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101.

`int_(0)^(oo)((tan^(-1)x)/(x(1+x^(2))))dx`

Answer» Correct Answer - `(pi)/(2)ln2`
Discussion
102.

Evaluate:`int_1^oo(e^(x+1)+e^(3-1))^(-1)dx`

Answer» Correct Answer - `(pi)/(4e^(2))`
`I=int_(1)^(oo)(dx)/((ee^(x)+e^(3)e^(-x)))`
`=int_(1)^(oo) (e^(x)dx)/(e(e^(2x)+e^(2)))` (multiply `N^(R)` and `D^(r)`by `e^(x)`)
Put `e^(x)=t` or `e^(x) dx=dt`
`:. I=1/e int_(e)^(oo) (dt)/(t^(2)+e^(2))`
`=1/(e^(2))"tan"(-1)t/e|_(e)^(oo)`
`=1/(e^(2))[(pi)/2-(pi)/4]=(pi)/(4e^(2))`
Discussion
103.

`int_(0)^(1) lnsin(pi/2x) dx`

Answer» Correct Answer - `-ln 2`
Discussion
104.

Let `f(x)` be a function satisfying `f(x) f(x+2) = 10 AA x in R`, thenA. `f(x)` is a periodic functionB. `f(x)` is aperiodic functionC. `underset(1)overset(7)intf(x) dx = 20`D. `underset(1)overset(7)intf(x) dx = 20`

Answer» Correct Answer - A::D
Discussion
105.

Evaluate:`int_0^(1/(sqrt(2)))(sin^(-1)x)/((1-x^2)sqrt(1-x^2))dx`

Answer» Correct Answer - `(pi)/4-1/2 log2`
Put `x=sin theta`.So `dx=cos theta d theta`
When `x=0, theta=0`, when `x=1/(sqrt(2)),theta=(pi)/4`
`:.` Given integral
`=int_(0)^(pi//4)(sin^(-1)(sin theta)cos theta d theta)/((1-sin^(2) theta)^(3//2))`
`=int_(0)^(pi//4)(theta cos theta)/(cos^(3) theta) d theta= int_(1)^(pi//4) underset(I)(theta).underset(II)(sec^(2)) theta d theta`
`=|theta tan theta |_(0)^(pi//4)-int_(0)^(pi//4)1.tan d theta`
`=(pi)/4 "tan" (pi)/4+log cos theta|_(0)^(pi//4)`
`=(pi)/4+"log cos"(pi)/4-"log cos"=(pi)/4+"log"1/(sqrt(2))`
`=(pi)/4+log1-log(2)^(1//2)=(pi)/4-1/2 log 2`
Discussion
106.

Let `f(x) = int_(0)^(pi)(sinx)^(n) dx, n in N` thenA. `I_(n)` is rational if n is oddB. `I_(n)` is irrational if n is evenC. `I_(n)` is an increasing sequenceD. `I_(n)` is a decreasing sequence

Answer» Correct Answer - A::B::D
Discussion
107.

`int_(-1)^(41//2)e^(2x-[2x])dx`, where `[*]` denotes the greatest integer function.

Answer» Correct Answer - `43/2(e-1)`
Discussion
108.

`int_(0)^(2pi)|sqrt(15) sin x + cos x|dx`

Answer» Correct Answer - 16
Discussion
109.

If `f(x)={{:(x+3 : x lt 3),(3x^(2)+1:xge3):}`, then find `int_(2)^(5) f(x) dx`.

Answer» `underset(2)overset(5)intf(x)dx=underset(2)overset(3)intf(x)dx+underset(5)overset(3)intf(x)dx=underset(2)overset(3)int(x+3)dx+overset(5)underset(3)int(3x^(2)+1)dx=[(x^(2))/(2)+3x]_(2)^(3)+[x^(3)+x]_(3)^(5)`
`= (9-4)/(2)+3(3-2)+5^(3)-3^(3)+5-3=(211)/(2)`
Discussion
110.

Evaluate:`int_0^1(2-x^2)/((1+x)sqrt(1-x^2))dx`

Answer» Correct Answer - `(pi)/2`
Put `x=sintheta`. So `dx=cos theta d theta`
`:.` Given integral
`=int_(0)^(pi//2)((2-sin^(2)theta)cos theta d theta)/((1+sin theta) cos theta)`
`=int_(0)^(pi//2)(1-sin theta+1/(1+sin theta))d theta`
`=|theta + cos theta|_(0)^(pi//2)+int_(0)^(pi//2)(d theta)/(1+sin theta)`
`=(pi)/2-1+int_(0)^(pi//2)(1-sin theta)/(cos^(2)theta) d theta`
`(pi)/2-1+int_(0)^(pi//2) (sec^(2) theta -sec theta tan )d theta`
`=(pi)/2-1+|tan theta -sec theta|_(0)^(pi//2)`
`=(pi)/2-1+lim_(theta to pi//2)(sin theta-1)/(cos theta)-(sin 0-1)/(cos 0)`
`=(pi)/2-1+lim_(theta to pi//2) (cos theta)/(sin theta) +1`
`=(pi)/2`
Discussion
111.

Evaluate:`int_(-1)^3(tan^(-1)d/(x^2+1)+(x^2+1)/x)dx`

Answer» Correct Answer - `2pi`
`int_(1)^(3)("tan"^(-1)x/(x^(2)+1)+"tan"^(-1)(x^(2)+1)/x)dx`
`=int_(-1)^(0)("tan"^(-1)x/(x^(2)+1)+"tan"^(-1)(x^(2)+1)/x)dx`
`+int_(0)^(3)("tan"^(-1)x/(x^(2)+1)+"tan"^(-1)(x^(2)+1)/x)dx`
`int_(-1)^(0)-(pi)/2 dx+int_(0)^(3)(pi)/2 dx`
`=[-(pi)/2 x]_(-1)^(0)+[(pi)/2x]_(0)^(3)`
`=2pi`
Discussion
112.

Let `f(x) = int_(0)^(x)|2t-3|dt`, then f isA. continuous at `x = 3//2`B. continuous at `x = 3`C. differentiable at `x = 3//2`D. differentiable at `x = 0`

Answer» Correct Answer - A::B::C::D
Discussion
113.

`int_(0)^((14pi)/3) |sinx|dx`

Answer» Correct Answer - `19/2`
Discussion
114.

Evaluate `int_(2)^(8)|x-5|dx`.

Answer» `overset(8)underset(2)int|x-5|dx=overset(5)underset(2)int(-x+5)dx+overset(8)underset(5)int(x-5)dx = 9`
Discussion
115.

Evaluate the following : `int_(0)^(pi//2)(dx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)`

Answer» Correct Answer - `(pi)/(2ab)`
`I=int_(0)^(pi//2) (dx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)`
`=int_(0)^(pi//2) (sec^(2)x dx)/(a^(2)+b^(2)tan^(2)x)=1/(b^(2)) int_(0)^(pi//2) (sec^(2) xdx)/((a/b)^(2)+tan^(2)x)`
Put `tan x=z` or `sec^(2) x dx=dz`
When `x=0, z=0, xto (pi)/2, zto oo`
`:.I=1/(b^(2)) int_(0)^(oo) (dz)/((a/b)^(2)+z^(2))`
`=1/(b^(2))1/(a/b)|"tan"^(-1)z/(a//b)|_(0)^(oo)`
`=1/(ab)[tan^(-1)oo-tan^(-1)0]`
`=1/(ab) (pi)/2`
`=(pi)/(2ab)`
Discussion
116.

Evaluate `int_(1)^(5)sqrt(x-2)sqrt(x-1)dx`.

Answer» Correct Answer - `2`
`int_(1)^(5)sqrt(x-2sqrt(x-1))dx=int_(1)^(5)sqrt((sqrt(x-1)-1)^(2))dx`
`=int_(1)^(5)|sqrt(x-1)-1|dx`
`=int_(1)^(2)(1-sqrt(x-1))dx+int_(2)^(5)(sqrt(x-1)-1)dx`
`=|x-(2x(x-1)^(3//2))/3|_(1)^(2)+|(2(x-1)^(3//2))/3-x|_(2)^(5)`
`=(2-2/3-1+0)+(16/3-5-2/3+2)=2`
Discussion
117.

`P rov et h a t4lt=int_1^3sqrt(3+x^2)lt=4sqrt(3)`

Answer» Correct Answer - NA
Since `1lexle3`
or `1lex^(2)le9`
or `4lex^(2)+3le12`
or `2lesqrt(3+x^(2))le2sqrt(3)`
or `2(3-1)leint_(1)^(3)sqrt(3+x^(2))dx le 2sqrt(3)(3-1)`
or `4le int_(1)^(3)sqrt(3+x^(2))le 4 sqrt(3)`
Discussion
118.

Evaluate `int_(0)^(2pi) (dx)/(1+3cos^(2)x)`

Answer» `I=int_(0)^(2pi)(dx)/(1+3cos^(2)x)`
`=int_(0)^(2pi)(sec^(2) xdx)/(sec^(2)x+3)`
`=int_(0)^(2pi)(sec^(2) xdx)/(4+tan^(2)x)`
Here we cannot substitute `tan x=t` as `tanx` is discontinuous in `(0,2pi)`
Let `f(x)=(sec^(2)x)/(4+tan^(2)x)`
`:f(2pix)=f(x)`
`:.I=2 int_(0)^(pi)(sec^(2) xdx)/(4+tan^(2)x)`
Also `f(pi-x)=f(x)`
`:. I=4 int_(0)^(pi//2)(sec^(2) xdx)/(4+tan^(2)x`
`=4 int_(0)^(oo) (dt)/(4+t^(2))` (Putting `tanx=t` as `tanx` is continuous in `(0,pi//2)`
`=4/2[tan^(-1)t/2]_(0)^(oo)`
`=2[ta ^(-1)oo-tan^(-1)0]`
`=pi`
Discussion
119.

`int_(0)^(pi//4)(x.sinx)/(cos^(3)x)` dx equal to :A. `(pi)/(4)+1/2`B. `(pi)/(4)-1/2`C. `(pi)/(4)`D. `(pi)/(4)+1`

Answer» Correct Answer - B
Discussion
120.

`int_(-1)^(2)[([x])/(1+x^(2))]dx`, where [.] denotes the greatest integer function, is equal toA. `-2`B. `-1`C. zeroD. none of these

Answer» Correct Answer - B
`[x]=0,AAxepsilon[0,1)`
For `x epsilon[1,2),[x]=1`
`:.([x])/(1+x^(2))=1/(1+x^(2))lt 1AAxepsilon[1,2)` or `[([x])/(1+x^(2))]=0`
For `x epsilon[-1,0),[x]=-1` or `([x])/((1+x)^(2))=-1/(1+x^(2))`
Clearly, `2ge1 +x^(2)gt1AAxepsilon[-1,0)`
or `1/2le 1(1+x^(2))lt 1`
or `-1/2ge - 1/(1+x^(2))gt-1`
or `[([x])/(1+x^(2))]=-1AAxepsilon[-1,0)`
Thus, the given integral `=-int_(-1)^(0)dx=-1`.
Discussion
121.

`int_(0)^(pi//4) log (1+tan x) dx =?`

Answer» Correct Answer - `(pi)/(8) (log2)`
Let `I=int_(0)^(pi//4) log (1+tanx )dx` . . . (i)
`I=int _(0)^(pi//4)log (1+tan((pi)/(4)-x))dx` `:. I= int _(0)^(pi//4)log (1+(1-tanx)/(1+tanx))dx`
`=int_(0)^(pi//4) log ((1+tanx+1-tanx)/(1+tanx))dx`
`I= int_(0)^(pi//4)log((2)/(1+tanx))dx rArrI=int _(0)^(pi//4)log 2 dx - I`
`rArr 2 I = (pi)/(4) log 2 rArr I=(pi)/(8) (log2)`
Discussion
122.

Evaluate `int_0^1(dx)/((5+2x-2x^2)(1+e^(2-4x))) dx`

Answer» Let `I=int_(0)^(1)(dx)/((5+2x-2x^(2))(1+e^(2-4x)))`
Also `I=int_(0)^(1)(dx)/([5+2(1-x)-2(1-x)^(2)][1+e^(2-4(1-x))])`
`=int_(0)^(1)(dx)/((5+2x-2x^(2))(1+e^(-2+4x)))`
`=int_(0)^(1)(e^(2-4x)dx)/((5+2x-2x^(2))(e^(2-4x)+1))`
Adding 1 and 2 we get
`2I=int_(0)^(1)((1+e^(2-4x))dx)/((5+2x-2x^(2))(1+e^(2-4x)))`
`int_(0)^(1)(dx)/(5-2(x^(2)-x))=int_(0)^(1)(dx)/(1/2+5-2(x-1/2)^(2))`
`=1/2int_(0)^(1)(dx)/(11/4-(x-1/2)^(2))`
`=1/(4sqrt(11)//2)|log(sqrt(11)//2+x-1/2)/(sqrt(11)//2-(x-1/2))|_(0)^(1)`
`=1/(2sqrt(11))["log" ((sqrt(11))2+1/2)/((sqrt(11))/2-1/2)-"log"((sqrt(11))/2-1/2)/((sqrt(11))/2+1/2)]`
`=1/(2sqrt(11))[2log((sqrt(11)+1)/(sqrt(11)-1))]`
`=1/(sqrt(11))log((sqrt(11)+1)/(sqrt(11)-1))`
`=1/(sqrt(11))"log"(sqrt(11)+1)/(sqrt(11)-1)(sqrt(11)+1)/(sqrt(11)+1)`
`=1/(sqrt(11))"log"((sqrt(11)+1)^(2))/10`
Discussion
123.

For `theta in (=,pi/2),p rov et h a tint_0^theta"log"(1+tanthetatanx)dx=thetalog(sectheta)`

Answer» `I=int_(0)^(theta)log(1+tan theta tan x) dx`
`=int_(0)^(theta)log(1+tan theta tan (theta-x))dx`
`=int_(0)^(theta)log(1+(tantheta(tan theta-tan x))/(1+tan theta tan x))dx`
`=int_(0)^(theta)"log"((1+tan^(2)theta))/(1+tan theta tan x)dx`
`=int_(0)^(theta)log(1+tan^(2)theta)dx-int_(0)^(theta)log(1+tan theta tan x)dx`
`=2theta log sec theta -I`
or `2I=2theta log sec theta`
or `I=thetalog sec theta`
Discussion
124.

Determine the value of `int_(-pi)^(pi) (2x(1+sinx))/(1+cos^(2)x)dx`.

Answer» `I=int_(-pi)^(pi)(2x(1+sinx))/(1+cos^(2)x) dx`
`=int_(-pi)^(pi)(2x)/(1+cos^(2)x)dx+2int_(-pi)^(pi)(x sin x)/(1+cos^(2)x)`……………1
`=0+4int_(0)^(4)(x sinx)/(1+cos^(2)x)dx`
`=4int_(0)^(pi)((pi-x)sin(pi-x))/(1+cos^(2)(pi-x))dx`
`=4int_(0)(pi)((pi-x)sinx)/(1+cos^(2)x)dx`
`=4pii int_(0)^(pi)(sinx)/(1+cos^(2)x)dx-4 int_(0)^(pi)(x sinx)/(1+cos^(2)x)dx`
or `2I=4pi int_(0)^(pi)(sinx)/(1+cos^(2)x) dx`
or `I=2i int_(0)^(pi)(sinx)/(1+cos^(2)x) dx`
Put `cosx=t` so that `-sinx dx=dt`
when `x=0, t=1`, when `x=pi,t=-1`,
`:.I=2pi int_(1)^(-1)(-dt)/(1+t^(2))=4pi int_(0)^(1)(dt)/(1+t^(2))`
`=4pi [tan^(-1)t]_(0)^(1)`
`=4pi(pi)/4=pi^(2)`
Discussion
125.

`int_(3)^(10)[log[x]]dx` is equal to (where [.] represents the greatest integer function)A. 9B. `16-e`C. `10`D. `10+e`

Answer» Correct Answer - A
When `ele[x]lt e^(2), 1le log [x]lt 2`
when `e^(2) le [x] lte^(3), 2lt log [x] lt 3`
`:.int_(3)^(8)1 dx+int_(8)^(10)2dx=9`
Discussion
126.

The value of `int_(-pi)^(pi)(2x(1+sinx))/(1+cos^(2)x)dx` is

Answer» Correct Answer - `pi^(2)`
Let `I=int _(-pi)^(pi)(2x(1+sinx))/(1+cos ^(2)x)dx`
`I=int_(-pi)^(pi)(2x)/(1+cos^(2)x)dx =int_(-pi)^(pi)(2x sinx)/(1+cos^(2)x)`
`rArr I=I_(1)_I_(2)` Now , `I _(1)= int_(-pi)^(pi) (2x)/(1+cos^(2)x)dx`
Let ` f(x) = (2x)/(1+cos ^(2)x)`
`rArr f(-x)= (-2x)/(1+cos ^(2)(-x))=(-2x)/(1+cos^(2)x)=-f(x)`
`rArr f(-x) =- f(x)` which shows that f (x) is an odd function.
`:. I_(1)=0`
Again . let ` g(x)= (2xsinx)/(1+cos^(2)x)`
`rArr g(-x)= (2(-x)sin(-x))/(1+cos^(2)(-x))= (2x sin x )/(1+ cos^(2)x)=g(x)`
`rArr g (-x) = g (x)` which shows that g (x) is an even function.
`:. I_(2)=int_(-pi)^(pi) (2x sin x)/(1+cos ^(2)x)dx = 2*2int _(0)^(pi)(x sinx)/(1+cos ^(2)x)dx`
`= 4 int _(0)^(pi) ((pi-x)sin(pi-x))/(1+[cos (pi-x)]^(2))dx = 4 int _(0)^(pi) ((pi-x)sinx)/(1+cos ^(2)x)dx`
`=4 int_(0)^(pi) (pisinx)/(1+cos^(2)x)dx -4 int _(0)^(pi)(xsinx)/(1+cos^(2)x)dx`
`rArr I_(2) = 4 pi int_(0)^(pi)(sinx)/ (cos ^(2)c)dx-I_(2)`
`rArr 2I_(2)=4piint _(0)^(pi) (sinx)/(1+cos^(2)x)dx`
Put `cos x= t rArr - sin x dx = dt`
`:. I_(2)=-2piint _(1)^(-1)(dt)/(1+t^(2))= 2 pi int _(-1)^(1)(dt)/(1+t^(2))=4pi int _(0)^(1)(dt)/(1+ t^(2))`
`=4pi [tan^(-1)t] _(0)^(1) =4pi [tan^(-1)1-tan^(-1)0]`
`4pi (pi//4-0)=pi^(2)`
`:. I-I_(1) =I_(2)=0+pi^(2)=pi^(2)`
Discussion
127.

Evaluate `int_(-pi//2)^(pi//2) sqrt(cosx-cos^(3)x)dx`.

Answer» `I=int_(-pi//2)(pi//2)ubrace(sqrt(cosx-cos^(3)x)dx)_(f(x))`
`2int_(0)^(pi/2)sqrt(cosx-cos^(3)x) dx` (as `f(x)` is an even function)
`=2int_(0)^(pi//2) sqrt(cosx sin^(2)x) dx`
`=2int_(0)^(pi//2)sqrt(cosx) sin xdx`
`=2[-((cosx)^(3//2))/(3//2)]_(0)^(pi/2)`
`=2xx2/3`
`=4/3`
Discussion
128.

Find the value of `int_(-2)^(2)(sin^(-1)(sinx)+cos^(-1)(cosx))/((1+x^(2))(1+[(x^(2))/5]))dx`, where [.] represents the greatest integer function.

Answer» `I=int_(-2)(2)(sin^(-1)(sinx)+cos^(-1)(cosx))/((1+x^(2))(1+[(x^(2))/5]))`
`sin^(-1)(sinx)` is an odd function.
Also for `-2lt xlt 2, 0le x^(2)lt4`.
`:. [(x^(2))/5]=0`
`:.I=int_(-1)^(2)(cos^(-1)(cosx))/(1+x^(2))dx`
`=int_(0)^(2)(2cos^(-1)(cosx))/(1+x^(2))dx`
`=int_(0)^(2)(2x)/(1+x^(2)dx`
`=[log_(e)(1+x^(2))]_(0)^(2)`
`=log_(e)4`
Discussion
129.

Evaluate:`int_2^3(2x^5+x^4-2x^3+2x^2+1)/((x^2+1)(x^4-1))dx`

Answer» Correct Answer - `(1)/(2)log6-(1)/(10)`
Let `I= int_(2)^(3)(2x^(5)+x^(4)-2x^(3)+2x^(2)+1)/((x^(2)+1)(x^(4)-1))dx`
`=int_(2)^(3)(2x^(5)-2x^(3)+x^(4)+1+2x^(2))/((x^(2)+1)(x^(2)-1)(x^(2)+1))dx`
`=int_(2)^(3)(2x^(3)(x^(2)-1)+(x^(2)+1)^(2))/((x^(2)+1)^(2)(x^(2)-1))dx`
`int_(2)^(3)(2x^(3)(x^(2)-1))/((x^(2)+1)^(2)(x^(2)-1))dx+int_(2)^(3)((x^(2)+1)^(2))/((x^(2)+1)^(2)(x^(2)-1))dx`
`=int_(2)^(2)(2x^(3))/((x^(2)+1)^(2))dx+int_(2)^(3)(1)/((x^(2)-1))dx`
`rArr I=I_(1)+I_(2)`
Now , `I_(1)=int_(2)^(3)(2x^(3))/((x^(2)+1)^(2))dx`
Put `x^(2)=1=trArr2x dx = dt`
`:. I_(1)= int_(5)^(10)((t-1))/(t^(2))dt=int_(5)^(10)(1)/(t)dt-int(1)/(t^(2))dt`
`=[logt _(5)^(10)+[(1)/(t)]_(5)^(10)`
`=log10- log 5+(1)/(10)-(1)/(5)`
`= log2 -(1)/(10)`
Again , `I_(2)=int _(2)^(3)(1)/((x^(2)-1))dx =int_(2)^(3)(1)/((x-1)(x+1))dx`
`=(1)/(2)int_(2)^(3)(1)/((x-1))dx -(1)/(2) int _(2)^(3)(1)/((x-1)(x+1))dx`
`=[(1)/(2)log(x-1)]_(2)^(3)-(1)/(2)["log"(x+1)]_(2)^(3)`
`=(1)/(2)" log " (2)/(1)" log " (4)/(3)`
From Eq . (i) , `I=I_(1)+I_(2)`
` =" log " 2 - (1)/(10)+(1)/(2)" log " 2 -(1)/(2)" log " (4)/(3)`
`=log [2*2^(1//2)((4)/(3))^(-1//2)]-(1)/(10)=(1)/(2) log 6 - (1)/(10)`
Discussion
130.

Evaluate the definite integral:`int_(-1/(sqrt(3)))^(1/(sqrt(3)))((x^4)/(1-x^4))cos^(01)((2x)/(1+x^2))dxdot`

Answer» Let `I=int_(-1//sqrt(3))^(1//sqrt(3))(x^(4))/(1-x^(4))cos^(-1)((2x)/(1+x^(2)))dx`
`=int_(1//sqrt(3))^(1//sqrt(3))((-x)^(4))/(1-(-x)^(4))cos^(-1)((2(-x))/(1+(-x)^(2)))dx`
`=int_(-1//sqrt(3))^(1//sqrt(3))(x^(4))/(1-x^(4))[pi-cos^(-1)((2x)/(1+x^(2)))]dx`
`=pi int_(-1//sqrt(3))^(1/sqrt(3))(x^(4))/(1-x^(4))dx-I`
`:.2I=2pi int_(0)^(1//sqrt(3))(x^(4))/(1-x^(4))dx`
`:.I=pi int_(0)^(1//sqrt(3))[-1+1/(1-x^(4))]dx`
`=-pi int_(0)^(1//sqrt(3))dx+(pi)/2 int_(0)^(1//sqrt(3))[1/(1-x^(2))+1/(1+x^(2))]dx`
`=-(pi)/(sqrt(3))+(pi)/2[(-1/2 log_(e)|(1-x)/(1+x)|+tan^(-1)x)]_(0)^(1//sqrt(3))`
`=-(pi)/(sqrt(3))+(pi)/2(1/2log_(e)|(sqrt(3)+1)/(sqrt(3)-1)|+(pi)/6)`
Discussion
131.

`Lim_(n->oo)[1/n^2 * sec^2 (1/n^2)+2/n^2 * sec^2 (4/n^2)+..............+1/n * sec^2 1]`

Answer» Correct Answer - `1/2 tan1`
`lim_(n to oo) [1/(n^(2)) "sec"^(2)1/(n^(2))+2/(n^(2))"sec"^(2) 4/(n^(2))+……..+1/nsec^(2)1]`
`=lim_(nto oo) 1/n sum_(r=1)^(n)r/n"sec"^(2)(r/n)^(2)`
`=int_(0)^(1)x sec^(2) x^(2) dx`
Put `x^(2)=t` so that `2x dx=dt`
When `x=0, t=0`, When `x=1, t=1`
`:. `Required limit `=1/2 int_(0)^(1) sec^(2) t dt `
`=1/2 [tan t]_(0)^(1)=1/2[tan 1-0]`
`=1/2 tan 1`
Discussion
132.

`IfS_n=[1/(1+sqrt(n))+1/(2+sqrt(2n))++1/(n+sqrt(n^2))],t h e n(lim)_(nvecoo)S_n`is equal tolog 2 (b) log4log 8 (d) none of theseA. `log2`B. `log 4`C. `log 8`D. none of these

Answer» Correct Answer - B
`lim_(nto oo) S_(n)=lim_(nto oo) [1/(1+sqrt(n))+1/(2+sqrt(2n))+………..+1/(n+sqrt(n)^(2))]`
`=lim_(nto oo) 1/n [1/(1/n+1/(sqrt(n)))+1/(2/n+sqrt(2/n))+……..+1/(n/n+sqrt(n/n))]`
`=int_(0)^(1)(dx)/(sqrt(x)(sqrt(x)+1))`
Put `sqrt(x)=z` or `1/(2sqrt(x))dx=dz`
`:. lim_(n to oo) S_(n)=int_(0)^(1)(2 dz)/(z+1)=2|log(z+1)|_(0)^(1)`
`=2(log2-log1)`
`=2log2=log4`
Discussion
133.

`L e tJ=int_(-5)^(-4)(3-x^2)tan(3-x^2)dxa n dK=int_(-2)^(-1)(6-6x+x^2)``tan(6x-x^2-6)dxdotT h e n(J+K)`equals _____

Answer» We have `J=int_(-5)^(-4)(3-x^(2))tan(3-x^(2))dx`
Put `(x+5)=t`. Then
`J=int_(0)^(1)(3-(t-5)^(2))tan(3-(t-5)^(2))dt`
`=int_(0)^(1)(-22+10t-t^(2))tan(-22+10t-t^(2))dt`
Now `K=int_(-2)^(-1)(6-6x+x^(2))tan(6x-x^(2)-6)dx`
Put `(x+2)=z`. Then
`K=int_(0)^(1)(6-6(z-2)+(z-2)^(2))tan(6(z-2)-(z-2)^(2)-6)`
`=int_(0)^()(22-10z+z^(2))tan(-22+10z-z^(2))dz`
Hence `(J+K)=0`
Discussion
134.

Evaluate: `int_(pi//6)^(pi//4)(1+cotx)/(e^(x)sinx) dx`

Answer» Correct Answer - `2e^(-pi//6)-sqrt(2)e^(-pi//4)`
`I=int_(pi//6)^(pi//4) e^(-x)(cosecx+cot x cosec x)dx`
Put `x=timpliesdx=-dt`
`:. I=-int_(-pi//6)^(-pi//4) e^(t)(-cosect+cot.cosec)dt`
`=int_(-pi//6)^(-pi//4)e^(t)(cosect-cot t. cosec t)dt`
`=e^(t)cosec t|_(-pi//6)^(-pi//4)`
`=-sqrt(2)e^(-pi//4)+2d^(-pi//6)`
`=2e^(-pi//6)-sqrt(2)e^(-pi//4)`
Discussion
135.

Evaluate:`int_0^oolog(x+1/x)(dx)/(1+x^2)`

Answer» Putting `x=tan theta, dx=sec^(2)theta d theta`, given integral becomes
`I=int_(0)^(pi//2)(log(tan theta+cot theta))/(1+tan^(2) theta) sec^(2) theta d theta`
`=int_(0)^(pi//2) log (sin theta//cos theta +cos theta // sin theta )d theta`
`=int_(0)^(pi//2)log{1//(sin theta cos theta)}d theta`
`=-int_(0)^(pi//2) log sin theta d theta -int_(0)^(pi//2) log cos theta d theta`
`=-2(-1/2pi log 2)=pi log 2`
Discussion
136.

If `f`is an odd function, then evaluate`I=int_(-a)^a(f(sinx)/(f(cosx)+f(sin^2x)dx`

Answer» Let `phi(x)=(f(sinx))/(f(cosx)+f(sin^(2)x))`
`:.phi(-x)=(f(sin(-x)))/(f(cos(-x))+f(sin^(2)(-x)))`
`=(f(-sinx))/(f(cosx)+f(sin^(2)x))`
`=(-f(sinx))/(f(cosx)+f(sin^(2)x))=-phi(x)`
( `:.f` is an odd function)
`:.I=int_(a)^(a)(f(sinx))/(f(cosx)+f(sin^(2)x))dx=0`
Discussion
137.

Evaluate:`int_(-1)^4f(x)dx=4a n dint_2^4(3-f(x))dx=7,`then find the value of `int_2^(-1)f(x)dxdot`

Answer» Correct Answer - `-5`
We have `int_(2)^(4)(3-f(x))dx=7`
`:.6-int_(2)^(4)f(x)dx=7` or `int_(2)^(4)f(x)dx=-1`
Now, `int_(2)^(-1)f(x)dx=-int_(-1)^(2)f(x)dx=-[int_(-1)^(4)f(x)dx+int_(4)^(2)f(x)dx]`
`=-[int_(-1)^(4)f(x)dx-int_(2)^(4)f(x)dx]=-[4+1]=-5`
Discussion
138.

Show that `int_(0)^(2)(2x+1)dx = int_(0)^(5)(2x+1)+int_(5)^(2)(2x+1)`

Answer» L.H.S `= x^(2)+x]_(0)^(2)= 4+2=6` , R.H.S. `= 25+5-0+(4+2)-(25+5) = 6`
`:.` L.H.S = R.H.S
Discussion
139.

Evaluate:`5050(int0 1(1-x^(50))^(100)dx)/(int0 1(1-x^(50))^(101)dx)`

Answer» Correct Answer - 5051
`int_(0)^(1)(1-x^(50))^(101)dx`
`=[x(1-x^(50))^(101)]_(0)^(1)-int_(0)^(1)x.101(1-x^(50))^(100)(-50x^(49))dx`
`=0+5050 int_(0)^(1)x^(50)(1-x^(50))^(100)dx`
`=-5050int_(0)^(1)(1-x^(50)-1)(1-x^(50))^(100)dx`
`=-5050int_(0)^(1)(1-x^(50))^(101)dx+5050int_(0)^(1)(1-x^(50))^(100)dx`
`=5051int_(0)^(1)(1-x^(50))^(101)dx=5050int_(0)^(1)(1-x^(50))^(100) dx`
`implies 5050(int_(0)^(1)(1-x^(50))^(100)dx)/((1-x^(50))^(101)dx)=5051`
Discussion
140.

Evaluate `int_(0)^(1)(e^(-x)dx)/(1+e^(x))`

Answer» Correct Answer - `log(1+e)-1/3-log2`
`I=int_(0)^(1)(e^(-x) dx)/(1+e^(x))=int_(0)^(1)(dx)/(e^(x)(1+e^(x)))`
Put `e^(x)=z`
`:.e^(x)dx=dz`
`implies dx=(dz)/(e^(x))=(dz)/z`
`impliesI=int_(1)^(e)(dz)/(z^(2)(1+z))`
`=int_(1)^(e)(1/(1+z)-(z-1)/(z^(2)))dz`
`=|log_(e)(1+z)-log_(e)z-1/z|_(1)^(e)`
`=(log_(e)(1+e)-log_(e) e-1/e)-(log_(e)2-log_(e)1-1)`
`=log(1+e)-1/e-log2`
Discussion
141.

Evaluate:`("lim")_(nvecoo)(1/(sqrt(4n^2-1))+1/(sqrt(4n^2-2^2))++1/(sqrt(3n^2)))`

Answer» Correct Answer - `(pi)/6`
Given limit
`=lim_(nto oo) 1/n[n/(sqrt(4n^(2)-1))+n/(sqrt(4n^(2)-2^(2)))+…………+n/(sqrt(4n^(2)-n^(2)))]`
`=lim_(n to oo) 1/n[1/(sqrt(4-(1/n)^(2)))+1/(sqrt(4-(2/n)^(2)))+……………1/(sqrt(4-(n/n)^(2)))]`
`=int_(0)^(1)(dx)/(sqrt(4-x^(2)))=|sin^(-1)x/2|_(0)^(1)="sin"^(-1)1/2-0=(pi)/6`
Discussion
142.

`int_(0)^(pi) x log sinx dx`

Answer» Let `I=int_(0)^(pi)x log sin x dx`………………….1
`:.I=int_(0)^(pi)(pi-x)logsin(pi-x)dx`
or `I=int_(0)^(pi)(pi-x)logsinxdx`……………..2
Adding 1 and 2 we get
`2pi=piint_(0)^(pi)logsinxdx`
or `2I=2pi int_(0)^(pi//2)log sin x dx`
or `I=pi int_(0)^(pi//2)log sinx`
`=pi{1/2 pi log (1//2)}=1/2pi^(2)log(1//2)`
Discussion
143.

Evaluate:`int_0^(pi/2)xcotx dx`

Answer» Integrating by parts, taking `cotx` as second function given integral becomes.
`I=[x log sin x]_(0)^(pi//2)-int_(0)^(pi//2)log sinx dx`
`=0-lim_(xto oo) (xlog sin x)-int_(0)^(pi//2)log sin x dx=1/2 pilog 2`
as `lim_(xto oo) x log sinx=lim_(xto oo) ((log sinx)/(1//x))`
`=lim_(xto oo)((cotx))/(-1//(x^(2)))`
`=lim_(xto oo) ((-x^(2))/(tanx))`
`=lim_(xto oo) (-x xx x/(tanx))`
`=0xx1=0`
Discussion
144.

Prove that `int(g(sinx))/(g(sinx)+g(cosx))dx = int_(0)^(pi/2)(g(cosx))/(g(sinx)+g(cosx))dx = (pi)/(4)`.

Answer» Let `=overset(pi/2)underset(0)int(g(sinx))/(g(sinx)+g(cosx))dx rArr I = underset(0)overset(pi/2)int(g(sin(pi/2-x)))/(g(sin(pi/2-x))+g(cos(pi/2-x)))dx`
`=overset(pi/2)underset(0)int(g(cosx))/(g(cosx)+g(sinx))dx`
on adding, we obtain
` = 2I = overset(pi/2)underset(0)int((g(sinx))/(g(sinx)+g(cosx))+(g(cosx))/(g(cosx)+g(sinx)))dx =overset(pi/2)underset(0)int dx rArr I = (pi)/(4)`.
Discussion
145.

Evaluate `int_(-1)^(1)(3^(x)+3^(-x))/(1+3^(x))dx`

Answer» `overset(1)underset(-1)int(3^(x)+3^(-x))/(1+3^(x))dx = underset(0)overset(1)int((3^(x)+3^(-x))/(1+3^(x))+(3^(-x)+3^(x))/(1+3^(x)))dx = underset(0)overset(1)int((3^(x)+3^(-x))/(1+3^(x))+(3(3^(-x)+3^(x)))/(1+3^(x)))`
` = underset(0)overset(1)int(3^(x)+3^(-x))dx=((3^(x))/(ln3)-(3^(-x))/(ln3))_(0)^(1) = (3/(ln3)-(3^(-1))/(ln3)) - ((1)/(ln3) - (1)/(ln3))=(1)/(ln3)[3-(1)/(3)] = (8)/(3ln3)`
Discussion
146.

Evaluate `int_(0)^(pi//2)|sinx-cosx|dx`.

Answer» Correct Answer - `2sqrt(2)-1)`
`int_(0)^(pi//2)|sinx-cosx|dx`
`=int_(0)^(pi//4)-(sinx-cosx)dx+int_(pi//4)^(pi//2)(sinx-cosx)dx`
`=|cosx+sinx|_(0)^(pi//4)+|-cosx-sinx|_(pi//4)^(pi//2)`
`=(1/(sqrt(2))+1/(sqrt(2))-1-0)+(-0-1+1/(sqrt(2))+1/(sqrt(2)))`
`=4/(sqrt(2))-2=2sqrt(2)-2`
`2(sqrt(2)-1)`
Discussion
147.

Prove that `int_0^(102)(x-1)(x-2)(x-100)``x(1/((x-1)+1/((x-2))+1/((x-100))dx=101 !-100 !`

Answer» Correct Answer - NA
`I=int_(0)^(102)(x-1)(x-2)……(x-100)`
`xx(1/((x-1))+1/((x-2))+…+1/((x-100)))dx`
`-int_(0)^(102)d/(dx)((x-1)(x-2)………(x-100))dx`
`=[(x-1)(x-2)…………(x-100)]_(0)^(102)`
`=101!-100!`
Discussion
148.

If `int_0^1(e^t)/(1+t)dt=a ,`then find the value of `int_0^1(e^t)/((1+t)^2)dt`in terms of `a`.

Answer» Correct Answer - `a+1-e/2`
`a=int_(0)^(1)(e^(t))/(1+t)dt=(1/((1+t))e^(t))_(0)^(1)=int_(0)^(1)(e^(t))/((1+t)^(2))dt`
(Integrating by parts)
`=e/2-1+int_(0)^(1)(e^(t))/((1+t)^(2)) dt`
or `int_(0)^(1) (e^(t))/((1+t)^(2))dt=a+1-e/2`
Discussion
149.

Evaluate:`int_(-pi/2)^(pi/2)sin^2xcos^2x(sinx+cosx)dx`

Answer» Correct Answer - `4//15`
Let `=int_(-pi//2)^(pi//2)sin^(2)x cos^(2)x(sinx+cosx)dx`
`=int_(-pi//2)^(pi//2) sin^(3)x cos^(2)x dx+int_(-pi//2)^(pi//2) sin^(2) x cos^(3)x dx`………….1
Since `sin^(3)x cos^(2)x` is an odd functional and `sin^(2)x cos^(3)x` is an even function. Therefore `int_(-pi//2)^(pi//2) sin^(3)x cos^(2)x dx=0`
and `int_(-pi//2)^(pi//2) sin^(2) x cos^(3)x dx =int_(0)^(pi//2) sin^(2)x cos^(3)x dx`.
`:. I=2int_(0)^(pi//2) sin^(2)x cos^(3)x dx`
`=2int_(0)^(1)t^(2)(1-t^(2))dt`
`=int_(0)^(1)(t^(2)-t^(4))dt=2[1/3-1/5]=4/15`
Discussion
150.

Show that :`int_0^1(logx)/((1+x))dx=-int_0^1(log(1+x))/x dx`

Answer» Correct Answer - NA
Let `I=int_(0)^(1)(logx)/((1+x))dx`
`=[log x log (1+x)]_(0)^(1)-int_(0)^(1)(log(1+x))/x dx`
`=0-int_(0)^(1)(log(1+x))/x dx`
Discussion