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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.
| 101. |
The value of `int sqrt((1-cosx)/(cosalpha-cosx)) dx` where `0ltalphaltxltpi` is equal to:A. `2"ln"(cosalpha/2-cosx/2)+C`B. `sqrt(2)"ln"(cosalpha/2-cosx/2)+C`C. `2sqrt(2)"ln"(cosalpha/2-cosx/2)+C`D. `-2sin^(-1)(cosx/2)/(cosalpha/2)+C` |
| Answer» Correct Answer - d | |
| 102. |
Evaluate `int sec^(-1)xdx` |
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Answer» Put `sec^(-1)x=t` so that x =`sect` and `dx=sect tant dt` `therefore intsec^(-1)xdx=intt(sect tant) dt = t(sect)-int(1.sectdt)` `tsect=ln|sect+tant|+C` `tsect-ln|sect + sqrt(sec^(2)t-1)|+C=x(sec^(-1)x)-ln|x+sqrt(x^(2)-1)|+C` |
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| 103. |
If `int(e^(x-1))/((x^2-5x+4))2xdx=A F(x-1)+B F(x-4)+Ca n dF(x)=int(e^x)/xdx ,t h e n``A=-2/3`(b) `b=(4/3)e^3``A=2/3`(d) `d in R`A. `A= -2//3`B. `B=(4//3)e^(3)`C. `A=2//3`D. `B=(8//3)e^(3)` |
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Answer» Correct Answer - A::D `(2x)/((x-1)(x-4))=(C)/(x-1)+(D)/(x-4)` `2x=C(x-4)+D(x-1)` ` :. C= -2//3, D=8//3` ` :. int(e^(x-1))/((x^(2)-5x+4))2xdx=int e^(x-1)((-2//3)/(x-1)+(8//3)/(x-4))dx` `= -(2)/(3)F(x-1)+(8)/(3)e^(3)F(x-4)+C` ` :. A= -2//3, B=8//3e^(3)` |
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| 104. |
Evaluate `int tan^(2)x sin^(2)xdx` |
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Answer» Correct Answer - `tanx-(3)/(2)x+(sin2x)/(4)+C` `int tan^(2)x sin^(2)xdx=int(sin^(4)x)/(cos^(2)x)dx` `=int(sin^(2)x(1-cos^(2)x))/(cos^(2)x)dx` `=int(tan^(2)x-sin^(2)x)dx` `=int(sec^(2)x-1-(1-cos2x)/(2))dx` `=tanx-(3)/(2)x+(sin2x)/(4)+C` |
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| 105. |
If `I=int(sqrt(cotx)-sqrt(tanx))dx,` then `I` equalsA. `sqrt(2)log(sqrt(tanx)-sqrt(cotx))+C`B. `sqrt(2)log|sinx+cosx+sqrt(sin2x)|+C`C. `sqrt(2)log|sinx-cosx+sqrt(2)sinx cosx|+C`D. `sqrt(2)log|sin(x+pi//4)+sqrt(2)sinx cosx|+C` |
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Answer» Correct Answer - B `I=int(cosx-sinx)/(sqrt(cosx sinx))dx` Put `sinx+cosx=t, " so that " 2sinx cosx=t^(2)-1` ` :. I=sqrt(2)int(dt)/(sqrt(t^(2)-1))=sqrt(2)log|t+sqrt(t^(2)-1)|+C` `=sqrt(2)log|sinx+cosx+sqrt(sin2x)|+C` |
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| 106. |
`int(sec xdx)/(sqrt(sin(2x+A)+sinA)) " is equal to "`A. `(secA)/(sqrt(2))sqrt(tanx cosA-sinA)+c`B. `sqrt(2)secAsqrt(tanx cosA-sinA)+c`C. `sqrt(2)secAsqrt(tanx cosA+sinA)+c`D. non of these |
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Answer» Correct Answer - C `I=int(sec x dx)/(sqrt(2sin(x+A)cosx))` `=int(sec^(2)x dx)/(sqrt((2sin(x+A))/(cosx)))` `=(1)/(sqrt(2))int(sec^(2)x dx)/(sqrt(tanx cosA+sinA))` `=(secA)/(sqrt(2))int(2pdp)/(p) " " (tanx cosA+sinA=p^(2), " then " cosA sec^(2)x dx=2pdp)` `=sqrt(2)secA int dp ` ` =sqrt(2)secA sqrt(tanx cosA+sinA)+C` |
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| 107. |
`"If " int(cos 4x+1)/(cotx-tanx)dx=Acos4x+B, " then "`A. `A=-1//2`B. `A=-1//8`C. `A=-1//4`D. non of these |
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Answer» Correct Answer - B `int(cos4x+1)/(cotx-tanx)dx=int(2cos^(2)2x)/(cos^(2)x-sin^(2)x)sinx cosx dx` `=int cos2x sin2x dx` `=(1)/(4)int sin4xdx=-(1)/(8)cos4x+C` |
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| 108. |
If `int(dx)/(x^2(x^n+1)^((n-1)/n))=-(f(x))^(1/n)+C` then `f(x)` is (A) `1+x^n` (B) `1+x^-n` (C) `x^n+x^-n` (D) `x^n-x^-n`A. `(1+x^(n))`B. `1+x^(-n) `C. `x^(n)+x^(-n) `D. non of these |
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Answer» Correct Answer - B `"We have " int(dx)/(x^(2)(x^(n)+1)^((n-1)//n))=int(dx)/(x^(2)x^(n-1)(1+(1)/(x^(n)))^((n-1)//n))` `=int(dx)/(x^(n+1)(1+x^(-n))^((n-1)//n))` `"Put " 1+x^(-n)=t` ` :. -nx^(-n-1)dx=dt " or " (dx)/(x^(n+1))=-(dt)/(n)` ` :. int(dx)/(x^(2)(x^(n)+1)^((n-1)//n))=-(1)/(n)int(dt)/(t^((n-1)//n)) ` `=-(1)/(n)int t^(((1)/(n)-1))dt ` `=-(1)/(n)(t^((1)/(n)-1+1))/((1)/(n)-1+1)+C` `=-t^(1//n)+C` `=-(1+x^(-n))^(1//n)+C` |
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| 109. |
`int(cos4x-1)/(cotx-tanx)dx` is equal toA. `(1)/(2)In|sec2x|-(1)/(4)cos^(2)2x+c`B. `(1)/(2)In|sec2x|+(1)/(4)cos^(2)x+c`C. `(1)/(2)In|cos2x|-(1)/(4)cos^(2)2x+c`D. `(1)/(2)In|cos2x|+(1)/(4)cos^(2)x+c` |
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Answer» Correct Answer - C `I=int(cos4x-1)/(cotx-tanx)dx=int(-2sin^(2)2x(sinxcosx))/((cos^(2)x-sin^(2)x))dx` `=-int(sin^(2)2xsin2x)/(cos2x)x` `=int((cos^(2)2x-1)sin2x)/(cos2x)dx` `"Let " t=cos2x " or "dt=-2sin2xdx` ` :. I=(1)/(2)int((1-t^(2)))/(t)dt=(1)/(2)In|t|-(t^(2))/(4)+C ` `=(1)/(2)In|cos2x|-(1)/(4)cos^(2)2x+c` |
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| 110. |
If `int(3cot3x-cotx)/(tanx-3tan3x)dx = pf(x) + qg(x)+C`, then which of the following may be correct?A. `p=1,q=1/sqrt(3), f(x)=x,g(x)="ln"|(sqrt(3)-tanx)/(sqrt(3)+tanx)|`B. `p=1, q=-1/sqrt(3), f(x)=x,g(x)="ln"|(sqrt(3)-tanx)/(sqrt(3)+tanx)|`C. `p=1, q=-2/sqrt(3), f(x)=x,g(x)="ln"|(sqrt(3)+tanx)/(sqrt(3)-tanx)|`D. `p=1, q=-1/sqrt(3),f(x)=x,g(x)="ln"|(sqrt(3)+tanx)/(sqrt(3)-tanx)|` |
| Answer» Correct Answer - AD | |
| 111. |
What is `int (dx)/(a^(2) sin^(2)x + b^(2) cos^(2)x)` equal to ?A. `c + (1)/(ab) tan^(-1) ((a tanx)/(b))`B. `c - (1)/(ab) tan^(-1) ((b tanx)/(a))`C. `c + (1)/(ab) tan^(-1) ((b tan x)/(a))`D. None of these |
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Answer» Correct Answer - A `I = (1)/(a^(2)) int (sec^(2)x)/(tan^(2) x + ((b)/(a))^(2))` `= (1)/(a^(2)) xx (a)/(b) tan^(-1) ((a tan x)/(b)) + c` `= c + (1)/(ab) tan^(-1) ((a tan x)/(b))` |
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| 112. |
What is `int ln (x^(2)) dx` equal to ?A. `2x ln (x) - 2x +c`B. `(2)/(x) + c`C. `2x ln (x) + c`D. `(2 ln (x))/(x) - 2x + c` |
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Answer» Correct Answer - A `int ln (x^(2)).dx = 2.int ln x.dx` `= 2 int 1. ln x.dx = 2 [ln x.x - int (1)/(x) x.dx` `= 2 (x.ln x -x) + c = 2x lnx - 2x + c` |
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| 113. |
A curve `g(x)=intx^(27)(1+x+x^2)^6(6x^2+5x+4)dx`is passing through origin. Then`g(1)=(3^7)/7`(b) `g(1)=(2^7)/7``g(-1)=1/7`(d) `g(-1)=(3^7)/(14)`A. `g(1)=(3^(7))/(7)`B. `g(1)=(2^(7))/(7)`C. `g(-1)=(1)/(7)`D. `g(-1)=(3^(7))/(14)` |
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Answer» Correct Answer - A::C `g(x)=int x^(27)(1+x+x^(2))^(6)(6x^(2)+5x+4)dx` `=int(x^(4)+x^(5)+x^(6))^(6)(6x^(5)+5x^(4)+4x^(3))dx ` `"Let " x^(6)+x^(5)+x^(4)=t " or "(6x^(5)+5x^(4)+4x^(3))dx=dt ` ` :. g(x)=int t^(6)dt=(t^(7))/(7)+C=(1)/(7)(x^(4)+x^(5)+x^(6))^(7)+C ` `g(0)=0 impliesC=0 impliesg(1)=(3^(7))/(7) " and " g(-1)=(1)/(7)` |
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| 114. |
Evaluate:`int1/(x^2-x+1)dx` |
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Answer» `int(1)/(x^(2)-x+1)dx` `=int(1)/((x-1//2)^(2)+3//4)dx` `=int(1)/((x-1//2)^(2)+(sqrt(3)//2)^(2))dx` `=(1)/(sqrt(3)//2)tan^(-1)((x-1//2)/(sqrt(3)//2))+C` `=(2)/(sqrt(3))tan^(-1)((2x-1)/(sqrt(3)))+C` |
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| 115. |
Evaluate:`int1/([(x-1)^3(x+2)^5]^(1/4))dx` |
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Answer» `I=int(1)/(((x-1)/(x+2))^(3//4)(x+2)^(2))dx` Let `(x-1)/(x+2)=t " or " (3dx)/((x+2)^(2))=dt` ` :. I=(1)/(3)int(1)/(t^(3//4))dt=(1)/(3)((t^(1//4))/(1//4))+C` `=(4)/(3)t^(1//4)+C=(4)/(3)((x-1)/(x+2))^(1//4)+C` |
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| 116. |
Evaluate:`int((x-1)(x-2)(x-3))/((x-4)(x-5)(x-6))dx` |
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Answer» `int(overset(underset(darr)("Improper fraction"))((x-1))(x-2)(x-3))/((x-4)(x-5)(x-6))dx` `int[1+((x-1)(x-2)overset(underset(darr)("Proper fraction"))((x-3))-(x-4)(x-5)(x-6))/((x-4)(x-5)(x-6))]dx` ` " " `(Adding and subtracting 1) `=int[1+(3xx2xx1)/((x-4)(-1)(-2))+(4xx3xx2)/(1(x-5)(-1))+(5xx4xx3)/((2)(1)(x-6))]dx` `=1+3log|x-4|-24 log|x-5|+30 log|x-6|+C` |
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| 117. |
Evaluate:`int1/(2x^2-x-1)dx` |
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Answer» `int(1)/(2x^(2)+x-1)dx` `=(1)/(2)int(1)/(x^(2)+(x)/(2)-(1)/(2))dx` `=(1)/(2)int(1)/((x+1//4)^(2)-(3//4)^(2))dx` `=(1)/(2)*(1)/(2(3//4))log|(x+1//4-3//4)/(x+1//4+3//4)|+C` `=(1)/(3)log|(x-1//2)/(x+1)|+C=(1)/(3)log|(2x-1)/(2(x+1))|+C` |
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| 118. |
Evaluate `int(1+x^(4))/((1-x^(4))^(3//2))dx.` |
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Answer» `I=int(1+x^(4))/((1-x^(4))^(3//2))dx=int(x^(3)(x+1//x^(3))dx)/(x^(3)(x^(2)-(1)/(x^(2)))^(3//2))` `=int((x+1//x^(3))dx)/(((1)/(x^(2))-x^(2))^(3//2))` Let `(1)/(x^(2))-x^(2)=t or ((-2)/(x^(3))-2x)dx=dt` or `(x+(1)/(x^(3)))dx= -(1)/(2)dt` ` :. I= -(1)/(2)int(dt)/(t^(3//2))=(1)/(sqrt(t))+C=(1)/(sqrt((1)/(x^(2))-x^(2)))+C` |
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| 119. |
Evaluate `int(x^(2))/((x^(2)+1)(x^(2)+4))dx.` |
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Answer» `int(x^(2))/((x^(2)+1)(x^(2)+4))dx=(1)/(3)int[(4)/(x^(2)+4)-(1)/(x^(2)+1)]dx` `= -(1)/(3)int(1)/(x^(2)+1)dx+(4)/(3)int(1)/(x^(2)+4)dx` `= -(1)/(3)tan^(-1)x+(4)/(3)xx(1)/(2)tan^(-1)((x)/(2))+C` `= -(1)/(3)tan^(-1)x+(2)/(3)tan^(-1)((x)/(2))+C` |
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| 120. |
Evaluate:`int(cosx)/(sin(x-pi/6)sin(x+pi/6))dx` |
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Answer» ` I =int(cosx)/(sin(x-(pi)/(6))sin(x+(pi)/(6)))dx` `=int(cosx)/(sin^(2)x-"sin"^(2)(pi)/(6))dx` Let `sinx=t` or `dt=cosx dx` ` :. I=int(dt)/(t^(2)-(1)/(4))` `=(1)/(2(1)/(2))log|(t-(1)/(2))/(t+(1)/(2))|+C` `=log|(2t-1)/(2t+1)|+C` `=log|(2sinx-1)/(2sinx+1)|+C` |
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| 121. |
`int("In"((x-1)/(x+1)))/(x^(2)-1)dx` is equal toA. `(1)/(2)("In"((x-1)/(x+1)))^(2)+C`B. `(1)/(2)("In"((x+1)/(x-1)))^(2)+C`C. `(1)/(4)("In"((x-1)/(x+1)))^(2)+C`D. `(1)/(4)("In"((x+1)/(x-1)))` |
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Answer» Correct Answer - C `I=int("In"((x-1)/(x+1)))/(x^(2)-1)dx` Let `t="In"((x-1)/(x+1))` or `(dt)/(dx)=(x+1)/(x-1){(x+1-(x-1))/((x+1)^(2))}=(2)/((x^(2)-1))` or `(dx)/(x^(2)-1)=(dt)/(2)` ` :. I=(1)/(2)int t dt=(1)/(4)t^(2)+C=(1)/(4)("In"((x-1)/(x+1)))^(2)+C` |
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| 122. |
Evaluate:`int([sqrt(1+x^2)+x]^n)/(sqrt(1+x^2))dx` |
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Answer» Correct Answer - `(1)/(n)(sqrt(1+x^(2))+x)^(n)+C` ` "Let " (sqrt(1+x^(2))+x)^(n)=z` ` or n(sqrt(1+x^(2))+x)^(n-1)((x)/(sqrt(1+x^(2)))+1)dx=dz` ` or ((sqrt(1+x^(2))+x)^(n))/(sqrt(1+x^(2)))dx=(dz)/(n)` ` :. " Given integral "=int(dz)/(n)=(1)/(n)z+C=(1)/(n)(sqrt(1+x^(2))+x)^(n)+C` |
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| 123. |
Evaluate:`int(sinx)/(sin4x)dx` |
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Answer» `I=int(sinx)/(sin4x)dx=int(sinx)/(2sin2x cos2x)dx` `=int(sinx)/(4sinx cos x cos2x)dx` `=(1)/(4)int(1)/(cosx cos2x)dx=(1)/(4)int(cosx)/(cos^(2)x cos2x)dx` `=(1)/(4)int(cosx)/((1-sin^(2)x)(1-2sin^(2)x))dx` Putting `sinx =t` and `cosx dx=dt,` we get `I=(1)/(4)int(dt)/((1-t^(2))(1-2t^(2)))` `=(1)/(4)int[(2)/(1-2t^(2))-(1)/(1-t^(2))]dt` `= -(1)/(4)int(1)/(1-t^(2))dt+(2)/(4)int(1)/(1-(sqrt(2)t)^(2))dt` `= -(1)/(4)xx(1)/(2)log|(1+t)/(1-t)|+(1)/(2).(1)/(2sqrt(2))log|(1+sqrt(2)t)/(1-sqrt(2)t)| +C` `= -(1)/(8)log|(1+sinx)/(1-sinx)|+(1)/(4sqrt(2))log|(1+sqrt(2)sinx)/(1-sqrt(2)sinx)| +C` |
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| 124. |
Evaluate:`int(x^2+1)/((x-1)^2(x+3))dx` |
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Answer» `I=int(x^(2)+1)/((x-1)^(2)(x+3))dx` Let `int(x^(2)+1)/((x-1)^(2)(x+3))=(A)/(x-1)+(B)/((x-1)^(2))+(C)/(x+3) " " ` (1) or `x^(2)+1=A(x-1)(x+3)+B(x+3)+C(x-1)^(2) " " ` (2) Putting `x-1=0`, i.e., `x=1` in equation (2), we get `2=4B " or " B=(1)/(2).` Putting `x+3=0`, i.e., `x= -3` in equation (2), we get `10=16C " or " C=(5)/(8).` Equating the coefficients of `x^(2)` on both the sides of the identity of equation (2), we get `I=A+C " or " A=1-C=1-(5)/(8)=(3)/(8)` Substituting the values of A, B in equation (1), we get `(x^(2)+1)/((x-1)^(2)(x+3))=(3)/(8)(1)/(x-1)+(1)/(2)(1)/((x-1)^(2))+(5)/(8)(x+3)` or `I=(3)/(8)int(1)/(x-1)dx+(1)/(2)int(1)/((x-1)^(2))dx+(5)/(8)int(1)/(x+3)dx` `=(3)/(8)log|x-1|-(1)/(2(x-1))+(5)/(8)log|x+3|+C` |
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| 125. |
Evaluate ` int x sin 3x dx`. |
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Answer» Here, both the functions, viz., `x` and `sin 3x` are easily integrable and the derivative of x is one, a less complicated function. Therefore, we take x as the first function and `sin 3x` as the second function. Thus, `int underset (I)(x)underset(II)(sin)3x dx` `=x{intsin 3xdx}-int{(d)/(dx)(x)intsin 3x dx}dx` `= -x (cos 3x)/(3)-int1 {-(cos 3x)/(3)}dx` `= -(1)/(3)x cos 3x +(1)/(3) int cos3x dx` `= -(1)/(3)x cos 3x +(1)/(9) sin 3x +C` |
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| 126. |
`int sqrt(e^(x)-1)dx` is equal toA. `2[sqrt(e^(x)-1)-tan^(-1) sqrt(e^(x)-1)]+c`B. `sqrt(e^(x)-1)-tan^(-1) sqrt(e^(x)-1)+c`C. `sqrt(e^(x)-1)+tan^(-1) sqrt(e^(x)-1)+c`D. `2[sqrt(e^(x)-1)+tan^(-1) sqrt(e^(x)-1)]+c` |
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Answer» Correct Answer - A `I=int sqrt(e^(x)-1)dx` Let ` e^(x)-1=t^(2)" or " e^(x)dx=2t dt " or " dx=(2t)/(t^(2)+1)dt` ` :. I=int t(2t)/(t^(2)+1)dt=int (2t^(2))/(t^(2)+1)dt` `=int(2(t^(2)+1)-2)/(t^(2)+1)dt=int2dt-int(2t)/(t^(2)+1)` `=2t-2tan^(-1)t+C` `=2sqrt(e^(x)-1)-2tan^(-1) sqrt(e^(x)-1)+C` |
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| 127. |
`int sqrt((x^(2)+1)/(x^(2)(1-x^(2))))dx=`A. `(1)/(4) log_(e)|(1-sqrt(1-x^(4)))/(1+sqrt(1-x^(4)))|+(1)/(2)sin^(-1)(x^(2))+C`B. `(1)/(2) log_(e)|(1-sqrt(1-x^(4)))/(1+sqrt(1-x^(4)))|+(1)/(2)cos^(-1)(x^(2))+C`C. `(1)/(2) log_(e)|(1-sqrt(1-x^(4)))/(1+sqrt(1-x^(4)))|+sin^(-1)(x^(2))+C`D. ` log_(e)|(1-sqrt(1-x^(4)))/(1+sqrt(1-x^(4)))|+(1)/(2)cos^(-1)(x^(2))+C` |
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Answer» Correct Answer - A `int sqrt((x^(2)+1)/(x^(2)(1-x^(2))))dx=int sqrt((1+x^(2))/(x^(2)(1-x^(2)))) xx(sqrt(1+x^(2)))/(sqrt(1+x^(2)))dx` `=int(1+x^(2))/(xsqrt(1-x^(4)))dx` `=int(1)/(xsqrt(1-x^(4)))dx+int (x)/(sqrt(1-x^(4)))dx` `=(1)/(4) log_(e)|(1-sqrt(1-x^(4)))/(1+sqrt(1-x^(4)))|+(1)/(2)sin^(-1)(x^(2))+c` |
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| 128. |
`int(x(x-1))/((x^(2)+1)(x+1)sqrt(x^(3)+x^(2)+x))=(1)/(2)log|(sqrt(x+(1)/(x)+1-1))/(sqrt(x+(1)/(x)+1+1))|-A+c.` Then the value of A is equal toA. `cos^(-1)sqrt(1+(1)/(x))`B. `tan^(-1)sqrt(x+(1)/(x)+1)`C. `cot^(-1)sqrt(x+(1)/(x))`D. `sin^(-1)sqrt(x+(1)/(x)+1)` |
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Answer» Correct Answer - B We have `int(x(x-1))/((x^(2)+)(x+1)sqrt(x^(3)+x^(2)+x))dx` `=int(x(x^(2)-1))/((x^(2)+1)(x+1)^(2)sqrt(x^(3)+x^(2)+x))dx` `=int(x^(3)(1-(1)/(x^(2))))/(x^(3)(x+(1)/(x))(sqrtx+(1)/(sqrtx))^(2)sqrt(x+(1)/(x)+1))dx` `=int((1-(1)/(x^(2))))/((x+(1)/(x))(x+(1)/(x)+2)sqrt(x+(1)/(x)+1))dx` `I=int(2t)/((t^(2)-1)(t^(2)+1)sqrt(t^(2)))dt` where `x+(1)/(x)+1=t^(2)` `=int(2)/((t^(2)-1)(t^(2)+1))dt` `=int(1)/(t^(2)-1)dt-int(1)/(t^(2)+1)dt` `=(1)/(2)log|(t-1)/(t+1)|-tan^(-1)t+c` `=(1)/(2)log|(sqrt(x+(1)/(x)+1)-1)/(sqrt(x+(1)/(x)+1+1))|-tan^(-1)sqrt(x+(1)/(x)+1)+c` |
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| 129. |
`int(sqrt(x^2+10 x+24))/(x+5)dx` is equal toA. `sqrt(x^(2)+10x+24)+sec^(-1)(x+5)+c`B. `sqrt(x^(2)+10x+24)-"cosec"^(-1)(x+5)+c`C. `sec^(-1)(x+5)-sqrt(x^(2)+10x+24)+c`D. `sqrt(x^(2)+10x+24)-sec^(-1)(x+5)+c` |
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Answer» Correct Answer - D ` I=int(sqrt(x^(2)+10x+24))/(x+5)dx=int(sqrt((x+5)^(2)-1))/(x+5)dx` `"Put " x+5=sec theta implies dx=sec theta tan theta d theta ` `:. I=int(tan theta)/(sec theta)*sec theta tan theta d theta ` `=int(sec^(2) theta-1)d theta` `=tan theta-theta+C` `=sqrt(x^(2)+10x+24)-sec^(-1)(x+5)+c` |
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| 130. |
`"If "int x^(5)(1+x^(3))^(2//3)dx=A(1+x^(3))^(8//3)+B(1+x^(3))^(5//3)+c, " then " `A. `A=(1)/(4),B=(1)/(5)`B. `A=(1)/(8),B=-(1)/(5)`C. `A=-(1)/(8),B=(1)/(5)`D. non of these |
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Answer» Correct Answer - B `"Here, " int x^(5)(1+x^(3))^(2//3)dx` `"Let " 1+x^(3)=t^(2) " and " 3x^(2)dx=2tdt ` `:. int x^(5)(1+x^(3))^(2//3)dx =int x^(3)(1+x^(3))^(2//3)x^(2)dx ` `=int(t^(2)-1)(t^(2))^(2//3)x^(2)dx ` `=(2)/(3)int(t^(2)-1)t^(7//3)dt` `=(2)/(3)int(t^(13//3)-t^(7//3))dt ` `=(2)/(3){(3)/(16)t^(16//3)-(3)/(10)t^(10//3)}+C ` `=(1)/(8)(1+x^(3))^(8//3)-(1)/(5)(1+x^(3))^(5//3)+C` |
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| 131. |
Evaluate: `int(4x+1)/(x^2+3x+2) dx` |
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Answer» `I=int(4x+1)/(x^(2)+3x+2)dx` `=int(2(2x+3)-5)/(x^(2)+3x+2)dx` `=2int(2x+3)/(x^(2)+3x+2)dx-5int(1)/(x^(2)+3x+2)dx` `=2log|x^(2)+3x+2|-5int(1)/(x^(2)+3x+(9//4)-(9//4)+2)dx` `=2log|x^(2)+3x+2|-5int(1)/((x+3//2)^(2)-(1//2)^(2))dx` `=2log|x^(2)+3x+2|-5(1)/(2(1//2))log|(x+(3)/(2)-(1)/(2))/(x+(3)/(2)+(1)/(2))|+C` `=2log|x^(2)+3x+2|-5log|(x+1)/(x+2)|+C` |
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| 132. |
Evaluate: `int(x^(3)/(x+1)^(2))dx` |
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Answer» `intx^(3)/(x+1)^(2)dx=int(x^(3)+1-1)/(x+1)^(2)dx=int(x^(3)+1)/(x+1)^(2)dx-int(1/(x+1)^(2))dx` `=int(x-2+3/(x+1))dx-int1/(x+1)^(2)dx=x^(2)/2-2x+3ln(x+1)+1/(x+1)+C` |
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| 133. |
Evaluate : `int1/(4+9x^(2))`dx |
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Answer» We have `int1/(4+9x^(2))dx=dx1/9int1/(4/9+x^(2))=1/9int1/(2//3)^(2)+x^(2)dx` `=1/9.1/(2//3)tan^(-1)(x/(2//3))+C+1/6tan^(-1)(3x)/(2)+C` |
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| 134. |
`int(1-cosx)/(cosx(1+cosx))dx` |
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Answer» Correct Answer - `log|secx+tanx|-2tanx//2+C` `"Let " I=int(1-cosx)/(cosx(1+cosx))dx.` `"Let " cosx=y. " Then "` `(1-cosx)/(cosx(1+cosx))=(1-y)/(y(1+y))` `=(1)/(y)-(2)/(1+y)` `=(1)/(cosx)-(2)/(1+cosx)` `:. I=int(1-cosx)/(cosx(1+cosx))dx` `=int(1)/(cosx)dx-int(2)/(1+cosx)dx` `=int sec x dx-int(2)/(2cos^(2)x//2)dx` `=int secx-int sec^(2)x//2dx` `=log|secx+tanx|-2tanx//2+C` |
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| 135. |
Evaluate `int sin x cos x cos2x cos 4x cos 8x dx` |
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Answer» Correct Answer - `-(1)/(256)cos16x+C` `I= intsin x cos x cos2x cos 4x cos 8x dx` `=(1)/(2)int sin2x cos2x cos4x cos8x dx` `=(1)/(4)intsin4xcos4xcos8xdx` ` = (1)/(8)intsin8xcos8xdx` ` =(1)/(16)intsin16x dx = (-1)/(256)cos16x+C` |
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| 136. |
`intx2^(ln(x^(2)+1))dx` is equal toA. `(2^(ln(x^(2)+1)))/(2(x^(2)+1))+C`B. `(x^(2)+1)2^(ln(x^(2)+1))/(ln2+1)`C. `((x^(2)+1)^(ln2+1))/(2(ln2+1))+C`D. `((x^(2)+1)^(ln2))/(2(ln2+1))+C` |
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Answer» Correct Answer - C `I=intx2^(ln(x^(2)+1))dx` Let `x^(2)+1=t,rArr xdx=(dt)/(2)` Hence `I=(1)/(2)int2^(lnt)dt` `=(1)/(2)intt^(ln2)dt` `=(1)/(2).(t^(ln2+1))/(ln2+1)+C` `=+(1)/(2).((x^(2)+1)^(ln2+1))/(ln2+1)C` |
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| 137. |
`int(x^9)/((4x^2+1)^6) dx` is equal toA. `(1)/(5x)(4+(1)/(x^(2)))^(-5)+C`B. `(1)/(5)(4+(1)/(x^(2)))^(-5)+C`C. `(1)/(10)(1+4x^(2))^(-5)+C`D. `(1)/(10)(4+(1)/(x^(2)))^(-5)+C` |
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Answer» Correct Answer - D `I=int(x^(9)dx)/((4x^(2)+1)^(6))=int(dx)/(x^(3)(4+(1)/(x^(2)))^(6))` `= -(1)/(2)int(d(4+(1)/(x^(2))))/((4+(1)/(x^(2)))^(6))= -(1)/(2)((4+(1)/(x^(2)))^(-5))/(-5)+C` `=(1)/(10)(4+(1)/(x^(2)))^(-5)+C` |
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| 138. |
Evaluate:`int(cos2xsin4x dx)/(cos^4x(1+cos^2 2x)` |
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Answer» Correct Answer - `-(1)/(2)log|cos 2x|-(1)/(2(1+cos 2x))+(1)/(4)log(1+cos^(2)2x)+C` `I=int(cos2x sin4x dx)/(cos^(4)x(1+cos^(2)2x))` `=int(2cos^(2)2x sin2x dx)/(((1+cos2x)/(2))^(2)(1+cos^(2) 2x))` `"Let " cos2x=t. " then "dt=-2sin2x dx. ` `:. I=-int(t^(2)dt)/(((1+t)/(2))^(2)(1+t^(2)))=-4int(t^(2)dt)/((1+t)^(2)(1+t^(2)))` ` "Now ",(t^(2))/((1+t)^(2)(1+t^(2)))=(A)/(1+t)+(B)/((1+t)^(2)) +(Ct+D)/(1+t^(2))` `"or " t^(2)=A(1+t)(1+t^(2))+B(1+t^(2))+(Ct+D)(1+t)^(2) ` ` "Put " t=-1. " Then "B=1//2` `"Put "t=0. " Then "0=A+1//2+D" (1)" ` `"Put "t=1. " Then " 1=4A+1+4C+4D " or "A+C+D=0 " "(2)` `"From equations(1)and(2), " C=1//2` `"Comparing coefficients of "t^(3),A+C=0 " "(3)` `"or "A=-1//2` `"From equations(2)and(3)",D=0` `"Hence, " I=int(-(1//2)/(1+t)+(1//2)/((1+t)^(2))+((1//2)t)/(1+t^(2)))dt` `=-(1)/(2)log|t|-(1)/(2(1+t))+(1)/(4)log(1+t^(2))+C` `=-(1)/(2)log|t|-(1)/(2(1+t))+(1)/(4)log(1+t^(2))+C," where "t=cos2x` |
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| 139. |
`int(dx)/(cos(2x)cos(4x))=`A. `(1)/(2sqrt2)log|(1+sqrt2sin2x)/(1-sqrt2sin2x)|-(1)/(2)(log|sec2x-tan2x|)+C`B. `(1)/(2sqrt2)log|(1-sqrt2sin2x)/(1+sqrt2sin2x)|-(1)/(2)(log|sec2x-tan2x|)+C`C. `(1)/(sqrt2)log|(1+sqrt2sin2x)/(1-sqrt2sin2x)|-(1)/(2)(log|sec2x-tan2x|)+C`D. none of these |
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Answer» Correct Answer - B `int(sin(4x-2x)dx)/(sin(2x)cos(2x)cos(4x))` `=int(sin(4x)dx)/(sin(2x)cos(4x))-int sec2xdx` `=2int(cos2xdx)/(cos4x)-(1)/(2)(log|sec 2x-tan 2x|)` `=2int(cos2xdx)/(1-2sin^(2)2x)-(1)/(2)(log|sec2x-tan2x|)` `=int(dt)/(1-2t^(2))-(1)/(2)(log|sec 2x-tan2x|)` |
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| 140. |
Evaluate:`int(x^2+1)/(x^4+1)dx` |
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Answer» `I=int(x^(2)+1)/(x^(4)+1)dx = int(1+(1)/(x^(2)))/(x^(2)+(1)/(x^(2)))dx=int(1+(1)/(x^(2)))/((x-(1)/(x))^(2)+2)dx` Let `x-(1)/(x)=t " or " d(x-(1)/(x))=dt " or " (1+(1)/(x^(2)))dx=dt` `:. I=int(dt)/(t^(2)+(sqrt(2))^(2))=(1)/(sqrt(2))tan^(-1)((t)/(sqrt(2)))+C` `=(1)/(sqrt(2))tan^(-1)((x-1//x)/(sqrt(2)))+C` `=(1)/(sqrt(2))tan^(-1)((x^(2)-1)/(sqrt(2)x))+C` |
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| 141. |
Evaluate: `int1/(e^x+e^(-x)) dx` |
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Answer» Correct Answer - `tan^(-1)(e^(x))+C` `I=int(dx)/(e^(x)+e^(-x))=int(e^(x)dx)/(e^(2x)+1)` Let `e^(x)=t.` ` :. e^(x)dx=dt` ` :. I=int(dt)/(t^(2)+1)=tan^(-1)t+C=tan^(-1)(e^(x))+C` |
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| 142. |
Evaluate: `int(dx)/((x-4)sqrt(x+5))` |
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Answer» Let `I=int(dx)/((x-4)sqrt(x+5))` { put `x+5=t^(2) rArr dx=2t dt}` `therefore I=int(2dt)/(t^(2)-9)= 2/6ln|(t-3)/(t+3)|+C=1/3ln|(sqrt(x+5)-3)/(sqrt(x+5)+3)|+C` |
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| 143. |
Evaluate ` int(dx)/(x+sqrt(x))` |
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Answer» Correct Answer - `2log(sqrt(x)+1)+C` ` I=int(dx)/(x+sqrt(x))=int(dx)/(sqrt(x)(sqrt(x)+1)). ` ` " Put " sqrtx=z ` ` :. (1)/(2sqrt(x))dx=dz ` ` :. I=int(2dz)/(z+1)` `=2log|z+1|+C` ` =2log(sqrt(x)+1)+C` |
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| 144. |
Evaluate: `int(x^2-1)/(x^4+x^2+1) dx` |
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Answer» `I=int(x^(2)-1)/(x^(4)+x^(2)+1)dx` `=int(1-(1)/(x^(2)))/(x^(2)+1+(1)/(x^(2)))dx` `=int(1-(1)/(x^(2)))/((x+(1)/(x))^(2)-1^(2))dx` Let `x+(1)/(x)=u. " Then " d(x+(1)/(x))=du " or "(1-(1)/(x^(2)))dx=du` `" or " I=int(du)/(u^(2)-1^(2))` `=(1)/(2(1))log|(u-1)/(u+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` |
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| 145. |
`int(dx)/(xsqrt(x^(6)-16))=`A. `sec.(x^(3))/(4)+c`B. `(1)/(12)sec.(x^(3))/(4)+c`C. `(1)/(3)sec.(x^(3))/(4)+c`D. none of these |
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Answer» Correct Answer - B `int(dx)/(xsqrt(x^(6)-16))=int(x^(2)dx)/(x^(3)sqrt(x^(6)-16))` `=(1)/(3)int(dt)/(sqrt(t^(2)-16))" "("putting "x^(3)=t)` `=(1)/(48)int(dt)/((1)/(4)sqrt((t^(2))/(16)-1))` `=(1)/(12)sec.(t)/(4)+c` `=(1)/(12)sec. (x^(3))/(4)+c` |
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| 146. |
Evaluate:`int1/(sqrt((x-1)(x-2)))dx` |
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Answer» `I=int(1)/(sqrt(x^(2)-3x+2))dx` `=int(1)/(sqrt(x^(2)-3x+(9)/(4)-(9)/(4)+2))dx` `=int(1)/(sqrt((x-(3)/(2))^(2)-((1)/(2))^(2)))dx` `=log|(x-(3)/(2))+sqrt((x-(3)/(2))^(2)-((1)/(2))^(2))|+C` `=log|(x-(3)/(2))+sqrt(x^(2)-3x+2)|+C` |
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| 147. |
Evaluate: `int(sin^(-1)sqrt(x)-cos^(-1)sqrt(x))/(sin^(-1)sqrt(x)+cos^(-1)sqrt(x)) dx` |
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Answer» Correct Answer - `(2)/(pi)[sqrt(x-x^(2))-(1-2x)"sin"^(-1)sqrt(x)]-x+c` Let `I=int (sin^(-1)sqrt(x)-cos^(-1)sqrt(x))/(sin^(-1)sqrt(x)+cos^(-1)sqrt(x))dx` `=int(sin^(-1)sqrt(x)-((pi)/(2)-sin^(-1)sqrt(x)))/((pi)/(2))dx` `=(2)/(pi)int(2sin^(-1)sqrt(x)-(pi)/(2))dx=(4)/(pi)intsin^(-1)sqrt(x)dx - x + c" "...(i)` Now, `int sin^(-1) sqrt(x)dx` Put `x = sin^(2) theta rArr dx = sin 2 theta` `=int theta*sin2 theta d theta =-(theta cos 2 theta)/(2)+int(1)/(2)cos 2 theta d theta` `=-(theta)/(2)cos 2 theta + (1)/(4)sin 2 theta` `=-(1)/(2)theta(1-2sin^(2)theta)+(1)/(2)sin theta sqrt(1-sin^(2)theta)` `=-(1)/(2)sin^(-1)sqrt(x)(1-2x)+(1)/(2)sqrt(x)sqrt(1-x)" "...(ii)` From Eqs. (i) and (ii), `I=(4)/(pi)[-(1)/(2)(1-2x)sin^(-1)sqrt(x)+(1)/(2)sqrt(x-x^(2))]-x+c` `=(2)/(pi)[sqrt(x-x^(2))-(1-2x)sin^(-1)sqrt(x)]-x+c` |
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| 148. |
Evaluate:`int((x-1)e^x)/((x+1)^3)dx` |
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Answer» Correct Answer - `(e^(x))/((x+1)^(2))+c` Let `I=int((x-1)e^(x))/((x+1)^(3))dx` `I=int{(x+1-2)/((x+1)^(3))}e^(x) dx=int{(1)/((x+1)^(2))-(2)/((x+1)^(3))}e^(x) dx` `=int e^(x)(1)/((x+1)^(2))dx-2 int e^(x)*(1)/((x+1)^(3))dx` Applying integration by parts, `={(1)/((x+1)^(2))*e^(x)-int e^(x)*(-2)/((x+1)^(3))dx}-2int e^(x)*(1)/((x+1)^(3))dx=(e^(x))/((x+1)^(2))+c` |
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| 149. |
Evaluate:`int(e^(logx)+sinx)cosxdx` |
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Answer» Correct Answer - `"x sin x" + "cos x" -("cos"2x)/(4)+c` Let `I=int (e^(log x)+sin x)cos x dx` `=int (x+sin x) cos x dx` `=int x cos x dx +(1)/(2)int(sin 2x)dx` `=(x*sin x -int 1*sin x dx)-(cos 2 x)/(4)+c` `=x sin x + cos x -(cos 2 x)/(4)+c` |
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| 150. |
Evaluate `int(dx)/(sqrt(1-x^(2))(1+sqrt(1-x^(2))))`. |
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Answer» Put `x="sin"theta` ` :. dx="cos"theta"d"theta` ` :. I=int(dx)/(sqrt(1-x^(2))(1+sqrt(1-x^(2))))` `=int("cos"theta"d"theta)/("cos"theta(1+"cos"theta))` `=(1)/(2)int"sec"^(2)(theta)/(2)"d"theta` `="tan"(theta)/(2)+c` `=(2"sin"(theta)/(2)"cos"(theta)/(2))/(2"cos"^(2)(theta)/(2))+c` `=("sin"theta)/(1+"cos"theta)+c` `=(x)/(1+sqrt(1-x^(2)))+c` |
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