Mathematics Previous year questions Of Definite Integrals For NDA
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Previous year questions Of Definite Integrals For NDA

previous year questions Of Definite Integrals for NDA

Q 2753491344

What is `int_(0)^(pi/2) ( d theta )/(1+costheta) ` equal to?
NDA Paper 1 2017
(A)

`1/2`

(B)

`1`

(C)

`sqrt3`

(D)

None of the above

Solution:

`I= int_0^(pi/2) (d theta)/(1+ cos theta) = int_0^(pi/2) (d theta)/( 2 cos^2 theta//2)`

`=1/2 int_0^(pi/2) sec^2 (theta/2) d theta`

`=2/2 [tan theta//2]_0^(pi//2)`

`= [1-0] = 1`
Correct Answer is `=>` (B) `1`
Q 2753191044

If `f(x)` and `g(x)` are continuous functions satisfying `f(x) = f(a- x)` and `g(x) + g(a- x) = 2`, then what is
`int_0^a f(x) g(x) dx` equal to?
NDA Paper 1 2017
(A)

`int_0^a g(x) dx`

(B)

`int_0^a f(x) dx`

(C)

`2 int_0^a f(x) dx`

(D)

`0`

Solution:

` I= int_0^a f(x) g(x) dx = int_0^a f(a-x) g(a-x) dx`

`= int_0^a f(x) (2 - g(x)) dx`

`= int_0^a 2 f(x) dx - int_0^a f(x) g(x) dx`

`2I = 2 int_a^a f(x) g(x)= 2 int f(x) dx`

`=> I = int_0^a f(x) g(x) = f(x) dx`
Correct Answer is `=>` (B) `int_0^a f(x) dx`
Q 2714345250

What is `int_(e^(-1))^(e^2) |(lnx)/x|` dx is equal to ?
NDA Paper 1 2017
(A)

`3/2`

(B)

`5/2`

(C)

`3`

(D)

`4`

Solution:

`I = int_(e^(-1))^(e^2) |(lnx)/x|` dx

As both `e^(-1)` and `e^2` are positive `| x | = x`

Let `ln x = t`

`1/x dx = dt`

`I = int_(-1)^(2) |t| dt`

`= (int_(-1)^(0) -t + int_(0)^(2) t ) dt`

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

`= 3/2`
Correct Answer is `=>` (A) `3/2`
Q 2125391261

Consider the functions `f(x) = xg(x)`
and `g{x) = [1/x], ` where[.] is the greatest integer function.

What is ` int _(1//3)^(1//2) g(x) dx` equal to?
NDA Paper 1 2016
(A)

`1/6`

(B)

`1/3`

(C)

`5/18`

(D)

`5/36`

Solution:

Given, `g(x) = [1/x]`

When `1/3 < x < 1/2 => 2 < 1/x < 3`

`:. [1/x] = 2`

Now, we can find ` int _(1//3)^(1//2) g(x) dx`

`:. int _(1//3)^(1//2) g(x) dx = int _(1//3)^(1//2) [1/x] dx = int _(1//3)^(1//2) 2dx`


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

` = 2 xx (3-2)/6 = 2 xx 1/6 = 1/3 `
Correct Answer is `=>` (B) `1/3`
Q 2155691564

Consider the functions `f(x) = xg(x)`
and `g{x) = [1/x], ` where[.] is the greatest integer function.

What is ` int _(1//3)^(1) f(x) dx` equal to?
NDA Paper 1 2016
(A)

`37/72`

(B)

`2/3`

(C)

`17/72`

(D)

`37/144`

Solution:

Given, `f(x) = x g(x)`

and `g(x)= [1/x]`

We have to split given limit into two parts.

When `1/3 < x < 1/2 => 2 < 1/x < 3`

`=> [1/x ] = 2`

`f(x) = x g(x) = 2x`

When `1/2 < x < 1 => 1 < 1/x < 2 => [1/x] = 1`

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

Now ` int _(1//3)^(1) f(x) dx = int _(1//3)^(1//2) 2x dx + int_(1//2)^1 x dx`

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

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

` = [1/4 - 1/9] + 1/2 [1- 1/4]`

` = (9-4)/36 + 1/2 [(4-1)/4] = 5/ (36) = 3/8 = (10 +27)/72 = 37/72`
Correct Answer is `=>` (A) `37/72`
Q 2166701675

Given that `a_(n) = int _(0)^(pi) (sin^(2) {(2n + 1) x}) /(sin 2x) dx`
Consider the following statements
1. The sequence `{a _(2n) }` is in AP with common difference
zero.
2. The sequence `{a _(2n +1)}` is in AP with common
difference zero.

Which of the above statements is/are correct?
NDA Paper 1 2016
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

` ∵ a_(2n) = int _(0)^(pi) (sin^(2) {(2n + 1) x}) /(sin 2x) dx .............(i)`

By applying property

` int _(0)^(pi) f(x) dx = int _(0)^(pi) f(pi - x) dx`

` :. a_(2n) = int _(0)^(pi) (sin^(2) {(2n + 1) (pi - x)}) /(sin 2(pi - x) ) dx`

` = int _(0)^(pi) (sin^(2) {(2n + 1) x}) /(- sin 2x) dx...........(ii)`

`[ ∵` for any value of n, `sin^(2) {(2n + 1) (pi- x)}
= sin^(2) {(2n + 1) x)`

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

` 2a_(2n) = int _(0)^(pi) (sin^(2) {(2n + 1) x}) /( sin 2x) dx - int _(0)^(pi) (sin^(2) {(2n + 1) x}) /( sin 2x) dx`

` => 2a_(2n) = 0`

` => a_(2n) =0`

Similarly, value `a_(2n + 1) = 0`

:. The sequence `{a_(2n) }` is in AP with common difference
zero and the sequence `{a_(2n + 1)}` is in AP with common
difference zero.
Correct Answer is `=>` (C) Both 1 and 2
Q 2156001874

Given that `a_(n) = int _(0)^(pi) (sin^(2) {(2n + 1) x}) /(sin 2x) dx`
Consider the following statements
1. The sequence `{a _(2n) }` is in AP with common difference
zero.

What is `a_(n-1) - a _(n-4)` equal to?
NDA Paper 1 2016
(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`∵ a_(n-1) = int _(0)^(pi) (sin^(2){(n-1+1)}x )/(sin 2x) dx `

` = int _(0)^(pi) (sin^(2) {nx})/(sin 2x) dx ...........(1)`

By applying property ` int _(0)^(pi) f(x) dx = int _(0)^(pi) f(pi-x) dx`

` :. a_(n-1) = int _(0)^(pi) (sin^(2){n (pi - x)})/(sin 2 (pi - x) ) dx`

` = int _(0)^(pi) (sin^(2) {nx} )/(-sin2x) dx ........(2)`

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

` 2a_(n-1) =0`

` a_(n-1) =0`

Similarly, `a_(n-4) = 0`

`:. a_(n-1) - a_(n-4) = 0 - 0 = 0`
Correct Answer is `=>` (B) `0`
Q 2156891774

What is `int _(-2)^(2) x dx - int _(-2)^(2) [x] dx`

equal to, where [.] is the greatest integer function?
NDA Paper 1 2016
(A)

`0`

(B)

`1`

(C)

`2`

(D)

`4`

Solution:

`int _(-2)^(2) x dx - int _(-2)^(2) [x] dx`

We have, `[x] = {tt ((-2 , -2 <= x < -1) ,(-1 , -1 <= x < 0) , (0 , 0 <= x < 1), (1, 1 <= x < 2 ))`

` :. int _(-2)^(2) x dx - [int _(-2)^(-1) (-2) dx + int _(-1)^(0) (-1) dx + int _(0)^(1) 0 dx + int _(1)^(2) 1 dx ]`

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

`[ ∵ int _(-a)^(a) f(x) dx = 0 ,` if ` f(x)` is odd`]`

`=-[-2(-1+ 2)+ (0- 1)+ (2- 1)]`

`=-[-2-1+1]=2`
Correct Answer is `=>` (C) `2`
Q 2116191970

If `int_(-2)^5 f(x) dx = 4` and `int_0^5 {1 + f(x) } dx = 7`,
then what is `int_(-2)^0 f(x) dx` equal to?
NDA Paper 1 2016
(A)

`-3`

(B)

`2`

(C)

`3`

(D)

`5`

Solution:

Given,

`int_(-2)^5 f(x) dx = 4, int_0^5 {1 + f(x)} dx = 7`

Let `I = int_(-2)^5 f(x) dx = int_(-2)^0 f(x) dx + int_5^0 f(x) dx`

`=> 4 = int_(-2)^0 f(x) dx + int_0^5 [(1 + f(x)) - 1] dx`

`=> 4 = int_(-2)^0 f(x) dx + int_0^5 [1 + f(x)] dx - int_0^5 1 dx`

` => 4 = int_(-2)^0 f(x) dx + 7 - (x)_0^5`

`=> 4 = int_(-2)^0 f(x) dx + 7 - 5`

`=> 4 = int_(-2)^0 f (x) dx + 2`

`=> int_(-2)^0 f(x) dx = 2`
Correct Answer is `=>` (B) `2`
Q 2331580422

What is `int_0^(4 pi) | cos x | dx` equal to ?
NDA Paper 1 2016
(A)

`0`

(B)

`2`

(C)

`4`

(D)

`8`

Solution:

Let ` I = int _0^(4 pi) | cos x | dx = 2 int _0^(2 pi) | cos x | dx`

`[because int_0^(2a) f(x) dx = 2 int _0^a f(x) dx`, if `f(2a -x) = f(x) `, here

`I cos (4pi- x) I= I cos xI]`

`2 * 2 int_0^(2a) f(x) dx = 2 int_0^a | cos x |dx ` `[ because | cos(2 pi -x)| = | cos x|]`

`= 2 * 2 * 2 int _0^(pi//2) | cos x | dx` `\ \ \ \ [ | cos ( pi -x ) | = | - cos x | = | cos | ]`

`=8 int _0^(pi/2) cos x dx ` `[ because cos x > 0 , AA x in (0, pi/2)]`

`= 8 [sin ] _0^(pi/2) = 8 [ sin frac (pi)(2) - sin 0 ] =8`
Correct Answer is `=>` (D) `8`
Q 2761267125

What is `int e^(sinx) ( x cos^3 x - sin x)/( cos^2 x) dx` equal to ?
NDA Paper 1 2016
(A)

`( x+secx) e^(sinx)+C`

(B)

`(x- secx) e^(sinx)+C`

(C)

`(x+tanx)e^(sinx)+C`

(D)

`(x-tanx)e^(sinx)+C`

Solution:

`int e^(sinx) [ x cos x - -tan x sec x] dx`

`int e^(sin x) [ x cos x -1] + int e^(sin x) [1- sec x tan x]`

`int (e^(sin x) cos x [x- sec x] + e^(sin x) [ 1- sec x tan x]) dx`

`int [( x e^(sin x) - e^(sin x) ) +(e^(sin x) cos x * sec x- e^(sin x) sec x tan x)]+ C`

`=int d/(dx) ( x e^(sin x))- int d/(dx) ( sec x e^(sin x))+ C`

`= (x- sec x)e^(sin x) + C`
Correct Answer is `=>` (B) `(x- secx) e^(sinx)+C`
Q 2731267122

If `int_(0)^(pi/2) (dx)/(3 cosx+5) = k cot^(-1) 2` . then what is the value of k ?
NDA Paper 1 2016
(A)

`1/4`

(B)

`1/2`

(C)

`1`

(D)

`2`

Solution:

`I= int_0^(pi//2) (dx)/(3 cos x +5)`

`= int_0^(pi//2) (cos x dx)/(3 (2 cos ^2 \ (x)/2 -1)+5) = int_0^(pi//2) (dx)/(6 cos^2 \ x/2 +2)`

`=int_0^(pi//2) (sec^2 \ x/2)/(6+2 + 2 tan^2 \ x/2) = int_0^(pi//2) (sec^2 \ x/2)/( 2tan^2 \ x/2 + 8)`

`= int_0^1 (2 dt)/(2 t^2 + 8)= int_0^1 1/(t^2 +(2)^2) dt`

[Let `tan \ x/2 =t \ \ 1/2 sec^2 \ x/2 dx= dt`]

`=1/2[ (tan^(-1) \ t/2)]_0^1 `

`=1/2 cot^(-1) \ 2`

`:. k=1/2`
Correct Answer is `=>` (B) `1/2`
Q 2711167029

What is `int_(1)^(3) | 1-x^4|dx` equal to
NDA Paper 1 2016
(A)

`(-232)/5`

(B)

`(-116)/5`

(C)

`(116)/5`

(D)

`(232)/5`

Solution:

`I= int_1^3 |1-x^4| dx`

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

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

`=[(3^5/5 -9/2)-(1/5 - 1/2)]`

`=232/5`
Correct Answer is `=>` (D) `(232)/5`
Q 2271380226

` int _(-1)^1 x |x| dx` is equal to
NDA Paper 1 2015
(A)

`0`

(B)

` 2/3`

(C)

`2`

(D)

` -2`

Solution:

` int _(-1)^1 x |x| dx = int _(-1)^0 x (-x) dx + int _(0)^1 x (x) dx`

[when `a < c < b`, then ` int _(a)^b f( x) dx = int _(a)^c f( x) dx + int _(c)^b f( x) dx `]

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

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

` = 0 - (-1/3) + 1/3 = 2/3 `
Correct Answer is `=>` (B) ` 2/3`
Q 2211178029

The value of ` int_a^b (x^7 + sin x)/(cos x) dx`, where `a + b = 0`, is
NDA Paper 1 2015
(A)

`2b- a sin (b- a)`

(B)

`a+ 3bcos (b- a)`

(C)

`sin a- (b- a) cos b`

(D)

`0`

Solution:

We have, `a+ b = 0 => a = - b`

`:. int_a^b (x^7 + sin x)/(cos x) dx = int_(-b)^b (x^7/(cos x) + (sin x)/(cos x)) dx`

` = int_(-b)^b (x^7 . sec x + tan x) dx`

The given integrand is an odd function.

Hence, its answer is `0`.
Correct Answer is `=>` (D) `0`
Q 2251878724

If `0 < a < b`, then ` int _a^b (|x|)/x dx `is equal to
NDA Paper 1 2015
(A)

`| b|- |a|`

(B)

`| a|- |b|`

(C)

`(| b|) /(|a|)`

(D)

`0`

Solution:

` int _a^b (|x|)/x dx = int _a^b 1 dx quad [ ∵ 0 < a < b ]`

`= [x] _a^b = | b|- |a|`
Correct Answer is `=>` (A) `| b|- |a|`
Q 2271078826

`int_0^(2pi) sin^5 (x/4) dx` is equal to
NDA Paper 1 2015
(A)

`8/(15)`

(B)

`(16)/(15)`

(C)

`(32)/(15)`

(D)

`0`

Solution:

We have, `int_0^(2pi) sin^5 (x/4) dx`

Put ` x/4 = t => dx = 4 dt`

`:. int_0^(pi//2) sin^5 t . 4 dt = 4 dt int_0^(pi//2) sin^5 t dt`

By Walli's formula, `int_0^(pi//2) sin^n dx = int_0^(pi//2) cos^n xdx`

` = (n -1)/n . (n -3)/(n - 2) . (n -5)/(n - 4) ... 2/3 ` When `n` is odd

` = (n -1)/n . (n -3)/(n - 2) . (n -5)/(n - 4) ... 3/4 . 1/2 . pi/2` When `n` is even

` = 4 . (4.2)/(5.3) = (32)/(15)`
Correct Answer is `=>` (C) `(32)/(15)`
Q 1659612514

Consider the integral
`I_m = int_0^pi ( sin 2mx)/( sin x) dx` where `m` is a positive integer.

What is `I_m` equal to?
NDA Paper 1 2015
(A)

`0`

(B)

`1`

(C)

`m`

(D)

`2m`

Solution:

We have,

`I_m = int_0^pi ( sin 2mx)/( sin x) dx` where `m` is a positive integer.


We know that , if m is a positive integer then

`(sin2mx)/(sinx) = 2 [cosx + cos3x + ..... + cos(2m -1) x]`

`:. I_m = int_0^2 2 [cosx + cos3x + ... + cos(2m -1) x]dx`

` = 2 [sinx + (sin3x)/3 + ... + (sin (2m- 1) x)/(2m- 1) ]_0^pi`

`= 2(sinx + (sin 3x)/3 ... + (sin (2m- 1) x)/(2m- 1) )`

` =2 ( sin 0 + (sin 0)/3 + .... + ( Sin 0) /(2m -1) )`

`=2 [0- 0]`

`=> I_m = 0`

`∵ I_m = 0`

`:. I_m = 0`
Correct Answer is `=>` (A) `0`
Q 1619712610

Consider the integral
`I_m = int_0^pi ( sin 2mx)/( sin x) dx` where `m` is a positive integer.

Consider the following
I. `I_m- I_(m -1)` is equal to 0
II. `I_(2m) > I_m`

Which of the above is/are correct?
NDA Paper 1 2015
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

We have,

`I_m = int_0^pi ( sin 2mx)/( sin x) dx` where `m` is a positive integer.


We know that , if m is a positive integer then

`(sin2mx)/(sinx) = 2 [cosx + cos3x + ..... + cos(2m -1) x]`

`:. I_m = int_0^2 2 [cosx + cos3x + ... + cos(2m -1) x]dx`

` = 2 [sinx + (sin3x)/3 + ... + (sin (2m- 1) x)/(2m- 1) ]_0^pi`

`= 2(sinx + (sin 3x)/3 ... + (sin (2m- 1) x)/(2m- 1) )`

` - ( sin 0 + (sin 0)/3 + .... + ( Sin 0) /(2m -1) )`

`=2 [0- 0]`

`=> I_m = 0`

`∵ I_m = 0`

`:. I_m - I_(m-1) = 0 - 0 = 0`

` I_m = 0`

` => I_(2m) = 0`

` :. I_(2m) > I_m`

` => I_1 = 0 ; I_2 + I_3 = 0; I_m - I_(m -1) = 0`

and `I_(2m) = I_m`
Correct Answer is `=>` (A) Only I
Q 2261291125

Consider the integrals
` A = int _0^pi (sin x dx)/(sin x + cos x) ` and ` B = int _0^pi (sin x dx)/(sin x - cos x) `

Which one of the following is correct ?
NDA Paper 1 2015
(A)

`A = 2B`

(B)

`B = 2A`

(C)

`A= B`

(D)

`A = 3B`

Solution:

Given `A= int _0^pi (sin x dx)/(sin x + cos x) ` and ` B = int _0^pi (sin x dx)/(sin x - cos x) `

Now ` A = int _0^pi (sin (pi - x) dx)/(sin (pi - x) + cos (pi - x) ) `

` [ ∵ int _0^a f(x) dx = int _a^0 f(a - x) dx]`

` = int _0^pi (sin x dx)/(sin x - cos x) = B `
Correct Answer is `=>` (C) `A= B`
Q 2271391226

Consider the integrals
` A = int _0^pi (sin x dx)/(sin x + cos x) ` and ` B = int _0^pi (sin x dx)/(sin x - cos x) `

What is the value of `B`?

NDA Paper 1 2015
(A)

`pi/4`

(B)

`pi/2`

(C)

`(3pi)/4`

(D)

`pi`

Solution:

Let `I = A = int _0^pi (sin x dx)/(sin x + cos x) ` ..........(i)

Let `I = B = int _0^pi (sin x dx)/(sin x - cos x) ` .........(ii)

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

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

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

`=> 2I = int_0^(pi) ( sin x (sin x - cos x + sin x + cos x))/( sin^2 x- cos^2 x) dx`

`=> 2I = 4 int_0^(pi//2) (sin^2 x)/(sin^2 x- cos^2 x) ` .........(iii)

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

` => 2I = 4 int_0^(pi//2) (cos^2 x)/(cos^2 x- sin^2 x) ` .........(iv)

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

` => 4I = 4 int_0^(pi//2) ( ( sin^2 x - cos^2 x)/( sin^2 x - cos^2 x)) dx`

[On adding Eqs. (iii) and (iv)]

` => 4I = 4(x)_0^(pi//2) `

`=> 4I = 4 xx pi/2 => I = pi/2`
Correct Answer is `=>` (B) `pi/2`
Q 1722080831

What is `int _(-pi//2)^(pi//2) x` sin `x dx` equal to?
NDA Paper 1 2014
(A)

`0`

(B)

`2`

(C)

`-2`

(D)

`pi`

Solution:

Let `I = int _(-pi//2)^(pi//2) x sin x dx`

` I = 2 int _(0)^(pi//2) x sin x dx`

`[∵ f(x) = x sin x` is an even function.

` => f(- x) = (- x) sin(- x) = (- x) (-sin x) = x sin x = f(x)]`

` [∵ int_(-a)^(a) f(x) = {2 int_(-a)^(a) f(x)`, if `f(x)` is an even function

` = 0`, if `f(x)` is an odd function]
Correct Answer is `=>` (B) `2`
Q 1722691531

What is ` int_(0)^(pi//2) ln (tan x) dx` equal to?
NDA Paper 1 2014
(A)

`In 2`

(B)

`- In 2`

(C)

`0`

(D)

None of these

Solution:

Let ` I = int_0^(pi//2) log(tan x) dx` .........(1)

` I = int_0^(pi//2) log tan ( pi/2 - x)dx`

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

` = int_0^(pi//2) log cot xdx`

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

` I = int_0^(pi//2) - log (tan x) dx`

` I = - int_0^(pi//2) log (tan x) dx = -1` [from Eq. (1)]

`=> 2I = 0`

` => I = 0 ` or ` int_0^(pi//2) log (tan x)dx = 0`
Correct Answer is `=>` (C) `0`
Q 1753712644

What is ` int _0^1 (e^(tan^(-1)quad_x) dx)/(1 + x^2)` equal to?
NDA Paper 1 2014
(A)

` e^(pi/4) - 1`

(B)

` e^(pi/4) + 1`

(C)

` e - 1`

(D)

`e`

Solution:

Let `I = int _0^1 (e^(tan^(-1)quad_x) dx)/(1 + x^2)`

put ` t = tan^(-1) x `

` dt = 1/(1 + x^2) dx`

Lower Limit `-> t = tan^(-1) 0 = 0`

Upper Limit `-> t = tan^(-1) 1 = pi/4`

` :. I = int _0^(pi/4) e^t = [e^t] _0^(pi/4)`

` = [e^(pi/4) - e^0 ] = e^(pi/4) -1`
Correct Answer is `=>` (A) ` e^(pi/4) - 1`
Q 1743456343

Consider the integrals

`I_1 = int _(pi//6)^(pi//3) (dx)/(1 + sqrt(tan x))` and `I_2 = int _(pi//6)^(pi//3) ( sqrt(sin x) dx)/(sqrt(sin x) + sqrt(cos x))`

What is `I_ 1 - I_ 2` equal to?
NDA Paper 1 2014
(A)

`0`

(B)

` 2 I_1`

(C)

`pi`

(D)

None of these

Solution:

Given that, `I_1 = int _(pi//6)^(pi//3) (dx)/(1 + sqrt(tan x))`

and ` I_2 = int _(pi//6)^(pi//3) ( sqrt(sin x) dx)/(sqrt(sin x) + sqrt(cos x))` ......(1)

` I_2` can be written as

` I_1 = int _(pi//6)^(pi//3) (dx)/( 1 + sqrt(sin x ) /sqrt(cos x))`

` I_2 = int _(pi//6)^(pi//3) ( sqrt(cos x) dx)/(sqrt(sin x) + sqrt(cos x))` ......(2)

We have ` I_1 - I_2 = int _(pi//6)^(pi//3) ( sqrt(cos x) )/(sqrt(sin x) + sqrt(cos x)) - int _(pi//6)^(pi//3) ( sqrt(sin x) dx)/(sqrt(sin x) + sqrt(cos x))`

` = int _(pi//6)^(pi//3) ( (sqrt(cos x ) - sqrt(sin x) )/(sqrt(cos x ) + sqrt(sin x)) dx)`

` = int _(pi//6)^(pi//3) ( sqrt(cos(pi/3 + pi/6 - x)) - sqrt(sin(pi/3 + pi/6 - x)) //sqrt(cos(pi/3 + pi/6 - x)) + sqrt(sin(pi/3 + pi/6 - x)) ) .dx`

` = int _(pi//6)^(pi//3) ( ( cos ( pi/2 - x)) - sqrt (sin (pi/2 - x)) //( cos ( pi/2 - x)) - sqrt (sin (pi/2 - x)) ) dx `

` = int _(pi//6)^(pi//3) ( sqrt (sin x ) - sqrt (cos x) )/ ( sqrt (sin x) + sqrt(cos x) ) . dx`

` = - { int _(pi//6)^(pi//3) ( sqrt (cos x ) - sqrt (sin x) )/ ( sqrt (sin x) + sqrt(cos x) ) . dx }`

` = - ( I_1 - I_2)`

` => 2 ( I_1 - I_2) = 0 => I_1 - I_2 = 0`
Correct Answer is `=>` (A) `0`
Q 1753456344

Consider the integrals

`I_1 = int _(pi//6)^(pi//3) (dx)/(1 + sqrt(tan x))` and `I_2 = int _(pi//6)^(pi//3) ( sqrt(sin x) dx)/(sqrt(sin x) + sqrt(cos x))`

What is `I_1` equal to?
NDA Paper 1 2014
(A)

` pi/(24)`

(B)

` pi/(18)`

(C)

` pi/(12)`

(D)

` pi/(6)`

Solution:

Given that, `I_1 = int _(pi//6)^(pi//3) (dx)/(1 + sqrt(tan x))`

and ` I_2 = int _(pi//6)^(pi//3) ( sqrt(sin x) dx)/(sqrt(sin x) + sqrt(cos x))` ......(1)

` I_2` can be written as

` I_1 = int _(pi//6)^(pi//3) (dx)/( 1 + sqrt(sin x ) /sqrt(cos x))`

` I_2 = int _(pi//6)^(pi//3) ( sqrt(cos x) dx)/(sqrt(sin x) + sqrt(cos x))` ......(2)

From Eq. (1),

` I_2 = int _(pi//6)^(pi//3) ( sqrt(cos x) dx)/(sqrt(sin x) + sqrt(cos x))` .....(3)

` = int _(pi//6)^(pi//3) sqrt (cos ( pi/3 + pi/2 - x) ) / sqrt (sin ( pi/3 + pi/2 - x) + sqrt (cos ( pi/3 + pi/2 - x) ).dx`

` = int _(pi//6)^(pi//3) sqrt (cos ( pi/2 - x) ) / sqrt (sin ( pi/2 - x) + sqrt (cos ( pi/2 - x) ).dx`

` = int _(pi//6)^(pi//3) sqrt(sinx)/(sqrt(cos x) + sqrt(sin x) ) . dx` .............(4)

On adding Eqs. (1) and (2), we get

` 2 I_1 = int _(pi//6)^(pi//3) (( sqrt(cos x) + sqrt (sin x)) /( sqrt(cos x) + sqrt (sin x) )) dx`

` 2 I_1 = int _(pi//6)^(pi//3) (1) dx = [x]_(pi//6)^(pi//3)`

` = ( pi/3 - pi/6) = pi/6 => I_1 = pi/(12)`
Correct Answer is `=>` (C) ` pi/(12)`
Q 1741401323

What is ` int_0^(pi/2) (dx)/(a^2 cos^2 x + b^2 sin^2 x)` equal to?

NDA Paper 1 2014
(A)

`2 ab`

(B)

`2pi ab`

(C)

` pi/(2 ab)`

(D)

`pi/(ab)`

Solution:

Let `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)`

[divide numerator and denominator by `cos^2 x`]

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

When `x = 0`, then `t = 0` and when `x = pi/2` then `t = oo`

`:. I = int_0^(oo) (dt)/(a^2 + b^2t^2) = 1/b^2 int_0^(oo) (dt)/((a/b)^2 + t^2)`

` = 1/b^2 1/(a/b) [ tan^(-1) ((bt)/a) ]_0^(oo)`

` [ ∵ int (dx)/(a^2 + x^2) = 1/a tan^(-1) (x/a) + C]`

` 1/(ab) [tan^(-1)(oo) - tan^(-1)(0)]`

` 1/(ab) [pi/2 - 0] = pi/(2ab)`
Correct Answer is `=>` (C) ` pi/(2 ab)`
Q 1771701626

Consider the function `f' '(x) = sec^4 x + 4` with `f(0) = 0` and
`f'(0) =0`.

What is `f' (x)` equal to?
NDA Paper 1 2014
(A)

`tan x - (tan^3 x)/3 + 4x`

(B)

`tan x + (tan^3 x)/3 + 4x`

(C)

`tan x + (sec^3 x)/3 + 4x`

(D)

`- tan x - (tan^3 x)/3 + 4x`

Solution:

Clearly, `f ' (x) = int f' '(x) dx + C_1`

`= int (sec^4 x+ 4) dx + C_1`

` = int sec^2 x sec^2 x dx + int 4 dx + C_1`

`= int (1 + tan^2 x) sec^2 xdx + 4x + C_1`

` = I_1 + 4x + C_1`

Put `tan x = t` in the integra `I_1`, then

`sec^2 x dx = dt`

`:. I_1 =int (1+t^ 2) dt = t + (t^3)/3 + C'`

`= tan x + ( tan^3 x)/3 + C'`

`:. f'(x) = tan x + ( tan^3 x)/3 + 4x + C`

Where `C = C_1 + C'`

`∵ f' (x) = 0 => C = 0`

Thus `f'(x) = tan x + ( tan^3 x)/3 + 4x`
Correct Answer is `=>` (B) `tan x + (tan^3 x)/3 + 4x`
Q 1701101928

For the next three solutions consider `I = int_0^pi ( x dx)/(1 +sinx)`

What is `I` equal to?
NDA Paper 1 2014
(A)

`- pi`

(B)

`0`

(C)

`pi`

(D)

`2 pi`

Solution:

Given, `I = int_0^pi ( x dx)/(1 +sinx)` .......(1)

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

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

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

`[∵ sin (pi - x) = sin x]`

On adding Egs. (1) and (2), we get

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

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

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

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

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

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

Put `tan (x/2) + 1 = t`

`=> sec^2 (x/2) . 1/2 dx = dt`

`=> sec^2 (x/2) dx = 2 dt`

When `x = 0` then `t = 1` and when `x = pi/2 `. then `t = 2`

`:. I = 2pi int_1^2 ( dt)/t^2 = -2pi [1/t]_1^2= -2pi [1/2 -1]`

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

According to the explanation, `I = pi`
Correct Answer is `=>` (C) `pi`
Q 1771112026

For the next three solutions consider `I = int_0^pi ( x dx)/(1 +sinx)`

What is ` int_0^pi ((pi -x) dx)/(1 + sin x)` equal to?
NDA Paper 1 2014
(A)

`pi`

(B)

`pi/2`

(C)

`0`

(D)

`2 pi`

Solution:

Let `I _1 = int_0^pi ((pi -x) dx)/(1 + sin x)`

` = int_0^pi ( [ pi - (pi - x )] dx)/( 1 + sin ( pi - x) ) quad [ ∵ int_0^a f(x) dx = int_0^a f(a- x)dx]`

`I _1 = int_0^pi ( x dx)/( 1 + sin x) quad [ ∵ sin ( pi - x) = sin x]`

` = I = pi`
Correct Answer is `=>` (A) `pi`
Q 1711212120

For the next three solutions consider `I = int_0^pi ( x dx)/(1 +sinx)`

What is ` int_0^pi (dx)/(1 + sin x)` equal to?

NDA Paper 1 2014
(A)

`1`

(B)

`2`

(C)

`4`

(D)

`-2`

Solution:

Given, `I = int_0^pi ( x dx)/(1 +sinx)` .......(1)

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

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

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

`[∵ sin (pi - x) = sin x]`

On adding Egs. (1) and (2), we get

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

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

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

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

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

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

Put `tan (x/2) + 1 = t`

`=> sec^2 (x/2) . 1/2 dx = dt`

`=> sec^2 (x/2) dx = 2 dt`

When `x = 0` then `t = 1` and when `x = pi/2 `. then `t = 2`

`:. I = 2pi int_1^2 ( dt)/t^2 = -2pi [1/t]_1^2= -2pi [1/2 -1]`

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

From Eq. (3),

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

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

` => int_0^pi (dx)/(1 + sin x) = 2/pi xx pi =2 quad (∵ I = pi )`
Correct Answer is `=>` (B) `2`
Q 1741412323

Consider the integral `I = int_0^pi ln (sin x) dx`

What is ` int_0^(pi/2) ln (sin x) dx` equal to?
NDA Paper 1 2014
(A)

`4I`

(B)

`2I`

(C)

`I`

(D)

`I/2`

Solution:

Consider ` I = int_0^(pi) ln (sin x) dx`

` I = int_0^(pi) ln (sin x) dx`

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

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

` = 2 int_0^(pi/2) In [ sln ( pi/2 - x) ] dx`

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

` = 2 int_0^(pi/2) ln (cos x) dx ` .................(2)

From Eq. (1),

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

` => int_0^(pi/2) ln (sin x) dx = 1/2 I`
Correct Answer is `=>` (D) `I/2`
Q 1701412328

Consider the integral `I = int_0^pi ln (sin x) dx`

What is ` int_0^(pi/2) ln(cos x) dx` equal to?
NDA Paper 1 2014
(A)

`I/2`

(B)

`I`

(C)

`2I`

(D)

`4I`

Solution:

Consider ` I = int_0^(pi)ln(sin x) dx`

` I = int_0^(pi)ln(sin x) dx`

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

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

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

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

` = 2 int_0^(pi/2)ln(cos x) dx ` .................(2)

From Eq. (2),we have

` I = 2 int_0^(pi/2)ln(cos x) dx `

` => int_0^(pi/2)ln(cos x) dx = 1/2 I`
Correct Answer is `=>` (A) `I/2`
Q 2369180915

What is `int_0^2(dx)/(x^2+4)` equal to ?
NDA Paper 1 2013
(A)

`pi/2`

(B)

`pi/4`

(C)

`pi/8`

(D)

None of these

Solution:

Let `I = int_0^2(dx)/(x^2+4)`

`(because int_0^a (dx)/(x^2+a^2) =1/atan^(-1)(x/a) )`

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

` = 1/2[tan^(-1)(1)-tan^(-1)(0)] = 1/2(pi/4-0) = pi/8`
Correct Answer is `=>` (C) `pi/8`
Q 2319291110

What is `int_(-a)^a (x^3+sinx)dx` equal to ?
NDA Paper 1 2013
(A)

`a`

(B)

`2a`

(C)

`0`

(D)

`1`

Solution:

Let `I = int_(-a)^a (x^3+sinx)dx`

Here `f(x) = x^3+sinx => f(-x) = (-x)^3+sin(-x)`


`= -x^3-sinx = -(x^3+sinx) = -f(x)`


`f(x)` is an odd function

`therefore int_(-a)^a(x^3+sinx)dx = 0`

`[because int_(-a)^a f(x)dx = {2int_0^3f(x)dx` if f(x)is an even function , `0` if f(x) is an odd fuction`}]`
Correct Answer is `=>` (C) `0`
Q 2309291118

What is `int_0^1 xe^xdx` equal to ?
NDA Paper 1 2013
(A)

`1`

(B)

`-1`

(C)

`0`

(D)

`e`

Solution:

Let `I = int_0^1 underset(I)x underset(II)e^x dx`

Using integration by parts,

`I = [x*e^x]_0^1-int_0^1 1*e^xdx`


`=> I = [1*e -0]-[e^x]_0^1 => I = e-(e-1) = e-e+1`

`therefore I = 1`
Correct Answer is `=>` (A) `1`

Set 2

Q 2359691514

What is the value of `int_0^1 (tan^(-1)x)/(1+x^2)dx ?`
NDA Paper 1 2012
(A)

`pi/4`

(B)

`pi/8`

(C)

`pi^2/8`

(D)

`pi^2/32`

Solution:

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

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


`I = int_0^(pi/4) t*dt = [t^2/2]_0^(pi/4)`


` = 1/2 * (pi/4)^2 = pi^2/32`
Correct Answer is `=>` (D) `pi^2/32`
Q 2389691517

What is the value of `int_(-pi/2)^(pi/2) |sinx|dx?`

NDA Paper 1 2012
(A)

`2`

(B)

`1`

(C)

`pi`

(D)

`0`

Solution:

`int_(-pi/2)^(pi/2) |sinx|dx`


` = int_(-pi/2)^0 (-sinx)dx+int_0^(pi/2) (sinx)dx`



` = -[-cosx]_(-pi/2)^0-[cosx]_0^(pi/2)`


` = [cos0-cos(-pi/2)]-[cos(pi/2)-cos0]`


` = (1-0)-(0-1) = 1+1 = 2`
Correct Answer is `=>` (A) `2`
Q 2309091818

If `int_1^2{k^2+(4-4k)x+4x^3}dx le 12` then which one of the following is correct
NDA Paper 1 2011
(A)

`k = 3`

(B)

`0 le k < 3`

(C)

`k le 4`

(D)

`k = 0`

Solution:

Given, `int_1^2{k^2+(4-4k)x+4x^3}dx le 12`


`=> [k^2x+(4-4x)x^2/2+(4x^4)/4]_1^2 le 12`


`=> (2k^2+(2-2k) * 4+16)-(k^2+(2-2k)+1) le 12`


`=> (2k^2+8-8k+16-k^2-2+2k-1) le 12`


`=> k^2-6k+21 le 12`

` => k^2-6k+9 le0`

`=> (k-3)^2 le 0 => (k-3)^2 = 0` `[because (k-3)^2 ne 0]`

`=> k-3 = 0 => k = 3`
Correct Answer is `=>` (A) `k = 3`
Q 2319191919

If `I_n = int_0^(pi/4) tan^n x dx` then What is `I_n+I_(n-2)` equal to ?
NDA Paper 1 2011
(A)

`1/n`

(B)

`1/(n-1)`

(C)

`n/(n-1)`

(D)

`1/(n-2)`

Solution:

`because I_n = int_0^(pi/4) tan^n xdx`


`therefore I_n+I_(n-2) = int_0^(pi/4) tan^nxdx+int_0^(pi/4) tan^(n-2) xdx`


` = int_0^(pi/4) (1+tan^2x)tan^(n-2)xdx`

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

Let `t = tanx => dt = sec^2xdx`

` = int_0^1 t^(n-2)dt`

` = [t^(n-2+1)/(n-2+1)]_0^1 = 1/(n-1)`
Correct Answer is `=>` (B) `1/(n-1)`
Q 2430201112

What is `int_0^pi(dx)/(1+2sin^2x)` equal to ?


NDA Paper 1 2011
(A)

`pi`

(B)

`pi/3`

(C)

`pi/sqrt3`

(D)

`(2pi)/sqrt3`

Solution:

`I = int_0^pi (dx)/(1+2sin^2x) = 2 int_0^(pi/2) (dx)/(1+2sin^2x)` `[because int_0^(2a) f(x)dx = 2int_0^a f(x)dx]`

If `{f(2a-x) = f(x)}`


`I = 2int_0^(pi/2) (dx)/(1+1-cos2x) = 2int_0^(pi/2)(dx)/(2-cos2x)`


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

Let `sqrt3tanx = t => sqrt3sec^2xdx = dt`


`I = 2 int_0^(oo) (dt)/(sqrt3 (1+t^2)) = 2/sqrt3[tan^(-1)t]_0^(oo)`


` = 2/sqrt3[tan^(-1)(oo)-tan^(-1)(0)]`

` = 2/sqrt3(pi/2-0) = (2pi)/(2sqrt3) = pi/sqrt3`
Correct Answer is `=>` (C) `pi/sqrt3`
Q 2440301213

If `f(x)` is an even function then what is `int_0^(pi) f(cosx)dx` equal to ?
NDA Paper 1 2011
(A)

`0`

(B)

`int_0^(pi/2) f(cosx)dx`

(C)

`2int_0^(pi/2) f(cosx)dx`

(D)

`1`

Solution:

Since, f(x)is an even function.

`therefore int_0^(pi/2) f(cosx)dx = 2int_0^(pi/2) f(cosx)dx`


`[(because int_0^(2a)f(x) = {_0 ^2 int_0^af(x) , text(if , f(2a-x) = f(x) , even)) , (0 , text(if , f(2a-x) = -f(x) (odd)))]`
Correct Answer is `=>` (C) `2int_0^(pi/2) f(cosx)dx`
Q 2400301218

What is the value of `int_(pi/6)^(pi/4) (dx)/(sinx cosx)?`
NDA Paper 1 2010
(A)

`2 ln sqrt3`

(B)

`ln sqrt3`

(C)

`2 ln 3`

(D)

`4 ln 3`

Solution:

`int_(pi/6)^(pi/4) (dx)/(sinxcosx) = 2int_(pi/6)^(pi/4)(dx)/(sin2x)`


` = 2int_(pi/6)^(pi/4)cosec2xdx = 2[logtanx]_(pi/6)^(pi/4) * 1/2`


` = [logtanpi/4-logtanpi/4] = 0-log1/sqrt3`

`= logsqrt3`
Correct Answer is `=>` (B) `ln sqrt3`
Q 2410301219

What is the value of `int_1^2e^x(1/x-1/x^2)dx?`
NDA Paper 1 2010
(A)

`e(e/2-1)`

(B)

`e(e-1)`

(C)

`e-1/e`

(D)

`0`

Solution:

`int_1^2e^x(1/x-1/x^2)dx`

` = [e^x*1/x]_1^2+int_1^2 1/x^2e^xdx-int_1^2e^x*1/x^2dx`


` = e^2/2-e = e(e/2-1)`
Correct Answer is `=>` (A) `e(e/2-1)`
Q 2450401314

What is the value of integral `int_(-1)^(1) |x|dx?`
NDA Paper 1 2010
(A)

`1`

(B)

`0`

(C)

`2`

(D)

`-1`

Solution:

`int_(-1)^(1) |x|dx`

` = -int_(-1)^0xdx+int_0^1xdx`


` = -[x^2/2]_(-1)^0+[x^2/2]_0^1 = -[-1/2]+[1/2] = 1`
Correct Answer is `=>` (A) `1`
Q 2410401319

What is the value of `int_0^(pi/2) (sin^3 x)/(sin^3 x+cos^3 x)dx ?`
NDA Paper 1 2010
(A)

`pi`

(B)

`pi/2`

(C)

`pi/4`

(D)

`0`

Solution:

Let `I = int_0^(pi/2) (sin^3 x)/( sin^3 x+cos^3 x) dx` .......(i)

`= int_0^(pi/2) (sin^3 (pi/2-x))/(sin^3(pi/2-x)+cos^3(pi/2-x))dx`


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


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

`2I = int_0^(pi/2)1dx => 2I = [x]_0^(pi/2) = pi/2 => I = pi/4`
Correct Answer is `=>` (C) `pi/4`
Q 2319245119

What is `int (x^4+1)/(x^2+1) dx` equal to?
NDA Paper 1 2010
(A)

`x^3/3-x+4 tan^(-1) x+C`

(B)

`x^3/3+x+4 tan^(-1)+C`

(C)

`x^3/3-x+2 tan^(-1) x+C`

(D)

`x^3/3-x-4 tan^(-1) x+C`

Solution:

`int (x^4+1)/(x^2+1) dx =int ((x^4-1)/(x^2+1)+2/(x^2+1)) dx`

`int (x^2-1+2/(x^2+1))dx =x^3/3-x+2 tan^(-1) x+C`
Correct Answer is `=>` (C) `x^3/3-x+2 tan^(-1) x+C`
Q 2319445310

If `int x^2 ln x dx = x^3/m ln x +x^3/n + C`, then
the values of m and n, respectively?
NDA Paper 1 2010
(A)

`1/3` and `-1/9`

(B)

`3` and `-9`

(C)

`3` and `9`

(D)

`3` and `3`

Solution:

`int x^2 ln x dx=ln x x^3/3-int 1/x * x^3/3 dx`

`=x^3/3 ln x -int x^2/3 dx =x^3/3 ln x-1/3 * x^3/3 +C`

`=x^3/3 ln x-x^3/9+C`

But `int x^2ln x dx=x^3/m ln x+x^3/n +C` (on comparing)

`:. m=3` and `n=-9`
Correct Answer is `=>` (B) `3` and `-9`
Q 2359445314

What is `int 1/(1+e^x) dx` equal to?
NDA Paper 1 2010
(A)

`x-log x +C`

(B)

`x-log (tan x)+C`

(C)

`x-log(1+e^x)+C`

(D)

`log (1+e^x)+C`

Solution:

`int 1/(1+e^x) dx=int (e^(-x)/(e^(-x)+1) )dx`

`=-log(1+e^(-x)+C=-log ((1+e^x)/e^x)+C`

`=-{log(1+e^x)-log e^x}+C=x-log(1+e^x)+C`
Correct Answer is `=>` (C) `x-log(1+e^x)+C`
Q 2410001810

The value of `int_(-2)^(2) (ax^3+bx+c)dx` depends on which of the following?
NDA Paper 1 2008
(A)

Values of x only

(B)

Values of each of a, b and c

(C)

Value of c only

(D)

Value of b only

Solution:

`int_(-2)^(2) (ax^3+bx+c)dx = [(ax^4)/4+(bx^2)/2+cx]_(-2)^2`


` = [(a(16))/4+(b(4))/2+2c]-[(a(16))/4+(b(4))/2-2c]`

` = 4c`


So, the value of given integral depends on only `c.`
Correct Answer is `=>` (C) Value of c only
Q 2410101910

What are the values of `p` which satisfy the equation `int_0^p (3x^2+4x-5)dx = p^3-2?`
NDA Paper 1 2008
(A)

`1/2` and `2`

(B)

`-1/2` and `2`

(C)

`1/2` and `-2`

(D)

`-1/2` and `-2`

Solution:

`int_0^p (3x^2+4x-5)dx = p^3-2`


`=> [(3x^3)/3+(4x^2)/2-5x]_0^p = p^3-2`

`=> p^3+2p^2-5p = p^3-2`

`=> 2p^2-5p+2 = 0`

`=> 2p^2-4p-p+2 = 0`

`=> 2p(p-2)-1(p-2) = 0`

`=> p-2 = 0 , p-1 = 0`

`therefore p = 2 ` and `1/2`
Correct Answer is `=>` (C) `1/2` and `-2`
Q 2480101917

What is the value of `int_0^1 xe^(x^2) dx ?`
NDA Paper 1 2008
(A)

`(e-1)/2`

(B)

`e^2-1`

(C)

`2(e-1)`

(D)

`e-1`

Solution:

let `I = int_0^1 x e^(x^2) dx`


put `x^2 = t`

`=> 2xdx = dt => xdx = dt/2`


`therefore I = 1/2int_0^1e^tdt = 1/2[e^(x^2)]_0^1 = 1/2[e-e^0] = (e-1)/2`
Correct Answer is `=>` (A) `(e-1)/2`
Q 2339156012

What is `int (e^x+1)^(-1) dx` equal to?
NDA Paper 1 2008
(A)

`ln (ex + 1) + C`

(B)

`ln (e^(-x) + 1) + C`

(C)

`-ln (e^(-x) + 1) + C`

(D)

`-(ex + 1) + C`

Solution:

Let `I=int(e^x+1)^(-1) dx=int1/(e^x+1) dx =int (e^(-x))/(1+e^(-x)) dx`

Let `1+e^(-x)=t=> e^(-x) dx=dt`

`:. I=-int 1/t dt =-log t+C=-log (1+e^(-x))+C`
Correct Answer is `=>` (C) `-ln (e^(-x) + 1) + C`
Q 2319156019

What is `int (d theta)/(sin^2 theta +2 cos^theta-1)` equal to?
NDA Paper 1 2008
(A)

`tan theta+c`

(B)

`cot theta+C`

(C)

`1/2 tan theta+C`

(D)

`1/2 cot theta+C`

Solution:

Let `I=int (d theta)/(sin^2theta+2 cos^2 theta-1)`

`=(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`
Correct Answer is `=>` (A) `tan theta+c`
Q 2480112017

What is the value of integral `I = int_0^1 x(1-x)^9dx?`
NDA Paper 1 2007
(A)

`1/110`

(B)

`1/111`

(C)

`1/112`

(D)

`1/119`

Solution:

Let `I = int_0^1x(1-x)^9dx`


put `1-x = t => dx = -dt`

`therefore I = int_1^0(1-t)t^9(-dt)`


` = -int_1^0(-t^(10)+t^9)dt = int_0^1(-t^(10)+t^9)dt`


` = [(-t^(11))/11+t^(10)/10]_0^1 = -1/11+1/10`


`(-10+11)/110 = 1/110`
Correct Answer is `=>` (A) `1/110`
Q 2400312218

`int_0^(pi/2) cos^8 xdx ` is equal to ?


NDA Paper 1 2007
(A)

`(35 pi)/256`

(B)

`70/256`

(C)

`16/35`

(D)

`(8pi)/35`

Solution:

`I = int_0^(pi/2) sin^0 x * cos^8 xdx `

`(because int_0^(pi/2) sin^m x * cos^n xdx = (r((m+1)/2)r((n+1)/2))/(2sqrt((m+n+2)/2)))`



`I = (r((0+1)/2)r((8+1)/2))/(2r((0+8+2)/2))`


` = (r(1/2) r (9/2))/(2r 5) ` `(becausern = nrn)`


` = sqrtpi * (7/2 * 5/2 * 3/2 * 1/2 * sqrtpi )/(2 * 4 * 3* 2 * 1) = (7 * 5)/(16 * 16) pi = (35pi)/(256) ` `[because sqrt(1/2) = sqrtpi]`
Correct Answer is `=>` (A) `(35 pi)/256`
Q 2430712612

If `int_(ln2)^x (e^x-1)^(-1)dx = ln3/2` then what is the value of x?
NDA Paper 1 2007
(A)

`e^2`

(B)

`1/e`

(C)

`ln4`

(D)

`1`

Solution:

Given that `int_(ln2)^x 1/(e^x-1)dx = ln3/2`


Let `I = int_(ln2)^x 1/(e^x-1)dx`


put `e^x-1 = t` then `dx = (dt)/(1+t)`


`therefore I = int_(ln2)^x 1/(1+t)dt = int_(ln2)^x [1/t-1/(1+t)]dt`


` = [lnt-ln(1+t)]_(ln2)^x`

` = [ln(e^x-1)-lne^x]_(ln2)^x`


`= [ln((e^x-1)/e^x)]_(ln2)^x = [ln((e^x-1)/e^x)-ln((2-1)/2)]`

` = ln((e^x-1)/e^x)-ln1/2 = ln2((e^x-1)/e^x)`


But `I = ln3/2`

`therefore 2((e^x-1)/e^x) = 3/2`

`=> 4e^x-4 = 3e^x`

`=> e^x = 4`

`therefore x = ln4`
Correct Answer is `=>` (C) `ln4`
Q 2410012810

If `int_(-3)^(2) f(x)dx = 7/3` and `int_(-3)^9f(x)dx = -5/6` then what is the value of `int_2^9f(x)dx?`
NDA Paper 1 2007
(A)

`-19/6`

(B)

`19/6`

(C)

`3/2`

(D)

`-3/2`

Solution:

We know that

`int_(-3)^9f(x)dx = int_(-3)^2f(x)dx+int_2^9f(x)dx` (by property) ... (i)

given `int_(-3)^(2) f(x)dx = 7/3` and `int_(-3)^9f(x)dx = -5/6`

`[becauseint_a^bf(x)dx = int_a^cf(x)dx+int_c^bf(x)` where `alecleb]`

from eq(i)


`-5/6 = 7/3+int_2^9f(x)dx`

`int_2^9f(x)dx = -5/6-7/3 = (-5-14)/6 = -19/6`
Correct Answer is `=>` (A) `-19/6`
Q 2420112911


NDA Paper 1 2007

Assertion : (A) `int_0^(pi) sin^7 xdx = 2int_0^(pi/2) sin^7 xdx`

Reason : (R) `sin^7 x` is odd function

(A) Both A and R individually true and R is the correct explanation of A
(B) Both A and R are individually true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Solution:

`because f(pi-x) = sin^7 (pi-x)`

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

`therefore int_0^(pi) sin^7 xdx = int_0^((2pi)/2) sin^7 x dx`


` = 2int_0^(pi/2) sin^7xdx`


Also `sin^7 x` is an odd function `[because f(-x) = f(x)]`

Hence, both A and Rare individually true but R is not the correct
explanation of A.
Correct Answer is `=>` (B)
 
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