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Must Do Problems Of Definite Integrals For NDA

Must Do Problems Of Definite Integrals

Must Do Problems Of Definite Integrals
Q 2589191017

The value of ` int_0^1 ( log (1 + x))/(1 + x^2) dx` is
BCECE Mains 2015
(A)

` pi/8 log_e 2`

(B)

` pi/4 log_e 2`

(C)

` - pi/8 log_e 2`

(D)

`- pi/4 log_e 2`

Solution:

Let `I = int_0^1 ( log (1 + x))/(1 + x^2) dx`

On putting `x = tan theta`, we get

` I = int_0^(pi//4) log ( 1 + tan theta) d theta` ..........(i)

` => I = int_0^(pi//4) log { 1 + tan ( pi/4 - x) }`

` => I = int_0^(pi//4) log ( 1 + (1 - tan x)/(1 + tan x) ) dx `

` => I = int_0^(pi//4) log ( 2/(1 + tan x) ) dx` .......(ii)

On adding Eqs. (i) and (ii), we get

` 2I = int_0^(pi//4) log 2 dx = pi/4 log_e 2`

` => I = pi/8 log_e 2`
Correct Answer is `=>` (A) ` pi/8 log_e 2`
Q 1612412330

The value of the integral `int_0^(1/(sqrt2)) (sin^(-1)xdx)/((1-x^2))^(3/2)`
UPSEE 2016
(A)

`pi/2+1/2log2`

(B)

`pi-1/2log2`

(C)

`pi/2-log2`

(D)

`pi/4-1/2log2`

Solution:

Let ` x = sintheta =>`

` dx =costheta d theta => x =0, theta = 0 and x =1/sqrt2 theta = pi/4 `

`I = int_0^(pi/4) thetasec^2thetad(theta) = [thetatantheta]_0^(pi/4)-int_0^(pi/4).tanthetad(theta)`


= `[(pi/4)tan(pi/4)-0tan0]-[logsectheta]_0^(pi/4)`


=`[pi/4]-[logsec(pi/4)-logsec0] = pi/4-(logsqrt2-log1) = pi/4-1/2log2`
Correct Answer is `=>` (D) `pi/4-1/2log2`
Q 2503345248

The value of the integral `int_(0)^(pi/2) sin^5 x dx` is
WBJEE 2010
(A)

`4/15`

(B)

`8/5`

(C)

`8/15`

(D)

`4/5`

Solution:

`I = int_(0)^(pi/2) sin^4 x * sinx dx`

`= int_(0)^(pi/2) (1-cos^2 x)^2 sinxdx`


Put `cosx = t`

`=> -sinxdx = dt`


`-int_(1)^(0)(1-t^2)^2 dt = int_(0)^(1) (t^4-2t^2+1)dx` `[because -int_(a)^(b) f(x)dx = int_(a)^(b) f(x)dx]`


` = 1/5 (t^5)_0^1-2/3(t^3)_0^1+(t)_0^1`

` = 1/5-2/3+1 = (3-10+15)/15 = 8/15`
Correct Answer is `=>` (C) `8/15`
Q 2805812768

` 1/c int_(ac)^(bc) f (x/c) dx` is equal to

(A)

`1/c int_a^b f(x)dx`

(B)

`int_a^b f(x)dx`

(C)

`c int_a^b f(x)dx`

(D)

` int_(ac^2)^(bc^2) f (x) dx`

Solution:

Let `I = 1/c int_(ac)^(bc) f (x/c) dx`

Put `x/c = t`

At `x = ac , t = (ac)/c = a`

At `x = bc , t = (bc)/c = b` and `(dx)/c = dt`

`:. I = int_a^b f(t) dt` [replace `t` by `x`]

`= int_a^b f(x)dx`
Correct Answer is `=>` (B) `int_a^b f(x)dx`
Q 2313778649

The value of `int_(-pi/2)^(pi/2) (ax^3 cos x + b sin x +c ) dx`

ultimately depends upon
BITSAT Mock
(A)

`b`

(B)

`c`

(C)

`a`

(D)

`a` and `b`

Solution:

`x^3 cos x` and `b sin x` are odd functions.

`:. int_(-pi/2)^(pi/2) (ax^3 cos x + b sin x +c) dx`

`= 0 + 0 + 2c pi = 2c pi`
Correct Answer is `=>` (B) `c`
Q 2855534464

If `f (-x) = - f(x)`, then `f(x)` is odd function and if `f(-x) = f(x)`, then `f(x)` is even function.
Also , ` int_(-a)^a f(x) dx = { tt ((2 int_0^a f(x) , i f f(x) text (is even)) , ( 2 int_0^a f(x) , i f f(x) text (is odd)))`
The value of the integral ` int_(-pi//2)^(pi//2) sin^7 x dx ` is

(A)

`0`

(B)

`(-pi)/2`

(C)

`pi/2`

(D)

`1`

Solution:

`I = int_(-pi//2)^(pi//2) sin^7 x dx`

Here, `f(x) = sin^7 x`

`f(-x) = sin^7 (-x) = [ - sin x ]^7 = - sin^7 x`

`:. f (-x) = - f(x)`

So, `f( x)` is an odd function.

` int_(-pi//2)^(pi//2) sin^7 x dx = 0`
Correct Answer is `=>` (A) `0`
Q 2503123048

Suppose `M=int_0^(pi//2) (cos x)/(x+2) dx` `N=int_0^(pi//4) (sin x cos x)/((x+1)^2)dx` Then. the value of `(M-N)` equals
WBJEE 2014
(A)

`3/(pi+2)`

(B)

`2/(pi-4)`

(C)

`4/(pi-2)`

(D)

`2/(pi+4)`

Solution:

Given, `M= int_0^(pi//2) (cos x)/(x+2) dx`

and `N=int_0^(pi//4) (sin xcos x)/((x+1)^2) dx`

`=> N=int_0^(pi//4) 1/2 (sin 2x)/((x+1)^2) dx`

Put `2x=t=> dx=(dt)/2`

`:. N=int_0^(pi//2) (2/2sin t)/((t/2 +1)^2) dt`

`=int_0^(pi//2) (sin x)/((x+2)^2) dx`

`:. M-N= int_0^(pi//2) cos x . 1/((x+2)) dx- int_0^(pi//2) (sin x)/((x+2)^2) dx`

`=[(sin x)/(x+2)]_0^(pi//2) +int_0^(pi//2)(sin x)/((x+2)^2) dx-int_0^(pi//2) (sin x)/((x+2)^2) dx`

`= (sin pi/2)/(pi/2+2) =1/((pi+4)/2) =2/(pi+4)`
Correct Answer is `=>` (D) `2/(pi+4)`
Q 2561745625

`int_1^x (log(x^2))/x dx` is equal to :
BCECE Stage 1 2015
(A)

`(log x)^2`

(B)

`1/2 (log x)^2`

(C)

`(log x^2)/2`

(D)

None of these

Solution:

Let `I = int _1^x (logx^2)/x dx`

`= int _1^x (2 log x) /x dx`

Put, `log x =t => 1/x dx = dt`

`:. I =2 int_0^(logx) t dt = [t^2]_0^(logx)`

`= (logx)^2`
Correct Answer is `=>` (A) `(log x)^2`
Q 2522834731

If `f(x) = { tt ((2x^2+1, x le 1),(4x^3-1, x > 1))` then `int_0^2 f(x) dx` is
WBJEE 2014
(A)

`47//3`

(B)

`50//3`

(C)

`1//3`

(D)

`47//2`

Solution:

Given, `f(x) = { tt ((2x^2+1, x le 1),(4x^3-1, x > 1))`

`:. int_0^2 f(x) =int_0^1 f(x) dx +int_1^2 f(x) dx`

`=int_0^1 (2x^2+1) dx +int_1^2 (4x^3-1) dx`

`=[(2x^3)/3+x]_0^1 +[(4x^4)/4 -x]_1^2`

`=2/3 (1)^3 +1-(0+0) +{(2)^4-2-{(1)^4-1}]`

`=2/3 +1 +[16-2-0]`

`=2/3 +15 =(2+45)/3`

`=47/3 ` sq units
Correct Answer is `=>` (A) `47//3`
Q 2416423379

`int_0^1 |5x-3| dx` is equal to
UPSEE 2010
(A)

`10/13`

(B)

`31/10`

(C)

`13/10`

(D)

None of these

Solution:

`I=int_0^1 |5x-3| dx`

Now , `|5x-3|= { tt ((5x-3, x ge 3/5), (-5x-3, x le 3/5)`

`:. I=int_0^1 |5x-3| dx`

`= int_0^(3//5) -(5x-3)dx+int_(3//5)^1 (5x-3) dx`

`=- [(5x^2)/2-3x]_0^(3//5) +[(5x^2)/2-3x]_(3//5)^1`

`=-[(5/2 xx 9/25 -9/5)-(0-0)]+[(5/2-3)- (5/2 xx 9/25 -9/5)]`

`=-45/50+9/5 -1/2 -45/50 +9/5=13/10`
Correct Answer is `=>` (C) `13/10`
Q 2486323277

`int _0^1 (dx)/(sqrt (1+x) +sqrt x)` is equal to
UPSEE 2010
(A)

`4/3 (sqrt 2-1)`

(B)

`3/4 (sqrt 2-1)`

(C)

`4/3 (1-sqrt 2)`

(D)

`3/4 (1-sqrt 2)`

Solution:

`I=int_0^1 (dx)/(sqrt(1+x) +sqrt x)`

`=int_0^1 (sqrt(1+x) - sqrt x)/((sqrt (1+x)+sqrt x) (sqrt (1+x)-sqrt x)) dx`

`=int_0^1 ((sqrt (1+x)-sqrt x))/((1+x)-x) dx`

`=int _0^1 (1+x)^(1//2) dx -int_0^1 x^(1//2) dx`

`= [((1+x)^(3//2))/(3//2)]^1-[(x^(3//2))/(3//2)]_0^1`

`=2/3 [2^(3//2)-1] -2/3 [1-0]`

`=2/3 [2 sqrt 2-1-1]`

`=2/3 (2 sqrt 2-2)`

`=4/3(sqrt 2-1)`
Correct Answer is `=>` (A) `4/3 (sqrt 2-1)`
Q 2512178039

Let `[x]` denote the greatest integer less than or equal to `x`, then the value of the integral `int_(-1)^1 ( | x | - 2[x]) dx` is equal to
WBJEE 2012
(A)

`3`

(B)

`2`

(C)

`-2`

(D)

`-3`

Solution:

Let ` I = int_(-1)^1 ( | x | - 2[x]) dx`

` = int_(-1)^0 ( | x | - 2[x]) dx + int_(1)^0 ( | x | - 2[x]) dx`

` = int_(-1)^0 ( - x - 2(-1)) dx + int_(1)^0 ( x - 2(0)) dx`

` = int_(-1)^0 (- x + 2)dx + int_(1)^0 x dx`

` = ( - x^2/2 + 2x)^0 + [ x^2/2]_0^1`

` = - ( - 1/2 + 2(-1) ) + 1/2`

` = 1/2 + 2 + 1/2 = 1 + 2 = 3`
Correct Answer is `=>` (A) `3`
Q 2553212144

The value of `int_(0)^(pi) sin^(50) x cos^(49) x dx` is
WBJEE 2011
(A)

`0`

(B)

`pi/4`

(C)

`pi/2`

(D)

`1`

Solution:

Let `I= int_(0)^(pi) sin^(50) x cos^(49) x dx`

`= int_(0)^(pi) sin^(50) (pi-x) cos^(49) (pi- x) dx`

`= - int_(0)^(pi) sin^(50) x cos^(49) x dx`

`=> I = -I => I =0`
Correct Answer is `=>` (A) `0`
Q 2509191918

` int_0^(pi) (dx)/(1 + 2 sin ^2 x)` is equal to
BCECE Mains 2015
(A)

` pi/3`

(B)

`pi/(3sqrt3)`

(C)

`pi/sqrt3`

(D)

`0`

Solution:

`int_0^(pi) (dx)/(1 + 2 sin ^2 x) = 2 int_0^(pi//2) (dx)/(1 + 2 sin ^2 x)`

` [ ∵ int_0^(2a) f(x) dx = 2 int_0^a f(x) dx` if ` f ( 2a - x ) = f(x) ]`

` = 2 int_0^(pi//2) ( sec^2 x)/( sec^2 x + 2 tan^2 x) dx = 2 int_0^(pi//2) ( sec^2 x)/( 1 + 3 tan^2 x) dx`

` = 2 int_0^(oo) (dt)/(1 + 3 t^2)` [put tan x = t]

` = 2 xx 1/sqrt3 [ tan^(-1) t sqrt3 )_0^(oo) = 2/sqrt3 xx pi/2 = pi /sqrt3`
Correct Answer is `=>` (C) `pi/sqrt3`

Problems Set 2

Q 2419591410

`int_(0)^(pi/2) (d theta)/(1+tan theta) ` is equal to
BCECE Stage 1 2012
(A)

`pi`

(B)

`pi/2`

(C)

`pi/3`

(D)

`pi/4`

Solution:

Let `I = int_(0)^(pi/2) (d theta)/(1+tan theta) = int_(0)^(pi/2) (d theta)/(1+tan(pi/2-theta))`



`= int_(0)^(pi/2) (d theta)/(1+cot theta)`


On adding, we get


`2I = int_(0)^(pi/2) (1/(1+tantheta)+1/(1+cot theta))d theta`



` = int_(0)^(pi/2) d theta = [theta]_(0)^(pi/2) = pi/2`


`=> I = pi/4`
Correct Answer is `=>` (B) `pi/2`
Q 2416656570

Let `g (x) = int_(0)^(x) f(t) dt`, where `f` is such that `1/2 le f (x) le 1` for `t in [0,1]` and `0 le f(t) le 1/2` for

`t in [1,2]`.Then, `g(2)` satisfies the inequality
UPSEE 2013
(A)

`-3/2 le g(2) < 1/2`

(B)

`0 le g(2) < 2`

(C)

`1/2 le g(2) le 3/2`

(D)

`2 < g(2) < 4`

Solution:

`g(2) = int_(0)^(1) f(t) dt = int_(0)^(1) f(t) dt + int_(1)^(2) f(x) dt`

`:. 1/2 le f(t) le ` for `t in [0,1]`

`:. 1/2 1 le l_2 le 1`

Agian, `0 le f(t) le 1/2, t in [1,2]`

`=> 1/2 le l_1 + l_2 le 1+1/2`

`=> 1/2 le g(2) le 3/2`
Correct Answer is `=>` (C) `1/2 le g(2) le 3/2`
Q 2314801750

`∫_0^1 (dx)/(e^x + e^(−x)` is equal to :
BITSAT Mock
(A)

`π/4`

(B)

`π/2`

(C)

`tan^(−1) ((e − 1)/( e + 1) )`

(D)

`tan^(−1) (( 1 - e)/( 1 + e) )`

Solution:

`∫_0^1 (dx)/(e^x + e^(−x))`

`= ∫_0^1 (dx)/(e^x + 1 // e^x)`

`= ∫_0^1 (e^x dx)/(1 + e^(2x))`

Put `e^x = t ⇒ e^x dx = dt`

`⇒ ∫_1^e (dt)/(1 + t^2)`

`[ ∵ e^0 = 1 = t`, when `x = 0` and `e^1 = t` when
`x = 1`]

`= [ tan^(−1) t ]_1^e`

`= tan^(−1) e − tan^(−1) 1`

`= tan^(−1) ((e − 1)/(1 + e))`

` [∵ tan^(-1) x − tan^(−1) y = tan^(−1) \ (x − y)/(1 + xy) ]`
Correct Answer is `=>` (C) `tan^(−1) ((e − 1)/( e + 1) )`
Q 2249180913

`int_(0)^(pi^2) sin sqrt(x) dx =`
BITSAT Mock
(A)

`pi`

(B)

`2pi`

(C)

`2 pi^2`

(D)

`pi^2`

Solution:

`int_(0)^(pi^2) sin sqrt(x) dx= int_(t=0)^(pi) 2t sin t dt`,

by putting `x = t^2`

`= 2 {t (-cos t) }_(0)^(pi) - int_(0)^(pi) (- cos t) (+ 1) st`

` = 2 [pi + (sin t)_(0)^(pi)] = 2 pi`
Correct Answer is `=>` (B) `2pi`
Q 2540191013

The value of `int_0^1 tan^(-1) (( 2x - 1)/(1 + x - x^2 )) dx` is

BCECE Stage 1 2013
(A)

`1`

(B)

`0`

(C)

`-1`

(D)

`pi/4`

Solution:

Let `I = int_0^1 tan^(-1) (( 2x - 1)/(1 + x - x^2 )) dx`

` = int_0^1 tan^(-1) { (x + (x -1))/(1 - x(x -1))} dx`

`= int_0^1 {tan^(-1) x + tan^(-1) (x - 1) }dx`

`{ ∵ tan^(-1) x + tan^(-1) y = tan^(-1) ((x + y)/(1 - xy) )}`

` => I = int_0^1 {tan^(-1) x - tan^(-1) (1-x) }dx` .....(i)

Also, `I = int_0^1 {tan^(-1) (1-x) - tan^(-1) {1 - (1 - x) } } dx`

`[ ∵ int_0^a f(x)dx = int_0^a f(a- x) dx ]`

`I = int_0^1 [tan^(-1) (1 - x) - tan^(-1) x]dx` .. (ii)

On adding Eqs. (i) and (ii), we get

` 2 I = 0`

`=> I = 0`
Correct Answer is `=>` (B) `0`
Q 2553445344

If `d/(dx) { f(x)} = g(x)` then `int_(a)^(b) f(x) g(x) dx` is equal to
WBJEE 2010
(A)

`1/2[ f^2(b)-f^2(a)]`

(B)

`1/2[g^2(b)-g^2(a)]`

(C)

`f(b)-f(a)`

(D)

`1/2[f(b^2)-f(a^2)]`

Solution:

`f(x) = int g(x)dx`

`int_(a)^(b) f(x)g(x)dx = (f(x) * f(x))_a^b-int_(a)^(b) g(x) * f(x)dx`


`I = f^2 (b)-f^2(a)`


`I = 1/2(f^2(b)-f^2(a))`
Correct Answer is `=>` (A) `1/2[ f^2(b)-f^2(a)]`
Q 2835012862

`int_0^(pi//2) ( sin 2 theta)/( sin 2 theta + cos 2 theta) d theta` is equal to

(A)

`pi/2`

(B)

`(2pi)/3`

(C)

`pi/4`

(D)

`pi/6`

Solution:

`I = int_0^(pi//2) ( sin 2 theta)/( sin 2 theta + cos 2 theta) d theta`

Put ` x = 2 theta , dx = 2 d theta`

`:. I = 1/2 int_0^(pi) (sin x)/(sin x + cos x) dx`

` = 1/(2 sqrt2) int_0^pi (sin x)/( cos (x - pi/4)) dx`

Put ` x - pi/4 = z`

`=> dx = dz`

` :. I = 1/(2 sqrt2) int_(-pi//4)^(3pi//4) ( sin( z - pi/4))/(cos z) dz`

` = 1/(2 sqrt2) int_(-pi//4)^(3pi//4) ( 1/sqrt2 tan z + 1/sqrt2 ) dz = pi/4`
Correct Answer is `=>` (C) `pi/4`
Q 1573191046

`\int_0^{\pi/2} \frac{\sin^n\theta}{\sin^n\theta +\cos^n\theta}d\theta` is equal to:
BITSAT 2012
(A)

`1`

(B)

`0`

(C)

`\frac{\pi}{2}`

(D)

`\frac{\pi}{4}`

Solution:

Let `I= \int_0^{\pi/2} \frac{\sin^n\theta}{\sin^n\theta +\cos^n\theta}d\theta` .........(1)

Also

`I= \int_(0)^{\pi/2} \frac{\sin^n (\frac{\pi}{2}-\theta\ )}{\sin^n(\frac{\pi}{2}-\theta ) +\cos^n (\frac{\pi}{2}-\theta )}d\theta `

`= \int_0^{\pi/2}\frac{\cos^n\theta}{\sin^n\theta +\cos^n\theta}d\theta` ......(2)

Now add (1) and (2)

`\Rightarrow 2I =\int_0^{\frac{\pi}{2}}d\theta=\ frac{\pi}{2}`

`\therefore I =\ frac{\pi}{4}`
Correct Answer is `=>` (D) `\frac{\pi}{4}`
Q 2419856719

Find the value of `int_0^(pi//2) (dx)/(1+tan^3 x)`
BCECE Stage 1 2011
(A)

`0`

(B)

`1`

(C)

`pi//2`

(D)

`pi//4`

Solution:

let `I=int_0^(pi//2) (dx)/(1+tan^3 x)`............(i)

`I=int_0^(pi//2) (dx)/(1+tan^3 (pi/2-x))`

`=int_0^(pi//2) (dx)/(1+cot^3 x)`

`=> I= int_0^(pi//2) (tan^2 x)/(1+tan^3 x) dx`.............(ii)

On adding Eqs. (i) and (ii), we get

`2I=int_0^(pi//2) ((1+tan^3x)/(1+tan^3 x))dx`

`=int_0^(pi//2) dx=[x]_0^(pi//2)`

`=> I=pi//4`
Correct Answer is `=>` (D) `pi//4`
Q 2512734639

The value of integral `int_(pi/6)^(pi/3) (sinx-xcosx)/(x(x+sinx))dx` is

WBJEE 2013
(A)

`log_e {(2(pi+3)/(2pi+3sqrt3)}`

(B)

`log_e {(pi+3)/(2pi+3sqrt3)}`

(C)

`log_e{(2pi+3sqrt3)/(2(pi+3))}`

(D)

`log_e{(2pi+3sqrt3)/(pi+3)}`

Solution:

Let `I = int_(pi/6)^(pi/3) (sinx-xcosx)/(x(x+sinx))dx`


`I = int_(pi/6)^(pi/3) ((x+sinx)-x(1+cosx))/(x(x+sinx))dx`


`I = int_(pi/6)^(pi/3) ((1/x-(1+cosx)/(x+sinx))dx`


`=> I = [logx]_(pi/6)^(pi/3) -int_(pi/6)^(pi/3) (1+cosx)/(x+sinx) dx`

put`t = x+sinx`

`dt = (1+cosx)dx`


`I = (log(pi/3)-log(pi/6) -int_(pi/6+1/2)^(pi/3+sqrt3/2) (dt)/t`


`I = log2[logt]_(pi/6+1/2)^(pi/3+sqrt3/2)`


`I = log2-[log(pi/3+sqrt3/2)-log(pi/6+1/2)]`


`I = log2-log((2pi+3sqrt3)/(pi+3))` `(because logm-logn = log (m/n))`


`I = log ((2(pi+3))/(2pi+3sqrt3))`


` = log((2pi+6)/(2pi+3sqrt3))`
Correct Answer is `=>` (A) `log_e {(2(pi+3)/(2pi+3sqrt3)}`
Q 2582167937

The value of the integral ` int_1^5 [ | x - 3 | + | 1 - x | ] dx` is equal to
WBJEE 2012
(A)

`4`

(B)

`8`

(C)

`12`

(D)

`16`

Solution:

` int_1^5 [ | x - 3 | + | 1 - x | ] dx`

` = int_1^5 | x - 3 | dx + int_1^5 | 1 - x | dx`

` = int_1^3 | x - 3 | dx + int_3^5 | x - 3 | dx + int_1^5 | 1 - x | dx`

` = int_1^3 -( x - 3 ) dx + int_3^5 ( x - 3 ) dx + int_1^5 -( 1 - x ) dx`

` = int_1^3 ( x - 3 ) dx + int_3^5 ( x - 3 ) dx + int_1^5 ( 1 - x ) dx`

` = [ 3x - x^2/2 ]_1^3 + [ x^2/2 - 3x]_3^5 + [x^2/2 - x]_1^5`

` = ( 3 xx 3 - 9/2) - ( 3 xx 1 - 1/2) + ( (5xx5)/2 - 3 xx 5)`

` - ( (3 xx 3)/2 - 3 xx 3) + ( ( 5 xx 5)/2 - 5) - ( 1/2 - 1)`

` = ( 9 - 9/2) - (3 - 1/2) + ( (25)/2 - 15) - ( 9/2 - 9)`

` + ( (25)/2 - 5) - ( - 1/2)`

` = 9/2 - 5/2 - 5/2 + 9/2 + (15)/2 + 1/2`

` = ( 9 - 5 - 5 + 9 + 15 + 1)/2 = (24)/2 = 12`
Correct Answer is `=>` (C) `12`
Q 2561380225

The value of the integral ` int_0^(pi//2) 1/(1 +(tan x)^(101)) dx`
WBJEE 2012
(A)

`1`

(B)

`pi/6`

(C)

`pi/8`

(D)

`pi/4`

Solution:

Let `I = int_0^(pi//2) 1/(1 +(tan x)^(101)) dx`

` = int_0^(pi//2) (dx)/( 1 + { tan (pi/2 - x)}^(101)) dx`

` = int_0^(pi//2) (dx)/(1 +(cot x)^(101))`

` = int_0^(pi//2) ( tan x^(101))/( tan x^(101) + 1) dx`

` = int_0^(pi//2) ( 1 + tan x^(101) -1)/(1 + tan x^(101)) = [x]_0^(pi//2) - I`

` => I = pi/2 - I => 2I = pi/2 => I = pi/4`
Correct Answer is `=>` (D) `pi/4`
Q 2805323268

If `I_n = int_0^(pi//4) tan^n x dx `, then what is `I_n + I_(n-2)` equal to?

(A)

`1/n`

(B)

`1/(n-1)`

(C)

`n/(n-1)`

(D)

`1/(n-2)`

Solution:

`I_n = int_0^(pi//4) tan^n x dx`

`= int_0^(pi//4) tan^(n-2) x tan^2 x dx `

`= int_0^(pi//4) tan^(n-2) x ( sec^2 x - 1 ) dx`

` = int_0^(pi/4) tan^(n -2) x sec^2 x dx`

` - int_0^(pi/4) tan^(n -2) x dx`

Put `t = tan x => dt = sec^2 x dx`

`:. I_n = int_0^1 t^(n-2) dt - I_(n-2)`

` = [ t^(n-1)/(n-1)]_0^1 - I_(n-2) = 1/(n-1) - I_(n-2)`

`=> I_n + I_(n-2) = 1/(n-1)`
Correct Answer is `=>` (B) `1/(n-1)`
Q 2385480367

`int_(0)^(4) {sqrt (x) } dx` equals (where `{x}` denotes
the fractional part of `x`)
BITSAT Mock
(A)

`16/3`

(B)

`7/3`

(C)

`4/3`

(D)

`8/3`

Solution:

` {sqrt (x) } = sqrt (x) - [sqrt (x)]`

`=> int_(0)^(4) { sqrt (x) }dx`

`= int_(0)^(4) (sqrt (x) - [sqrt (x)] ) dx`

`= (4^(3/2))/(3/2) - ( int_(0)^(1) [sqrt (x) ]dx + int_(1)^(4) [sqrt (x) ] dx )`

`= 16/3 - (0+ int_(1)^(4) dt) = 7/3`
Correct Answer is `=>` (B) `7/3`
Q 2513012840

Let `f(x) =max {x +| x|, x- [x]},` where `[x]` denotes the greatest value of `int_(-3)^3 f(x) dx` is
WBJEE 2014
(A)

`0`

(B)

`51//2`

(C)

`21//2`

(D)

`1`

Solution:

Given, `f(x) =max {x +| x|, x- [x]}`

`={ tt((2x, x ge 0),( x-[x], x le 0))`

`:. int_(-3)^3 f(x) dx =int_(-3)^0 x - [x] dx + int_0^3 2x dx`

`=3 int_(-3)^0 (1+x) dx +2 int_0^3 x dx`

[`x - [x]` is a periodic function at `x = 1`]

`=3[x+x^2/2]_(-1)^0 + 2[x^2/2]_0^3`

`=3[0-0-(-1+1/2)]+3^2-0`

`=3[1/2]+9 =3/2+9`

`=21/2`
Correct Answer is `=>` (C) `21//2`
Q 2560191015

If `int_( log2)^x 1/sqrt(e^x - 1) dx = pi/6` , then the value of x is
BCECE Stage 1 2013
(A)

log2

(B)

log3

(C)

log4

(D)

None of these

Solution:

Let `I = log_(log2)^x 1/sqrt(e^x - 1) dx `

Put `e^x - 1 = t^2`

`=> e^x dx = 2t dt`

`:. I = sqrt2 int_1^(sqrt(e^x) -1) 1/(1 + t^2) dt`

` = [ tan^(-1) sqrt(e^x - 1) - tan^(-1) 1]`

`= 2 tan^(-1) sqrt(e^x - 1) - pi/4`

But ` int_(log_e^2)^x = 1/ sqrt(e^x - 1) dx pi/6`

` 2 tan^(-1) sqrt(e^x - 1)`

` = (2 pi)/3`

` => sqrt(e^x - 1) = tan pi/3`

` = sqrt3`

` = e^x -1`

` = 3`

` => e^x = 4`

`:. x = log 4`
Correct Answer is `=>` (C) log4
Q 2552180034

Evaluate the following integral `int_(-1)^(2) |xsinpix|dx`
WBJEE 2010
(A)

`pi/5`

(B)

`-pi/5`

(C)

`5/pi`

(D)

`-5/pi`

Solution:

`I = int_(-1)^(2) |xsinpix|dx = int_(-1)^(1) |xsinpix|dx+ int_(1)^(2) |xsinpix|dx`


` = 2 int_(0)^(1) |x sinpix|dx+int(1)^(2) |x . sinpix|dx`


` = 2int_(0)^(1) x . sinpixdx-int_(1)^(2) x . sinpixdx = 2I_1-I_2`


`I_1 = int_(0)^(1) x sinpixdx = [-x (cospix)/pi]_(0)^(1)+int_(0)^(1) (cospix)/pi dx`


` = [-x (cospix)/pi+(sinpix)/pi^2]_(0)^(1) = 1/pi`



and `I_2 = int_(1)^(2) x sinpixdx = [-x (cospix)/pi+(sinpix)/pi^2]_(1)^(2)`


` = -2/pi+0(-1/pi) = -3/pi`


so `2I_1-I_2 = 2/pi+3/pi = 5/pi`
Correct Answer is `=>` (C) `5/pi`
 
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