2) x)/((sin^(3)x + cos^(3)x)^(2))dx=int(sin^(2)x cos^(2)x)/(cos^(6)x(1+tan^(3)x)^(2))dx` `=int(tan^(2)x sec^(2)x)/((1+tan^(3)x)^(2))dx` Put `tan^(3) x = t rArr 3 tan^(2)x sec^(2) xdx = dt` `therefore" "I = (1)/(3)int(dt)/((1+t)^(2))` `rArr" "I = (-1)/(3(1+t))+C rArr I=(-1)/(3(1+tan^(3)x))+C` |
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| 291. |
If `x^2!=n pi-1, n in N`. Then, the value of `int x sqrt((2sin(x^2+1)-sin2(x^2+1))/(2sin(x^2+1)+sin2(x^2+1)))dx` is equal to:A. `ln|(1)/(2)sec(x^(2)+1)|+C`B. `ln|sec((x^(2)+1)/(2))|+C`C. `(1)/(2)ln|sec(x^(2)+1)|+C`D. `(1)/(2)ln|(2)/(sec(x^(2)+1))|+C` |
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Answer» Correct Answer - B `I=(1)/(2)int2xsqrt((2sin(x^(2)+1)-sin2(x^(2)+1))/(2sin(x^(2)+1)+sin2(x^(2)+1)))dx` `x^(2)+1=t rArr 2x = dt` `I=(1)/(2)intsqrt((2sint-sin 2t)/(2sin t +sin 2r))dt` `=(1)/(2)intsqrt((2-2cost)/(2+2cost))dt` `=(1)/(2)int tan.(t)/(2)dt` `=(1)/(2)(ln|sec.(t)/(2)|)/((1)/(2))+c` `=ln|sec((x^(2)+1)/(2))|+c` |
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| 292. |
If `x^2!=n pi-1, n in N`. Then, the value of `int x sqrt((2sin(x^2+1)-sin2(x^2+1))/(2sin(x^2+1)+sin2(x^2+1)))dx` is equal to:A. `(1)/(2)log_(e)|sec(x^(2)-1)|+C`B. `log_(e)|sec((x^(2)-1)/(2))|+C`C. `log_(e)|(1)/(2)sec^(2)(x^(2)-1)|+C`D. `(1)/(2)log_(e)|sec^(2)((x^(2)-1)/(2))|+C` |
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Answer» Correct Answer - B Let `I = intx sqrt((2 sin(x^(2)-1)-sin 2(x^(2)-1))/(2 sin(x^(2)-1)+sin 2(x^(2)-1)))dx` Put `(x^(2)-1)/(2)=theta rArr x^(2)-1 = 2theta rArr 2x dx = 2 d theta` `rArr" ""x dx"=d theta` Now, `I = intsqrt((2sin 2 theta - sin 4 theta)/(2 sin 2 theta + sin 4 theta))d theta` `=int sqrt((2sin 2 theta - 2 sin 2 theta cos 2 theta)/(2sin 2 theta + 2 sin 2 theta cos 2 theta))d theta" "(therefore sin 2 A = 2 sin A cos A)` `=int sqrt((2 sin 2 theta(1-cos 2 theta))/(2 sin 2 theta(1+cos 2 theta)))d theta` `=int sqrt((1-cos 2 theta)/(1+cos 2 theta))d theta = int sqrt((2sin^(2)theta)/(2 cos^(2)theta))d theta` `[therefore 1-cos 2 A = 2 sin^(2)A and 1 + cos 2 A = 2 cos^(2) A]` `=int sqrt(tan^(2)theta) d theta = int tan theta d theta` `=log_(e)|sec theta|+C=log_(e)|sec((x^(2)-1)/(2))|+C[therefore theta = (x^(2)-1)/(2)]` |
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| 293. |
`int(sin2x)/(sin^4x+cos^4x)d x`A. `cot^(-1)(tan^(2)x)+c `B. `tan^(-1)(tan^(2)x)+c `C. `cot^(-1)(cot^(2)x)+c `D. `tan^(-1)(cot^(2)x)+c ` |
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Answer» Correct Answer - B ` I= int(sin2x)/(sin^(4)x+cos^(4)x)dx` `=int(2sinx cosx)/(sin^(4)x+cos^(4)x)dx` `=int(2tanx sec^(2)x)/(1+tan^(4)x)dx ` `"Let " tan^(2)x=t " or " 2tanxsec^(2)xdx=dt` ` :. I=int(dt)/(1+t^(2))=tan^(-1)t+C=tan^(-1)(tan^(2)x)+C` |
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| 294. |
`int(dx)/(x(x^n+1))` is equal toA. `(1)/(n)log((x^(n))/(x^(n)+1))+c`B. `(1)/(n)log((x^(n)+1)/(x^(n)))+c`C. `log((x^(n))/(x^(n)+1))+c`D. non of these |
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Answer» Correct Answer - A `I=int(dx)/(x(x^(n)+1))=int(x^(n-1))/(x^(n)(x^(n)+1))dx` `"Putting " x^(n)=t " so that " nx^(n-1)dx=dt, i.e.,` ` "we get " x^(n-1)dx=(1)/(n)dt` `I=int((1)/(n)dt)/(t(t+1))=(1)/(n)int((1)/(t)-(1)/(t+1))dt` `=(1)/(n)(log t-log(t+1))+C` `=(1)/(n)log((x^(n))/(x^(n)+1))+C` |
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| 295. |
Evaluate: `int(dx)/(1+3cos^(2)x)` |
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Answer» Multiply Nr. And Dr. of given integral by `sec^(2)x`, we get I = `int(sec^(2)x dx)(tan^(2)x+4) = 1/2tan^(-1)(tanx)/(2)+C` |
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| 296. |
`int(sin2x)/(sin5xsin3x)dx` is equal toA. `log sin3x-log sin5x+c`B. `(1)/(3)log sin3x+(1)/(5)log sin5x+c`C. `(1)/(3)log sin3x-(1)/(5)log sin5x+c`D. `3log sin3x-5log sin5x+c` |
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Answer» Correct Answer - C `int(sin2x)/(sin5xsin3x)dx=int(sin(5x-3x))/(sin5xsin3x)` `=int(sin5xcos3x-cos5xsin3x)/(sin5xsin3x)dx` `=(1)/(3)logsin3x-(1)/(5)logsin5x+C` |
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| 297. |
Evaluate: `intcos^(5)cosxsin^(4)xdx` |
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Answer» Let I`=intcos^(5)xsin^(4)dx` put `sinx=t rArr cosxdx=dt` `rArr I=int(1-t^(2))^(2).t^(4).dt=int(t^(4)-2t^(2)+1)t^(4)dt= intt^(B)-2t^(6)+t^(4)dt` `=t^(9)/9-(2t)^(7)/7+t^(5)/5+C`, where t=sinx |
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| 298. |
Evaluate: `intsin^(4)xcos^(2)xdx` |
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Answer» `intsin^(4)xcos^(2)dx=1/8int4sin^(2)xcos^(2)x.2sin^(2)xdx=1/8intsin^(2)2x(1-cos2x)dx` `=1/8intsin^(2)2dx-1/8intsin^(2)2xcos2xdx=1/16 int(1-cos4x)dx-1/48(sin2x)^(3)` `=x/16 -(sin4x)/(64) -1/48 (sin2x)^(3)+C` |
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| 299. |
`intcos^(3)x" " dx=` |
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Answer» `intcos^(3)xdx=(1)/(4)int(3cosx+cos3x)dx` `=(3)/(4)intcosxdx+(1)/(4)intcos3xdx` `=(3)/(4)sinx+(1)/(12)sin3x+c` |
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| 300. |
`int(sin2x)/(sin^4x+cos^4x)d x`A. `cot^(-1)cot^(2)x+C`B. `-cot^(-1)(tan^(2)x)+C`C. `tan^(-1)(tan^(2)x)+C`D. `-tan^(-1)(cos2x)+C` |
| Answer» Correct Answer - ABCD | |