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51.

Evaluate `intcos^(3)xdx.`

Answer» Correct Answer - `(3)/(4) sin x +(1)/(12) sin 3x+c`
Discussion
52.

Evaluate: `intcos^(-1)((1-x^2)/(1+x^2)) dx`

Answer» Correct Answer - `2x tan^(-1) x-log (1+x^(2))+c`
Discussion
53.

`intcos(logx)dx`A. `(x)/(2)[sin (log x) +cos (log x)]+c`B. `(x)/(2) [sin (log x) -cos (log x) ]+c`C. None of the aboveD.

Answer» Correct Answer - B
Discussion
54.

`intcos^(2) x dx`

Answer» Correct Answer - `(1)/(2) x+(1)/(4) sin 2x+c`
Discussion
55.

`int(1)/(3+(2-3x)^(2)) dx`

Answer» Correct Answer - `(-1)/(3sqrt(3)) tan^(-1) ((2-3x)/(sqrt(3)))+c`
Discussion
56.

`int(1)/(3sin^(2) x+4 cos^(2) x)dx`

Answer» Correct Answer - `(1)/(2sqrt(3))tan^(-1) ((sqrt(3)tanx)/(2))+c`
Discussion
57.

`intsqrt(1+cos x )dx` is equal toA. `2sqrt2 cos.(x)/(2)+C`B. `2sqrt2 sin.(x)/(2)+C`C. `sqrt2 cos .(x)/(2)+C`D. `sqrt2sin.(x)/(2)+C`

Answer» Correct Answer - B
`intsqrt(1+cosx)dx=intsqrt(2cos^(2).(x)/(2))dx`
`=sqrt2intcos.(x)/(2)dx=2sqrt2.sin.(x)/(2)+C`
Discussion
58.

`int b^(x+a) " dx "`

Answer» Correct Answer - `(b^(x+a))/(log b)+c`
Discussion
59.

`int(1)/(tsqrt(t^(2) -1))dt`

Answer» Correct Answer - `sec^(-1) t+c`
Discussion
60.

`int1/sqrt(1-y^2)dy`

Answer» Correct Answer - `sin^(-1) y+c`
Discussion
61.

`int(1)/(sin^(2)x-4 cos^(2)x)dx`

Answer» Correct Answer - `(1)/(4)log |(tan x-2)/(tan x+2)| +c`
Discussion
62.

`int_(0)^(pi//2) " cosec "(x-(pi)/(3)) " cosec "(x-(pi)/(6)) dx=?`A. `-log 3`B. `-2 log 3`C. `2 log 3`D.

Answer» Correct Answer - C
Discussion
63.

Evaluate:`int_0^(pi//2)1/((a^2cos^2x+b^2sin^2x)^2)dx`A. `(pi(a^(2) +b^(2)))/(4a^(3)b^(3))`B. `(pi(a^(2)+b^(2)))/(4a^(2)b^(2))`C. None of the aboveD.

Answer» Correct Answer - B
Discussion
64.

`int(1)/(5-4 sin x)dx`

Answer» Correct Answer - `(2)/(3) tan^(-1) [(5tan.(x)/(2) -4)/(3)]+c`
Discussion
65.

`int(1)/(cos^(2) x-3 sin^(2) x)dx`

Answer» Correct Answer - `(1)/(2sqrt(3))log |(1+sqrt(3) tanx)/(1-sqrt(3)tanx)|+c`
Discussion
66.

`int(1)/(5+2 cos x)dx`

Answer» Correct Answer - `(2)/(sqrt(21))tan^(-1) ((sqrt(3)tan.(x)/(2))/(sqrt(7))) +c`
Discussion
67.

`int(1)/((sin x-2 cos x) (2 sin x+c osx))dx`

Answer» Correct Answer - `(1)/(5)log |(tan x-2)/(2tan x+1)|+c`
Discussion
68.

`int(1+x)/(x+e^(-x))dx` is equal toA. `log|(x-e^(-x))|+C`B. `log|(x+e^(-x))|+C`C. `log|(1+xe^(x))|+C`D. `(1+xe^(x))^(2)+C`

Answer» Correct Answer - C
Let `l=int(1+x)/(x+e^(-x))dx=int(e^(x)(1+x))/(xe^(x)+1)dx`
Put`" "xe^(x)+1=t`
`rArr" "(xe^(x)+e^(x))dx=dt`
`rArr" "(x+1)e^(x)dx=dt`
`therefore" "l=int(dt)/(dt)=log|t|+C`
`=log|(xe^(x)+1)|+C`
Discussion
69.

The value of the integral `int(dx)/((e^(x)+e^(-x))^2)` isA. `(1)/(2)(e^(2x)+1)+C`B. `(1)/(2)(e^(-2x)+1)+C`C. `-(1)/(2)(e^(2x)+1)^(-1)+C`D. `(1)/(4)(e^(2x)-1)+C`

Answer» Correct Answer - C
Let `l=int(dx)/((e^(x)+e^(-x))^(2))=int(e^(2x)dx)/((e^(2x)+1)^(2))`
Put `e^(2x)+1-z rArr 2e^(2x)dx=dz`
`therefore" "i=(1)/(2)int(dz)/(z^(2))=-(1)/(2z)+C`
`=-(1)/(2(e^(2x)+1))+C`
`=-(1)/(2)(e^(2x)+1)^(-1)+C`
Discussion
70.

`intx^(-2//3)(1+x^(1//2))^(-5//3)` dx is equal toA. `3(1+x^(-1//2))^(-1//3)+C`B. `3(1+x^(-1//2))^(-2//3)+C`C. `3(1+x^(1//2))^(-2//3)+C`D. None of the above

Answer» Correct Answer - B
`intx^(-2//3).{(x^(1//2))(x^(-1//2)+1)}^(-5//3)dx`
`=int x^(-2//3).x^(5//5).(x^(-1//2)+1)^(-5//3)dx`
`=intx^(-3//2).(x^(-1//2)+1)^(-5//3)dx`
`"Put "x^(-1//2)+1=t`
`therefore" "-(1)/(2)x^(-3//2)dx=dt`
Then,`" "-2int t^(-5//3)dt=-2.(t^(-2//3))/(-2//3)+C`
`=3(x^(-1//2)+1)^(-2//3)+C`
Discussion
71.

`int(dx)/(sqrt(1-e^(2x)))` is equal toA. `log|e^(-x)+sqrt(e^(-2x)-1)|+C`B. `log|e^(x)+sqrt(e^(2x)-1)|+C`C. `-log|e^(-x)+sqrt(2x^(-2x)-1)|+C`D. `-log|e^(-2x)+sqrt(e^(-2x)-1)|+C`

Answer» Correct Answer - C
Let `l=int(dx)/(sqrt(1-e^(2x)))=int(e^(-x))/(sqrt(e^(-2x)-1))dx`
Put `e^(-x)=t rArr e^(-x)dx=-dt`
`therefore" "l=-int(dt)/(sqrt(t^(2)-1))=-log|t+sqrt(t^(2)-1)|+C`
`=-log|e^(-x)+sqrt(e^(-2)-1)|+C`
Discussion
72.

`int(x^(3)-1)^(1//3)x^(5)dx` is equal toA. `(1)/(7)(x^(3)+1)^(1//3)+(1)/(4)(x^(3)-1)^(3//4)+C`B. `(1)/(7)(x^(3)-1)^(7//3)+(1)/(4)(x^(3)-1)^(4//3)+C`C. `(3)/(7)(x^(3)-1)^(7//3)+(1)/(4)(x^(3)-1)^(4//3)+C`D. None of the above

Answer» Correct Answer - B
`int(x^(3)-1)^(1//3)x^(5)dx=int(x^(3)-1)^(1//3)x^(3).x^(2)dx`
Let `x^(3)-1=t rArr x^(3)=t+1`
Differentiating w.r.t. x, we get `3x^(2)=(dt)/(dx) rArr dx=(dt)/(3x^(2))`
`therefore" "intx^(3)-1^(1//3)x^(3).x^(2)dx=intt^(1//3)(t+1)x^(2)(dt)/(3x^(2))`
`=(1)/(3)int(t^(4//3)+t^(1//3))dt`
`=(1)/(3)[(t^(7//3))/((7)/(3))+(t^(4//3))/((4)/(3))]+C=(1)/(3)[(3)/(7)t^(7//5)+(3)/(4)t^(4//3)]+C`
`=(1)/(7)t^(7//3)+(1)/(4)t^(4//3)+C`
`=(1)/(7)(x^(3)-1)^(7//3)+(1)/(4)(x^(3)-1)^(4//3)+C`
Discussion
73.

`int(dx)/(x(x^(2)+1))` is equal toA. `log|x|-(1)/(2)log(x^(2)+1)+C`B. `log|x|+(1)/(2)log(x^(2)+1)+C`C. `-log|x|+(1)/(2)log(x^(2)+1)+C`D. `(1)/(2)log|x|+log(x^(2)+1)+C`

Answer» Correct Answer - A
Let `(1)/(x(x^(2)+1))=(A)/(x)+(Bx+C)/(x^(2)+1)`
`rArr" "1=A(x^(2)+1)+(Bx+C)x=(A+B)x^(2)+Cx+A`
On equating the coefficients of `x^(2),x` and constant term on both sides, we get
`A+B=0, C=0` and `A=1`
On solving these equations, we get
`A=1, B=-1 and C=0`
`therefore" "(1)/(x(x^(2)+1))=(1)/(x)+(-x)/(x^(2)+1)`
`therefore" "int(1)/(x(x^(2)+1))dx=int{(1)/(x)-(x)/(x^(2)+1)dx}`
`=log|x|-(1)/(2)log(x^(2)+1)+C`
Discussion
74.

`int(x^(2)+1)sqrt(x+1)dx` is equal toA. `((x+1)^(7//2))/(7)-2((x+1)^(5//2))/(5)+2((x+1)^(3//2))/(3)+C`B. `2[((x+1)^(7//2))/(7)-2((x+1)^(5//2))/(5)+2((x+1)^(3//2))/(3)]+C`C. `((x+1)^(7//2))/(5)-2((x+1)^(5//2))/(5)+5`D. `((x+1)^(7//2))/(7)-3((x+1)^(5//2))/(5)+11(x+1)^(1//2)+C`

Answer» Correct Answer - B
Put `x+1=t^(2)`
`rArr" "dx=2t dt`
`rArr" "x^(2)+1=(t^(2)-1)^(2)+1=t^(4)-2t^(2)+2`
`therefore" Given integral "=int(t^(4)-2t^(2)+2)t.2t dt`
`=2int(t^(6)-2t^(4)+2t^(2))dt`
`=2[(t^(7))/(7)-2(t^(5))/(5)+2(t^(3))/(3)]+C`
`=2[(x+1)^((7)/(2))/(7)-2(x+1)^((5)/(2))/(5)+2(x+1)^((3)/(2))/(3)]+C`
Discussion
75.

`int |x|ln|x|dx` equals `(x ne 0)`A. `(x^(2))/(2)ln|x|-(x^(2))/(4)+C`B. `(1)/(2)x|x|lnx+(1)/(4)x|x|+C`C. `-(x^(2))/(2)ln|x|+(x^(2))/(4)+C`D. `(1)/(2)x|x|ln|x|-(1)/(4)x|x|+C`

Answer» Correct Answer - D
Case I If x gt 0, then `|x|=x`
On applying integration by parts, we get
`therefore" "=intx ln x dx=ln x.(x^(2))/(2)-int(1)/(x).(x^(2))/(2)dx`
`=(x^(2))/(2).lnx-(x^(2))/(4)+C`
Case II If x lt 0, then `|x|=-x`
`therefore" "int |x|ln |x| dx = - int x ln (-x) dx`
`-{ln(-x).(x^(2))/(2)-(x^(2))/(4)}+C`
`-(x^(2))/(2)ln|x|+(x^(2))/(4)+C`
Combining both cases, we get
`(1)/(2)x|x|ln|x|-(1)/(4)x|x|+C`
Discussion
76.

`int(xdx)/((x-1)(x-2))` is equals toA. `log|((x-1)^(2))/(x-2)|+C`B. `log|((x-2)^(2))/(x-1)|+C`C. `log|((x-1)/(x-2))^(2)|+C`D. `log|(x-1)(x-2)|+C`

Answer» Correct Answer - B
Let `(x)/((x-1)(x-2))=(A)/((x-1))+(B)/((x-2))`
`rArr" "x=A(x-2)+B(x-1)" …(i)"`
On substituting x = 1 in Eq (i), we get `A=-1` and on substituting x = 2 in eq (i), we get B = 2
`therefore " "(x)/((X-1)(x-2))=-(1)/((x-1))+(2)/((x-2))`
`therefore int (x)/((x-1)(x-2))dx=int((-1))/(x-1)dx+int(2)/(x-2)dx`
`=-log|x-1|+2log|x-2|+C`
`=-log|x-1|+log(x-2)^(2)+C`
`=log|((x-2)^(2))/(x-1)|+C`
Discussion
77.

`int(e^(x)(1+x))/(cos^(2)(e^(x)x))dx` equal toA. `-cot(ex^(x))+C`B. `tan(xe^(x))+C`C. `tan(e^(x))+C`D. `cot(e^(x))+C`

Answer» Correct Answer - B
`int(e^(x)(1+x))/(cos^(2)(e^(x)x))dx`
Let`" "xe^(x)=t`
`rArr" "(xe^(x)+e^(x))=(dt)/(dx) rArr dx=(dt)/(e^(x)(x+1))`
`therefore int(e^(x)(1+x))/(cos^(2)(e^(x)x))dx=int(e^(x)(1+x))/(cos^(2)t)xx(dt)/(e^(x)(1+x))`
`=int(1)/(cos^(2)t)dt=intsec^(2)tdt`
`=tant +C tan(xe^(x))+C`
Discussion
78.

`int(logsqrtx)/(3sqrtx)dx` is equal toA. `(1)/(3)(log sqrtx)^(2)+C`B. `(2)/(2)(log sqrtx)^(2)+C`C. `(2)/(3)(logx)^(2)+C`D. `(1)/(3)(logx)^(2)+C`

Answer» Correct Answer - A
Let `l=int(logsqrtx)/(3x)dx`
Put`" "sqrtx=t`
`(1)/(sqrtx)dx=2dt`
`therefore" "l=int(2logt)/(3t)dt=(2)/(3)int(logt)/(t)dt`
Again, put `log t = mu rArr (1)/(t)dt= d mu`
`therefore" "l=(2)/(3)int mu d mu=(2)/(3)(mu^(2))/(2)+C`
`=(2)/(3).((logt)^(2))/(2)+C=(1)/(3)(log sqrtx)^(2)+C`
Discussion
79.

`int(sqrtx)/(sqrt(a^(3)-x^(3)))dx` equal toA. `(1)/(3)sin^(-1)sqrt((x^(3))/(a^(3)))+C`B. `(2)/(3)sin^(-1)sqrt((x^(3))/(a^(3)))+C`C. `(2)/(3)sin^(-1)sqrt((x)/(a))+C`D. None of these

Answer» Correct Answer - B
We have, `l=int(sqrtx)/(sqrt(a^(3)-x^(3)))dx`
`l=int(sqrtx)/(sqrt((a^(3//2))^(2)-(x^((3)/(2)))^(2)))dx`
Put`" "x^((3)/(2))=t`
`rArr" "(3)/(2)x^((1)/(2))dx=dt" "rArr" "sqrtxdx=(2)/(3)dt`
`therefore" "l=(2)/(3)int(dt)/(sqrt((a^((3)/(2)))^(2)-t^(2)))=(2)/(3)sin^(-1)((t)/(a^((3)/(2))))+C`
`=(2)/(3)sin^(-1)((x^((3)/(2)))/(a^((3)/(2))))+C=(2)/(3)sin^(-1)sqrt((x^(3))/(a^(3)))+C`
Discussion
80.

The value of `int(x^(2))/(1+x^(6))dx` is equal toA. `x^(3)+C`B. `(1)/(3)tan^(-1)(x^(3))+C`C. `log(1+x^(3))`D. None of these

Answer» Correct Answer - B
Let `l=int(x^(2))/(1+x^(6))dx`
Put`" "x^(3)=t rArr x^(2)dx=(1)/(3)dt`
`therefore" "l=(1)/(3)int(1)/(1+t^(2))dt=(1)/(3)tan^(-1)t+C`
`=(1)/(3)tan^(-1)(x^(3))+C`
Discussion
81.

If `int(dx)/((x+100)sqrt(x+99))=f(x)+C`, then `f(x)` isA. `2(x+100)^(1//2)`B. `3(x+100)^(1//2)`C. `2tan^(-1)(sqrt(x+99))`D. `2tan^(-1)(sqrt(x+100))`

Answer» Correct Answer - C
Let `l=int(dx)/([(sqrt(x+99))^(2)+1]sqrt(x+99))`
Put`" "sqrt(x+99)= t rArr (1)/(sqrt(x+99))dx=2dt`
`therefore" "=int(2dt)/(t^(2)+1)=2 tan^(-1)t+C`
`=2tan^(-1)sqrt((x+99))+C`
But`" "2tan^(-1)sqrt(x+99)+C=f(x)+C" [given]"`
`therefore" "f(x)=2 tan^(-1)(sqrt(x+99))`
Discussion
82.

If `d/(dx)f(x)=4x^3-3/(x^4)`such that `f(2)=0.`Then f(x) is(A) `x^4+1/(x^3)-(129)/8` (B) `x^3+1/(x^4)+(129)/8`(C) `x^4+1/(x^3)+(129)/8` (D) `x^3+1/(x^4)-(129)/8`A. `x^(3) +(1)/(x^(4)) +(129)/(8)`B. `x^(4) +(1)/(x^(3)) - (129)/(8)`C. `x^(3) +(1)/(x^(4)) + (129)/(8)`D. None Of These

Answer» Correct Answer - B
`(d)/(dx)f(x) =4x^(2) -(3)/(x^(4))`
`rArr f(x) =4 int x^(3) dx-3 int x^(-4) dx`
` = x^(4) +x^(-3) +c`
`" Given that " f(2) =0`
`rArr 16+ (1)/(8) + c = 0`
`rArr c=-(129)/(8)`
`:. f(x) =x^(4) +(1)/(x^(3))-(129)/(8)`
Discussion
83.

`int(dx)/(x(logx)^(m))` is equal to thenA. `((logx)^(m))/(m)+C`B. `((logx)^(m-1))/(m-1)+C`C. `((logx)^(1-m))/(1-m)+C`D. `((logx)^(1-m))/(m)+C`

Answer» Correct Answer - C
`int(1)/(x(logx)^(m))dx`
Let `log x-t rArr (1)/(x)=(dt)/(dx) rArr dx=xdx`
`therefore" "int(1)/(x(logx)^(m))dx=int(1)/(x(t)^(m))xdt int t^(-m)dt`
`=(t^(-m+1))/(-m+1)+C=((logx)^(1-m))/(1-m)+C`
Discussion
84.

`int(1+tanx)/(e^(-x)cosx)dx` is equal toA. `e^(-x)tanx+C`B. `e^(-x)secx+C`C. `e^(x)sec x+C`D. `e^(x)tanx+C`

Answer» Correct Answer - C
Let `l=int e^(x) sec x dx +int e^(x) sec x tan x dx`
On integration by parts, we get
`e^(x)secx - int e^(x) sec x tan x dx+int e^(x)sec x tan x dx+C`
`=e^(x)secx+C`
Discussion
85.

`int(1)/(x^(2)-9)dx` is equal toA. `log|(x-3)/(x+3)|+C`B. `(1)/(6)log|(x-3)/(x+3)|+C`C. `(1)/(6)log|(x+3)/(x-3)|+C`D. `log|(x+3)/(x-3)|+C`

Answer» Correct Answer - B
`int(1)/(x^(2)-9)dx=int(1)/(x^(2)-3^(2))dx=int(1)/((x+3)(x-3))dx`
Let`" "(1)/((x+3)(x-3))=(A)/((x+3))+(B)/((x-3))`
`" "1=A(x-3)+B(x+3)`
`rArr" "1=x(A+B)+(-3A+3B)`
On equation the coefficients of x and constant term on both sides, we get
`A+B=0 and -3A+3B=1`
On solving, we get `A=-(1)/(6) and B=(1)/(6)`
`therefore int(1)/((x+3)(x-3))dx=int((-1))/(6(x+3))dx+int(1)/(6(x-3))dx`
`=-(1)/(6)log|x+3|+(1)/(6)log|x-3|+C`
`=(1)/(5)log|(x-3)/(x+3)|+C`
Discussion
86.

`intsqrtx e^(sqrtx)dx` is equal toA. `2sqrtx-e^(sqrtx)-4sqrt(xe^(sqrtx))+C`B. `(2x-4sqrtx+4)e^(sqrtx)+C`C. `(2x+4sqrtx+4)e^(sqrtx)+C`D. `(1-4sqrtx)e^(sqrtx)+C`

Answer» Correct Answer - B
Let`" "l=intsqrt(x e^(sqrtx)dx`
Put`" "x=t^(2)" "rArr" "dx=2tdt`
Now, integration by parts, we get
`l=2int t^(2)e^(y)dt=2[t^(2)e^(t)-(2t)e^(t)+2e^(t)]+C`
`=(2x-4sqrtx+4)e^(sqrtx)+C`
Discussion
87.

Evaluate `int(1-cosx)"cosec"^(2)xdx`A. `tan.(x)/(2)+C`B. `-cot.(x)/(2)+C`C. `2tan.(x)/(2)+C`D. `-2cot.(x)/(2)+C`

Answer» Correct Answer - A
(a) `int(1-cos x)"cosec"^(2)x dx=int(1-cosx)/(1-cos^(2)x)dx`
`=int(1)/(1+cosx)dx=(1)/(2)int sec^(2).(x)/(2)dx=tan.(x)/(2)+C`
Discussion
88.

`int((sin theta+cos theta))/(sqrt(sin 2theta))d theta` is equal toA. `log |cos theta-sin theta+sqrt(sin 2 theta)|+C`B. `log|sin theta - cos theta+sqrt(sin 2 theta)|+C`C. `sin^(-1)(sin theta+cos theta)+C`D. `sin^(-1)(sin theta +cos theta)+C`

Answer» Correct Answer - C
Let `l=int(sin theta+cos theta)/(sqrt(1+sin 2 theta-1))d theta`
`=int (sin theta+cos theta)/(sqrt(1-(sin theta-cos theta)^(2)))d theta`
`"Put "sin theta - cos theta=t`
`rArr" "l=int(1)/(sqrt(1-t^(2)))dt = sin^(-1)t+C`
`=sin^(-1)(sin theta - cos theta)+C`
Discussion
89.

`int(x^(49)tan^(-1)(x^(50)))/((1+x^(100)))dx=k[tan^(-1)(x^(50))]^(2)+C,` , then k is equal toA. `(1)/(50)`B. `-(1)/(50)`C. `(1)/(100)`D. `-(1)/(100)`

Answer» Correct Answer - C
Let `l=intx^(49)(tan^(-1)(x^(50)))/(1+(x^(50))^(2))dx`
Put`" "x^(50)=t" "rArr" "50x^(49)dx=dt`
`therefore" "l=(1)/(50)int(tan^(-1)t)/(1+t^(2))dt`
Again, put `tan^(-1)t=u rArr (1)/(1+t^(2))dt=du`
`therefore" "l=(1)/(50)int u du =(u^(2))/(100)+C=((tan^(-1)x^(50)))/(100)+C`
`"But "l=k(tan^(-1)x^(50))^(2)+C" [given]"`
`therefore" "k=(1)/(100)`
Discussion
90.

What is the value of `int(sqrtx + x)^(-1) dx`?A. `(1)/(2)log(1+sqrtx)+C`B. `2log (1+sqrtx)+C`C. `(1)/(2)log(1+sqrtx)+C`D. `3log(1+sqrtx)+C`

Answer» Correct Answer - B
Let `l=int(dx)/(x+sqrtx)=int(1)/(sqrtx(1+sqrtx))dx`
`"Put "1+sqrtx-t rArr (1)/(sqrtx)dx=2dt`
`therefore" "l=2 int(1)/(t)dt=2og t+C`
`" "=2 log(1+sqrtx)+C`
Discussion
91.

Let `f(x)=int x^2/((1+x^2)(1+sqrt(1+x^2)))dx` and `f(0)=0` then `f(1)` isA. `log(1+sqrt2)`B. `log(1+sqrt2)-(pi)/(4)`C. `log(1+sqrt2)+pi//4`D. None of these

Answer» Correct Answer - B
On putting `x=tan theta rArr dx=sec^(2) theta d theta`, we get
`f(x)=int(tan^(2) theta.sec^(2)theta)/(sec^(2) theta(1+sec theta))d theta`
`=int(sec^(2) theta-1)/(1+sec theta)d theta=int(sec theta-1) d theta`
`=log (sec theta tan theta)- theta+C`
`rArr" "f(x)=log(sqrt(1+x^(2))+x)-tan^(-1)x+C`
At `x=0," "f(0)=log(1+0)-0+C rArr C=0`
At `x=1," "f(1)=log(1+sqrt2)-(pi)/(4)`
Discussion
92.

` int e^(x). " cos"^(2) " x dx"`

Answer» Correct Answer - `(1)/(2) e^(x) +(1)/(10)e^(x) (2 sin 2x +cos 2x)+c`
Discussion
93.

`intsqrt(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`

Answer» Correct Answer - A
Let `l=int sqrt(e^(x)-1)dx=int((sqrt(e^(x)-1))e^(x))/(1+(sqrt(e^(x)-1))^(2))dx`
`"Put "e^(x)-1=t^(2)" "rArr" "e^(x)dx=2t dt`
`therefore" "l=2int(t^(2)dt)/(1+t^(2))=2 int ((1+t^(2))/(1+t^(2)))dt-2int(1)/(1+t^(2))dt`
`=2 [t-tan^(-1)t]+C`
`=2[sqrt(e^(x)-1)-tan^(-1)sqrt(e^(x)-1)]+C`
Discussion
94.

`int cos^(-3//7)x sin^(-11//7)x dx` is equal toA. `log|sin^(4/7)x|+C`B. `(4)/(7)tan^(4//7)x+C`C. `-(7)/(4)tan^(-4//7)x+C`D. `log|cos^(3//7)x|+C`

Answer» Correct Answer - C
(c) `intcos^(-3//7)xsin^(-11//7)xdx=int(sin^(-11//7)x)/(cos^(-11//7)x).sec^(2)xdx`
`=int tan^(-11//7)x sec^(2)xdx`
`"Put "tanx=t rArr sec^(2)x dx=dt`
`therefore" "l=int t^(-11//7)dt=-(7)/(4)tan^(-4//7)x+C`
Discussion
95.

`intsqrt((x)/(1-x^(3)))dx` is equal toA. `(2)/(3)sin^(-1)(x^(2//3))+C`B. `(3)/(2)sin^(-1)(x^(2//3))+C`C. `(3)/(2)sin^(-1)(x^(2//3))+C`D. `(2)/(3)sin^(-1)(x^(2//3))+C`

Answer» Correct Answer - A
`l=intsqrt((x)/(1-x^(3)))dx=int(sqrtx)/(sqrt(1-x^(3)))dx=int(sqrtx)/(sqrt(1-(x^(3//2))^(2)))dx`
`"Put "x^(3//2)=t rArr sqrtx dx =(2)/(3)dt`
`"So, "l=(2)/(3)int(dt)/(sqrt(1-t^(2)))=(2)/(3)sin^(-1)(x^(3//2))+C`
Discussion
96.

`int (x^(2))/(1-2x^(3))dx`

Answer» Correct Answer - `-(1)/(6)log (1-2x^(3)) +c`
Discussion
97.

`int(x-.^(11)C_(1)x^(2)+.^(11)C_(2)x^(3)-.^(11)C_(3)x^(4)+...-.^(11)C_(x^(12)))dx` equal toA. `((1-x)^(12))/(12)-((1-x)^(11))/(11)+C`B. `((1-x)^(13))/(13)-((1-x)^(12))/(12)+C`C. `((1-x)^(11))/(11)-((1-x)^(12))/(12)+C`D. `((1-x)^(12))/(12)-((1-x)^(13))/(13)+C`

Answer» Correct Answer - B
Let `l=int x (1-.^(11)C_(1)x+.^(11)C_(2)x^(2)-.^(11)C_(3)x^(3)+…..-.^(11)C_(11)x^(11))dx`
`=int x(1-x)^(11)dx`
`"Put "1-x=t`
`therefore" "dx=-dt`
`"Then, "l=- int(1-t)t^(11)dt=-(t^(12))/(12)+(t^(13))/(13)+C`
`=((1-x)^(13))/(13)-((1-x)^(12))/(12)+C`
Discussion
98.

`int e^(" sin x "). " sin 2x dx "`

Answer» Correct Answer - `2.e^(sin x) (sin x-1)+c`
Discussion
99.

`int(1)/(sqrt(2x^(2)+3x-2))dx`

Answer» Correct Answer - `(1)/(sqrt(2)) log |4x+3+2sqrt(2(2x^(2)+3x-2))|+c`
Discussion
100.

if `I=int(log(t+sqrt(1+t^2)))/sqrt(1+t^2)dt=1/2(g(t))^2+c` then `g(2)` is (A) `2 log(2+sqrt5)` (B) `log (2 + sqrt 5)` (C) `1/sqrt5 log (2 + sqrt5)` (D) `1/2 log ( 2 + sqrt 5)`A. `(1)/(sqrt5)log(2+sqrt5)`B. `(1)/(1)log(2_sqrt5)`C. `2log(2+sqrt5)`D. `log(2+sqrt5)`

Answer» Correct Answer - D
`intlog(t+sqrt(1+t^(2)))/(sqrt(1+t^(2)))dt=(1)/(2)(g(t))^(2)+C" …(i)"`
`"Let "l=intlog(t+sqrt(1+t^(2)))/(sqrt(1+t^(2)))dt`
`"Put "u=log(t+sqrt(1+t^(2)))`
`rArr" "du=(1)/(t+sqrt(1+t^(2)))xx(1+(1)/(2sqrt(1+t^(2)))xx2t)dt`
`rArr" "du=(1)/(t+sqrt(1+t^(2)))xx(sqrt(1+t^(2))+1)/(sqrt(1+t^(2)))dt`
`rArr" "du=(dt)/(sqrt(1+t^(2)))`
`therefore" "l=int udu`
`rArr" "l=(u^(2))/(2)+C`
`rArr" "l=(1)/(2)[log(t+sqrt(1+t^(2)))]^(2)+C" ...(ii)"`
From Eps. (i) and (ii), we get
`g(t)=log(t+sqrt(1+t^(2)))`
`therefore" "g(2)=log(2+sqrt5)`
Discussion