Mathematics Tricks & Tips OF INDEFINITE INTEGRAL FOR NDA
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Topic Covered in This Lecture

In this lecture we will cover following topics :


1. Solving indefinite integration problem by differentiating the options intelligently

2. Solving by simplification of integrand

3. Integrals by application of mind and methods of integration

4. When some values are to be found

5. Straight formulaes based questions

6. Integration by substitution

7. Integration by parts

8. No need to do following type of integration
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 2329723611

What is `int (x cos x+ sin x) dx` equal to?
NDA Paper 1 2013
(A)

`x sin x+C`

(B)

`x cos x+C`

(C)

`-x sin x+C`

(D)

`-x cos x+C`

Solution:

Let `I=int (x cos x+sin x) dx=int cos x dx+ int sin x dx`

`=x sin x- int sin x dx+ int sin x dx`

`= x sin x +C`
Correct Answer is `=>` (A) `x sin x+C`
Q 2379723616

What is `int (dx)/(x ln x)` equal to?


NDA Paper 1 2013
(A)

`ln (ln x)+C`

(B)

`ln x +C`

(C)

`(ln x)^2+C`

(D)

None of these

Solution:

Let `I=int (dx)/(x log x)`

Let `{ tt((t=logx),(dt=1/x dx))=> I=int 1/t * dt=log|t|+C`

`:. I=log (log x)+C`
Correct Answer is `=>` (A) `ln (ln x)+C`
Q 2329323211

`5` books are to be chosen from a lot of `10` books. If
m is the number of ways of choice when one
specified book is always included and n is the
number of ways of choice when a specified book
is always excluded, then which one of the
following is correct?
NDA Paper 1 2011
(A)

m > n

(B)

m = n

(C)

m = n -1

(D)

m = n - 2

Solution:

Number of ways when one specified book is included

`= text()^9C_4 = m => m = 126`

and number of ways when one specified book is excluded

`= text()^9C_5 = n`

`=> n = 126 => m = n`
Correct Answer is `=>` (B) m = n
Q 2379134016

What is the value of `int (x^2+1)^(5//2) x dx`?
NDA Paper 1 2012
(A)

`(x^2+1)^(7//2)+C`

(B)

`2/7 (x^2+1)^(7//2)+C`

(C)

`1/7 (x^2+1)^(7//2)+C`

(D)

None of these

Solution:

Let `I=int (x^2+1)^(5//2) x dx` (Let `t=x^2+1)=> dt=2xdx`)

`:. I=int t^(5//2) *(dt)/2 =1/2 [(t^(7//2))/(7//2)]+C=1/7(x^2+1)^(7//2)+C`
Correct Answer is `=>` (C) `1/7 (x^2+1)^(7//2)+C`
Q 2359534414

What is `int (dx)/(sin^2x cos^2 x)` equal to ?
NDA Paper 1 2011
(A)

`tan x+cot x+C`

(B)

`tan x-cot x+C`

(C)

`(tan x+cot x)^2+C`

(D)

`(tan x-cot x)^2+C`

Solution:

Given, `int (dx)/(sin^2 x * cos^2x)=4 int (dx)/((2 sin * cos x)^2)`

`=4 int (dx)/((sin^2x)^2) =4 int cosec^2 2x dx`

`=- 4 (cot 2x)/2 +c=-2 cot 2x+C`

`=-(2 cos 2x)/(sin 2x)=(-2(cos^2x-sin^2x))/(2 sin x * cos x)+C`

`=-(cot x -tan x)+C=(tan x-cot x)+C`
Correct Answer is `=>` (B) `tan x-cot x+C`
Q 2339634512

What is `int e^x(sqrt x+1/(2 sqrt x)) dx` equal to?
NDA Paper 1 2011
(A)

`xe^x+C`

(B)

`e^x(sqrt x)+C`

(C)

`2e^x(sqrt x)+C`

(D)

`2xe^x+C`

Solution:

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

`=int e^x * sqrt x dx+int e^x * 1/(2 sqrt x) dx`

`=e^x * sqrt x-int e^x * 1/(2 sqrt x) dx + int e^x 1/(2 sqrt x)=e^x * sqrt x +C`
Correct Answer is `=>` (B) `e^x(sqrt x)+C`
Q 2319634519

What is `int (sin sqrt x)/(sqrt x)dx` equal to?
NDA Paper 1 2011
(A)

`(cos sqrt x)/2 +C`

(B)

`2 cos sqrt x+C`

(C)

`-(cos sqrt x)/2 +C`

(D)

`-2 cos sqrt x+C`

Solution:

Let `I=int (sin sqrt x)/(sqrt x) dx`

Put `sqrt x=I=> 1/(2 sqrt x) dx=dt => 1/(sqrt x) dx =2 dt`

`:. I=2 int sin t dt =-2 cos t +C=-2 cos sqrt x+C`
Correct Answer is `=>` (D) `-2 cos sqrt x+C`
Q 2369034815

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

`2e^(sqrt x) (x -2sqrt x + 2)+ C`

(B)

`2e^(sqrt x) (x + 2sqrt x+ 2) + C`

(C)

`2e^(sqrt x) (x + 2 sqrt x -2)+ C`

(D)

`2e ^(sqrt x) (x- 2sqrt x- 2) + C`

Solution:

Let `I=int sqrt x e^(sqrt x) dx`

Put `sqrt x=t=> 1/(2 sqrt x) dx =dt => dx =2t dt`

`:. I=int t e^t 2t dt=2 int t^2 e^t dt =2 (t^2 e^t-int 2t e^t dt)`

`= 2 [t^ 2e^t- 2(te^t-int e^t* dt)]`

`= 2 (t^2 e^t -2te^t + 2e^t) + C`

`=2(xe^(sqrt x) -2sqrt x e^(sqrt x)+ 2e^(sqrt x))+ C`

`=2e^(sqrt x) (x- 2sqrt x + 2)+ C`
Correct Answer is `=>` (A) `2e^(sqrt x) (x -2sqrt x + 2)+ C`
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 2309445318

What is `int(a+b sin x)/(cos^2x)dx` equal to?
NDA Paper 1 2009
(A)

`a sec x + b tan x + C`

(B)

`a tan x + b sec x+C`

(C)

`a cot x + b cosec x + C`

(D)

`a cosec x + b cot x+C`

Solution:

`int (a+b sin x)/(cos^2 x) dx =int(a sec^2 x+b tan x sec x) dx`

`=a tan x+b sec x+C`
Correct Answer is `=>` (B) `a tan x + b sec x+C`
Q 2319545419

What is `int (log x)/((1+log x)^2) dx` equal to
NDA Paper 1 2009
(A)

`1/((1+log x)^3)+C`

(B)

`1/((1+log x)^2)+C`

(C)

`x/((1+log x))+C`

(D)

`x/((1+log x)^2)+C`

Solution:

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

Put `log x=t=> 1/x dx=dt` and `x=e^t`

`:. I=int (e^t *t)/((1+t)) dt-int(e^t)/((1+t)^2)dt`

`=e^t/(1+t)-int e^t 1/((1+t)^2) dt - int e^t/((1+t)^2) dt =x/(1+log x)+C`
Correct Answer is `=>` (C) `x/((1+log x))+C`

Solving Indefinite Integration Problem By Differentiating The Options Intelligently

Q 2773591446

What is `int (( x^(e-1) + e^(x-1) )dx)/(x^e +e^x)`
NDA Paper 1 2017
(A)

`x^2/2+c`

(B)

`ln (x+e) +c`

(C)

`ln (x^e+e^x)+c`

(D)

`1/e ln (x^e+e^x) +c`

Solution:

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

Let `t= x^e + e^x`

`dt = (ex^(e-1) + e^x ) dx`

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

`=> I= int 1/(te) dt => 1/e ln (x^e+ e^x)`
Correct Answer is `=>` (D) `1/e ln (x^e+e^x) +c`
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 2231380222

` int (dx)/(1 + e^(-x))` is equal to

where, `C` is the constant of integration.
NDA Paper 1 2015
(A)

`1 +e^x + C`

(B)

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

(C)

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

(D)

` 2ln (1 + e^(-x)) + C`

Solution:

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

Put `t = 1 + e^x`

`=> dt = e^x dx`

Now ` I = int (dt)/t = ln (t) _C =ln (1+e^x) + C quad [∵ t = 1 + e^x]`
Correct Answer is `=>` (C) `ln(1 + e^(x)) + C`
Q 1649134913

What is `int ( xe^x dx)/(x+1)^2 ` equal to?

where, `C` is the constant of integration.
NDA Paper 1 2015
(A)

`(x + 1)^2 e^x + C`

(B)

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

(C)

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

(D)

` e^x/(x+1)^2 + C`

Solution:

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

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

` => e^x ( 1/( x + 1) ) + C quad ( :. int e^x (f(x) + f'(x))dx = e^x f(x) +C)`
Correct Answer is `=>` (C) ` e^x/(x+1) + C`
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 2329723611

What is `int (x cos x+ sin x) dx` equal to?
NDA Paper 1 2013
(A)

`x sin x+C`

(B)

`x cos x+C`

(C)

`-x sin x+C`

(D)

`-x cos x+C`

Solution:

Let `I=int (x cos x+sin x) dx=int cos x dx+ int sin x dx`

`=x sin x- int sin x dx+ int sin x dx`

`= x sin x +C`
Correct Answer is `=>` (A) `x sin x+C`
Q 2379723616

What is `int (dx)/(x ln x)` equal to?


NDA Paper 1 2013
(A)

`ln (ln x)+C`

(B)

`ln x +C`

(C)

`(ln x)^2+C`

(D)

None of these

Solution:

Let `I=int (dx)/(x log x)`

Let `{ tt((t=logx),(dt=1/x dx))=> I=int 1/t * dt=log|t|+C`

`:. I=log (log x)+C`
Correct Answer is `=>` (A) `ln (ln x)+C`
Q 2329323211

`5` books are to be chosen from a lot of `10` books. If
m is the number of ways of choice when one
specified book is always included and n is the
number of ways of choice when a specified book
is always excluded, then which one of the
following is correct?
NDA Paper 1 2011
(A)

m > n

(B)

m = n

(C)

m = n -1

(D)

m = n - 2

Solution:

Number of ways when one specified book is included

`= text()^9C_4 = m => m = 126`

and number of ways when one specified book is excluded

`= text()^9C_5 = n`

`=> n = 126 => m = n`
Correct Answer is `=>` (B) m = n
Q 2379134016

What is the value of `int (x^2+1)^(5//2) x dx`?
NDA Paper 1 2012
(A)

`(x^2+1)^(7//2)+C`

(B)

`2/7 (x^2+1)^(7//2)+C`

(C)

`1/7 (x^2+1)^(7//2)+C`

(D)

None of these

Solution:

Let `I=int (x^2+1)^(5//2) x dx` (Let `t=x^2+1)=> dt=2xdx`)

`:. I=int t^(5//2) *(dt)/2 =1/2 [(t^(7//2))/(7//2)]+C=1/7(x^2+1)^(7//2)+C`
Correct Answer is `=>` (C) `1/7 (x^2+1)^(7//2)+C`
Q 2359534414

What is `int (dx)/(sin^2x cos^2 x)` equal to ?
NDA Paper 1 2011
(A)

`tan x+cot x+C`

(B)

`tan x-cot x+C`

(C)

`(tan x+cot x)^2+C`

(D)

`(tan x-cot x)^2+C`

Solution:

Given, `int (dx)/(sin^2 x * cos^2x)=4 int (dx)/((2 sin * cos x)^2)`

`=4 int (dx)/((sin^2x)^2) =4 int cosec^2 2x dx`

`=- 4 (cot 2x)/2 +c=-2 cot 2x+C`

`=-(2 cos 2x)/(sin 2x)=(-2(cos^2x-sin^2x))/(2 sin x * cos x)+C`

`=-(cot x -tan x)+C=(tan x-cot x)+C`
Correct Answer is `=>` (B) `tan x-cot x+C`
Q 2339634512

What is `int e^x(sqrt x+1/(2 sqrt x)) dx` equal to?
NDA Paper 1 2011
(A)

`xe^x+C`

(B)

`e^x(sqrt x)+C`

(C)

`2e^x(sqrt x)+C`

(D)

`2xe^x+C`

Solution:

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

`=int e^x * sqrt x dx+int e^x * 1/(2 sqrt x) dx`

`=e^x * sqrt x-int e^x * 1/(2 sqrt x) dx + int e^x 1/(2 sqrt x)=e^x * sqrt x +C`
Correct Answer is `=>` (B) `e^x(sqrt x)+C`
Q 2319634519

What is `int (sin sqrt x)/(sqrt x)dx` equal to?
NDA Paper 1 2011
(A)

`(cos sqrt x)/2 +C`

(B)

`2 cos sqrt x+C`

(C)

`-(cos sqrt x)/2 +C`

(D)

`-2 cos sqrt x+C`

Solution:

Let `I=int (sin sqrt x)/(sqrt x) dx`

Put `sqrt x=I=> 1/(2 sqrt x) dx=dt => 1/(sqrt x) dx =2 dt`

`:. I=2 int sin t dt =-2 cos t +C=-2 cos sqrt x+C`
Correct Answer is `=>` (D) `-2 cos sqrt x+C`
Q 2369034815

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

`2e^(sqrt x) (x -2sqrt x + 2)+ C`

(B)

`2e^(sqrt x) (x + 2sqrt x+ 2) + C`

(C)

`2e^(sqrt x) (x + 2 sqrt x -2)+ C`

(D)

`2e ^(sqrt x) (x- 2sqrt x- 2) + C`

Solution:

Let `I=int sqrt x e^(sqrt x) dx`

Put `sqrt x=t=> 1/(2 sqrt x) dx =dt => dx =2t dt`

`:. I=int t e^t 2t dt=2 int t^2 e^t dt =2 (t^2 e^t-int 2t e^t dt)`

`= 2 [t^ 2e^t- 2(te^t-int e^t* dt)]`

`= 2 (t^2 e^t -2te^t + 2e^t) + C`

`=2(xe^(sqrt x) -2sqrt x e^(sqrt x)+ 2e^(sqrt x))+ C`

`=2e^(sqrt x) (x- 2sqrt x + 2)+ C`
Correct Answer is `=>` (A) `2e^(sqrt x) (x -2sqrt x + 2)+ C`
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 2309445318

What is `int(a+b sin x)/(cos^2x)dx` equal to?
NDA Paper 1 2009
(A)

`a sec x + b tan x + C`

(B)

`a tan x + b sec x+C`

(C)

`a cot x + b cosec x + C`

(D)

`a cosec x + b cot x+C`

Solution:

`int (a+b sin x)/(cos^2 x) dx =int(a sec^2 x+b tan x sec x) dx`

`=a tan x+b sec x+C`
Correct Answer is `=>` (B) `a tan x + b sec x+C`
Q 2319545419

What is `int (log x)/((1+log x)^2) dx` equal to
NDA Paper 1 2009
(A)

`1/((1+log x)^3)+C`

(B)

`1/((1+log x)^2)+C`

(C)

`x/((1+log x))+C`

(D)

`x/((1+log x)^2)+C`

Solution:

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

Put `log x=t=> 1/x dx=dt` and `x=e^t`

`:. I=int (e^t *t)/((1+t)) dt-int(e^t)/((1+t)^2)dt`

`=e^t/(1+t)-int e^t 1/((1+t)^2) dt - int e^t/((1+t)^2) dt =x/(1+log x)+C`
Correct Answer is `=>` (C) `x/((1+log x))+C`

Solving By Simplification Of Integrand

`e^(ln x) = x , a^(lnx)= a^((log_a x)/(log_a e))`

`= a (x) ^(1/(ln _a e))`
Q 2379523416

What is `int sin^2 xdx + int cos^2 xdx` equal to?

NDA Paper 1 2013
(A)

`x+C`

(B)

`x^2/2 +C`

(C)

`x^2+C`

(D)

None of these

Solution:

Let `I=int sin^2 x dx + int cos^2 x dx`

`=int (sin^2 x+ cos^x)dx`

`=int 1 * dx=x+C` `(sin^2 theta +cos^2 theta=1)`
Correct Answer is `=>` (A) `x+C`
Q 2379823716

What is `int e^(ln x) dx` equal to?
NDA Paper 1 2013
(A)

`x e^(|ln x|)+C`

(B)

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

(C)

`x+C`

(D)

`x^2/2+C`

Solution:

Let `I=int e^(log x) dx` (by logarithm proptrty, `e^(loga) =a`)

`:. I=int x dx=[x^2/2]+C`
Correct Answer is `=>` (D) `x^2/2+C`
Q 2369234115

Consider the following statements

I. `int log 10 dx= x + C`

II. `int 10^x dx=10x+C`

Which of the above statements is/are correct?
NDA Paper 1 2012
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. `int 10 dx =int 1 * dx =x+C`

II. `int 10^x dx =(10^x)/(log_e 10)+C`
Correct Answer is `=>` (A) Only I
Q 2359734614

What is `int sin^(-1)(cos x) dx` equal to?
NDA Paper 1 2011
(A)

`(x pi)/2-x^2/2+C`

(B)

`pi/2+x^2/2+C`

(C)

`-(x pi)/2-x^2/2+C`

(D)

`pi/2-x^2/2+C`

Solution:

Let `I=int sin^(-1)(cos x) dx`

`=int sin^(-1)[sin ( pi/2-x)]dx=int(pi/2-x) dx=(pi x)/2-x^2/2+C`
Correct Answer is `=>` (A) `(x pi)/2-x^2/2+C`
Q 2319145019

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

`e^(ln x) (sin x-cos x)+C`

(B)

`(sin x-x cos x)+C`

(C)

`(x sin x+cos x)+C`

(D)

`(sin x+x cos x)-C`

Solution:

`int e^(lnx) sin x dx=int x sin x dx` (`e^(log a)=a`)

`=- x cos x + int 1 * cos xdx = sin x - x cos x + C`
Correct Answer is `=>` (B) `(sin x-x cos x)+C`
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 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`

Integrals By Application Of Mind and Methods Of Integration

When Some Values are to be Found

Q 1711312220

Consider
`int x tan^(-1) x dx = A(x^2 + 1) tan^(-1) x + Bx + C`
where, `C` is the constant of integration.

What is the value of `A?`
NDA Paper 1 2014
(A)

`1`

(B)

`1/2`

(C)

`-1/2`

(D)

`1/4`

Solution:

Given, `int x tan^(-1) x dx = A(x^2 + 1)tan^(-1) x + Bx + C`

where, `C` is the constant of Integration

Consider, ` int x tan^(-1) x dx`

` = tan^(-1) x . x^2/2 - int d/(dx) (tan^(-1) x) . x^2/2 dx`

(using integration by parts)

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

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

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

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

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

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

Clearly, `A = 2`, hence option `(b)` is correct.
Correct Answer is `=>` (B) `1/2`
Q 1751312224

Consider
`int x tan^(-1) x dx = A(x^2 + 1) tan^(-1) x + Bx + C`
where, `C` is the constant of integration.

What is the value of `B?`
NDA Paper 1 2014
(A)

`1`

(B)

`1/2`

(C)

`- 1/2`

(D)

`1/4`

Solution:

Given, `int x tan^(-1) x dx = A(x^2 + 1)tan^(-1) x + Bx + C`

where, `C` is the constant of Integration

Consider, ` int x tan^(-1) x dx`

` = tan^(-1) x . x^2/2 - int d/(dx) (tan^(-1) x) . x^2/2 dx`

(using integration by parts)

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

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

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

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

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

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

Clearly, `B = - 1/2`, hence option `(c)` is correct.
Correct Answer is `=>` (C) `- 1/2`
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`

Straight Formulaes Based Questions

Q 2329523411

What is `int (dx)/(sqrt(4+x^2))` equal to?

where, `C` is an arbitrary constant
NDA Paper 1 2013
(A)

`log | sqrt(4+x^2)+x|+C`

(B)

`log | sqrt(4+x^2)-x|+C`

(C)

`sin^(-1)(x/2)+C`

(D)

None of the above

Solution:

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

`=log | {x+sqrt(x^2+4)}|+C`
Correct Answer is `=>` (A) `log | sqrt(4+x^2)+x|+C`
Q 2329045811

What is `int sec x ^o dx` equal to?
NDA Paper 1 2009
(A)

`log (sec x^o +tan x ^o ) + C`

(B)

`(180^o log tan (pi/4+(pi x)/(360^o)))/(180^o) +C`

(C)

`(180^o log tan (pi/4+( x)/(2)))/pi +C`

(D)

`(180^o log tan (pi/4+(pi x)/(360^o)))/pi +C`

Solution:

Let `I=int sec x^o dx`

`=int sec(pi x)/(180^o) dx`

Put `(pi x)/(180^o)=t=>dx=(180^o)/pi dt`

`:. I=int sec t dt * (180^o)/pi`

`=(180^o)/pi log tan (pi/4+t/2)+C`

`=(180^o)/pi log (pi/4+(pi x)/(360^o))+C`
Correct Answer is `=>` (D) `(180^o log tan (pi/4+(pi x)/(360^o)))/pi +C`

Integration By Substitution

Q 2339623512

What is `int e^(e^x) e^x dx` equal to?
NDA Paper 1 2013
(A)

`e^(e^x)+C`

(B)

`2e^(e^x)+C`

(C)

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

(D)

`2e^(e^x) e^x+C`

Solution:

Let `I=int e^(e^x) * e^x dx` (put `t=e^x=> dt=e^x dx`)

`= int e^t * dt =e^t+C=e^(e^x)+C`
Correct Answer is `=>` (A) `e^(e^x)+C`
Q 2359134914

What is `int sec^n x tan x dx` equal to?
NDA Paper 1 2010
(A)

`(sec^n x)/n+C`

(B)

`(sec^(n-1))/(n-1)+C`

(C)

`(tan^n x)/n+C`

(D)

`(tan^(n-1) x)/(n-1)+C`

Solution:

Let `I= int sec^n x tan x dx`

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

`:. I = int t^( n-1) dt = t^n/n + C = (sec^n x)/n + C`
Correct Answer is `=>` (A) `(sec^n x)/n+C`
Q 2379145016

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

`xe^x+C`

(B)

`cos (xe)^x+C`

(C)

`tan (xe^x)+C`

(D)

`x cosec(xe^x)+C`

Solution:

Let `I=int (e^x(1+x))/(cos^2(x e^x)) dx`

Put `xe^x=t=> e^x (1+x)dx=dt`

`:. I=int sec^2t dt=tan t+C=tan (xe^x)+C`
Correct Answer is `=>` (C) `tan (xe^x)+C`
Q 2379845716

What is `int tan^2 x sec^4 x dx` equal to?
NDA Paper 1 2009
(A)

`(sec^5 x)/5+(sec^3 x)/3+C`

(B)

`(tan^5x)/5+(tan^3 x)/3+C`

(C)

`(tan^5 x)/5+(sec^3 x)/3+C`

(D)

`(tan^5 x)/5+(sec^3 x)/3-C`

Solution:

Let `I=int tan^2 x sec ^4 x dx`

`=int tan^2 x(1+tan^2 x)sec^x dx`

Let `tan x=t` and `sec^2 x dt=int (t^2+t^4)dt`

`=t^5/5+t^3/3+C=(tan^5 x)/5+(tan^3 x)/3+C`
Correct Answer is `=>` (B) `(tan^5x)/5+(tan^3 x)/3+C`

Integration By Parts

Q 2329756611


Normals at P, Q, Rare drawn to y 2 = 4x which intersect at (3, 0). Then








Column IColumn II
(A) Area of `DeltaPQR` (P) 2
(B) Radius of circumcircle of `DeltaPQR` (Q) `5/2`
(C) Centroid of `DeltaPQR` (R) `(5/2, 0)`
(D) Circumcentre of `DeltaPQR` (S) `(2/3, 0)`

JEE 2006
(A)

`(A)->(P); (B)->(Q); (C)->(S); (D)->(R)`

(B)

`(A)-> (Q); (B)->(Q); (C)->(P); (D)->(R)`

(C)

`(A)->(R); (B)->(Q); (C)->(S); (D)->(P)`

(D)

`(A)->(P); (B)->(S); (C)->(Q); (D)->(R)`

Solution:

It is given that

`y^2 = 4x`

The equation of normal is

`y = mx -2 m - m^3` ................................(1)

which is passing through the point (3, 0). Therefore, Eq. (1) becomes

`0 =3m-· 2m- m^3`

`m^3 - m = 0 => m = 0, -1, 1`

That is, `(m_1^2- 2m_1 ), (m_2^2 -2m_2), (m_3^2, -2m_3)` ; hence, we have the points `P(0, 0), Q(l, 2)` and `R(1, ··2)` as shown in the following

Figure as shown :

(A) The area of triangle PQR is

`1/2 xx 1 xx 4 =2`

Therefore, `(A)-> (P)`.

(B) The radius of circumcircle is

`R = (abc) /(4 xx area of Delta PQR) = (4 sqrt5 sqrt5 )/(4 xx 2) = 5/2`

Therefore. `(B)->(Q)`.

(C) The centroid of `Delta PQR` is

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

Therefore, `(C)-> (S)`.

(D) The circumradius of `Delta PQR` is 5/2 and the circumcentre of `Delta PQR` is `(5/2 ,0)`

Therefore, `(D)->(R)`.
Correct Answer is `=>` (A) `(A)->(P); (B)->(Q); (C)->(S); (D)->(R)`

No Need to Do Following Type Of Integration


 
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