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RD Sharma Class 12 Exercise 18.28 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.28 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 10:17 AM IST

Class 12 RD Sharma chapter 18 exercise 18.28 solution is an accomplishment for each understudy. In any case, students can't get their optimal results without thorough practice. Therefore, the one book that has been recommended to students for preparing and alluding to for RD Sharma class 12 solutions chapter 18 exercise 18.28 arrangement. RD Sharma solutions To have phenomenal practice, you need solicitations, and RD Sharma Class 12 Solutions Indefinite Integrals Ex 18.28 meets a splendid fundamental of maths questions. RD Sharma Maths scrutinizing material is one of the most astounding course books for class 12 maths unequivocally for students planning for genuine tests.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.28


Indefinite Integrals Excercise 18.28 Question 1

Answer:-
12(x−1)3+2x−x2+2sin−1⁡(x−12)+c
Hint:-
Adding and subtracting with 1.
Given:-
∫3+2x−x2dx
Solution:-
=∫4+2x−x2−1dx=∫(2)2−(x2−2x+1)dx=∫(2)2−(x−1)2dx
=12(x−1)3+2x−x2+12×(2)2sin−1⁡(x−12)+c=12(x−1)3+2x−x2+2sin−1⁡(x−12)+c

Using the formula

∫a2+x2dx=12xa2+x2+12a2sin−1⁡(x2)+c

Indefinite Integrals Excercise 18.28 Question 2

Answer:-
(2x+14)x2+x+1+38log⁡|(2x+1)+x2+x+1|+c
Hint:-
∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c
Given:-
∫x2+x+1dx
Solution:-
∫x2+x+1dx =∫x2+x+14+34dx=∫x2+x+(15)2+(32)2dx
=∫(x+15)2+(32)2dx=(x+15)2(x+15)2+(32)2+(32)22log⁡|(x+15)+(x+15)2+(32)2|+c

Using the formula

∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c=(2x+14)x2+x+1+38log⁡|(x+15)+(x+15)2+(32)2|+c
=(2x+14)x2+x+1+38log⁡|(2x+12)+12x2+x+1|+c=(2x+14)x2+x+1+38log⁡|(2x+1)+x2+x+1|+c

Indefinite Integrals Excercise 18.28 Question 3

Answer:-
12(2x−1)x−x2+18sin−1⁡(2x−1)+c
Hint:-
Add and subtract 14
Given:-
∫x−x2dx
Solution:-
∫x−x2dx=∫14−14+x−x2dx

=∫(12)2−(12−x)2dx=(1−2x4)(12)2−(12−x)2−(12)22sin−1⁡(1−2x4)12+c
Using the formula
∫a2−x2=12xa2−x2+a22sin−1⁡xa+c=(2x−14)x−x2+18sin−1⁡(2x−1)+c






Indefinite Integrals Excercise 18.28 Question 4

Answer:-
18(4x−1)1+x−2x2+9232sin−1⁡(2x−1)+c

Multiplying by 2.

Given:-
∫1+x−2x2dx
Solution:-
∫1+x−2x2dx
By multiplying by 2
=2∫12+x2−x2dx=2∫916−(116−x2+x2)dx=2∫(34)2−(x−14)2dx

=2{((x−14)2)12+x2−x2+932sin−1⁡(x−143/4)2}+c=18(4x−1)1+x−2x2+9232sin−1⁡(4x−13)+c
Using the formula
∫a2−x2=12xa2−x2+a22sin−1⁡xa+c


Indefinite Integrals Exercise 18.28 Question 5

Answer:-
12sin⁡x4−sin2⁡x+2sin−1⁡sin⁡x2+c
Hint:-
Taking sin⁡x=t
Given:-
∫cos⁡x4−sin2⁡xdx
Solution:-
By taking sin⁡x=t
cos⁡xdx=dt
Hence
=∫4−t2dt=∫(2)2−(t)2dt=12t4−t2+12(2)2sin−1⁡t2+c

Using the formula
∫a2−x2=12xa2−x2+a22sin−1⁡xa+c=12sin⁡x4−sin2⁡x+2sin−1⁡sin⁡x2+c




Indefinite Integrals Exercise 18.28 Question 6

Answer:-
12exe2x+1+12log⁡{ex+e2x+1}+c
Hint:-
Let ex=z
Given:-
∫exe2x+1dx
Solution:-
By taking ex=z
exdx=zdx
Hence
=∫z2+1dz=∫(z)2+(1)2dz=12zz2+1+12(1)2log⁡{z+z2+1}+c

Using the formula
∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c=12exe2x+1+12log⁡{ex+e2x+1}+c


Indefinite Integrals Exercise 18.28 Question 7

Answer:-
12x9−x2+92sin−1⁡x3+c
Hint:-
Using the formula
∫a2−x2=12xa2−x2+a22sin−1⁡xa+c
Given:-
∫9−x2dx
Solution:-
By using the formula
=∫(3)2−(x)2dx=12x9−x2+12(3)2sin−1⁡x3+c
Using the formula
∫a2−x2=12xa2−x2+a22sin−1⁡xa+c=12x9−x2+92sin−1⁡x3+c

Indefinite Integrals Exercise 18.28 Question 8

Answer:-
2xx2+2516+258log⁡[x+x2+2516]+c
Hint:-
Taking common 4 and then use the formula.
Given:-
∫16x2+25dx
Solution:-
=4∫x2+2516dx=4∫(x)2+(54)2dx=4[12x(x)2+(54)2+12×2516log⁡[x+x2+2516]+c]

Using the formula
∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c=2xx2+2516+258log⁡[x+x2+2516]+c




Indefinite Integrals Exercise 18.28 Question 9

Answer:-
2xx2−54+52log⁡|x+x2−54|+c
Hint:-
Taking common 2 and then use the formula.
Given:-
∫4x2−5dx
Solution:-
=4∫x2−(52)2dx=4[12x(x)2−54−12×54log⁡[x+x2−54]+c]

By using the formula
∫x2−a2dx=x2x2−a2−a22log⁡|x+x2−a2|+c=2xx2−54+52log⁡|x+x2−54|+c




Indefinite Integrals Exercise 18.28 Question 10

Answer:-
4x+382x2+3x+4+23232log⁡|x+34x2−32+2|+c
Hint:-
Multiplying by 22
Given:-
∫2x2+3x+4dx
Solution:-
=122∫16x2+24x+32dx=122∫(4x)2+2.4x⋅3+(3)2+23dx=122∫(4x+3)2+232dx

By using the formula

∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c=122[12×14×(4x+3)(4x+3)2+2322×14log⁡[(4x+3)+(4x+3)2+23]+c]
=4x+382x2+3x+4+23232log⁡|(4x+3)+2x2+3x+4|+c=4x+382x2+3x+4+23232log⁡|(x+34)x2−32+2|+23232log⁡4+c=4x+382x2+3x+4+23232log⁡|x+34x2−32+2|+c

Indefinite Integrals Exercise 18.28 Question 11

Answer:-
The answer is
14(2x+1)3−2x−2x2+728sin−1⁡2x+17+c
Hints:-
∫a2−x2dx=x2a2−x2+a22sin−1⁡xa+c∫x2−a2dx=x2x2−a2−a22log⁡|x+x2−a2|+c∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c

Given:-

∫3−2x−2x2dx
Solution:-
Let,
I=∫3−2x−2x2dx∴∫3−2(x2+2(12)x)dx=∫3−2(x2+2(12)x+(12)2)+2(12)2dx
We have
I=∫74−2(x+12)2dx=∫2(72)2−(x+12)2dx
As I match with the form
∫a2−x2dx=x2a2−x2+a22sin−1⁡xa+cI=2{x+122(72)2−(x+12)2}+742sin−1⁡(x+1272)+c
I=14(2x+1)2{(72)2−(x+12)2}+728sin−1⁡(2x+17)+c
I=14(2x+1)3−2x−2x2+728sin−1⁡(2x+17)+c

Indefinite Integrals Exercise 18.28 Question 12

Answer:-
The answer is
I=x24x4+1+14log⁡|x+x2x4+1|+c
Hints:-
∫a2−x2dx=x2a2−x2+a22sin−1⁡xa+c∫x2−a2dx=x2x2−a2−a22log⁡|x+x2−a2|+c∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c
Given:-

I=∫xx4+1dx
Solution:-
Let,
I=∫xx4+1dx
Let, x2=t
Differentiating both sides,
⇒2xdx=dt⇒xdx=12dt
Substituting x2 with t, we have
I=12∫t2+1dtI=12∫t2+12dt
As I match with the form

∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+cI=12{t2t2+1+12log⁡|t+t2+1|}+cI=t4t2+1+14log⁡|t+t2+1|+c
Putting the value of t back.
I=x24(x2)2+1+14log⁡|x2+(x2)2+1|+cI=x24x4+1+14log⁡|x+x2x4+1|+c

Indefinite Integrals exercise 18.28 question 13

Answer:-
x36a6−x6+a66sin−1⁡(x3a3)+c
Hints:-
∫a2−x2dx=x2a2−x2+a22sin−1⁡xa+c∫x2−a2dx=x2x2−a2−a22log⁡|x+x2−a2|+c∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c

Given:-

∫x2a6−x6dx
Solution:-
Let,
I=∫x2a6−x6dxI=∫x2a6−(x3)2dx
Let, x3=t
Differentiating both sides,
⇒3x2dx=dt⇒x2dx=13dt

Substituting x3 with t, we have
I=13∫(a3)2−t2dtI=12∫(a3)2−t2dt


As I match with the form
∫a2−x2dx=x2a2−x2+a22sin−1⁡xa+c∴I=13{t2a6−t2+a62sin−1⁡(ta3)}+cI=t4t2+1+14log⁡|t+t2+1|+c


Putting the value of t i.e.t=x3

I=x36a6−x6+a66sin−1⁡(x3a3)+c

Indefinite Integrals exercise 18.28 question 14

Answer:-
The answer is
I=log⁡x2(log⁡x)2+16+8log⁡|log⁡x+(log⁡x)2+16|+c
Hints:-
∫a2−x2dx=x2a2−x2+a22sin−1⁡xa+c∫x2−a2dx=x2x2−a2−a22log⁡|x+x2−a2|+c∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c

Given:-

I=∫(log⁡x)2+16xdx
Solution:-
Let,
I=∫1x16+(log⁡x)2dx
Let, log⁡x=t
Differentiating both sides,
⇒1xdx=dt
Substituting logx with t, we have
I=∫t2+16dtI=∫t2+42dt

As I match with the form
∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+cI={t2t2+16+162log⁡|t+t2+16|}+c


Putting the value of t back.

I=log⁡x2(log⁡x)2+16+8log⁡|log⁡x+(log⁡x)2+16|+c

Indefinite Integrals exercise 18.28 question 15

Answer:-
12(x−a)2ax−x2+a22sin−1⁡(x−aa)+c
Hint:-
Using the formula
∫a2−x2=12xa2−x2+a22sin−1⁡(xa)+c∫x2−a2dx=x2x2−a2−a22log⁡|x+x2−a2|+c∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c

Given:-

∫2ax−x2dx
Solution:-
Let,
I=∫2ax−x2dx∴I=∫−(x2−2ax)dx=∫(a)2−(x2−2ax+a2)dx
We have
I=∫(a)2−(x−a)2dx
As I match with the form
∫a2−x2=12xa2−x2+a22sin−1⁡(xa)+cI=x−a2a2−(x−a)2+a22sin−1⁡(x−aa)+cI=12(x−a)2ax−x2+a22sin−1⁡(x−aa)+c

Indefinite Integrals Exercise 18.28 Question 16

Answer:-
I=x23−x2+32sin−1⁡(x3)+c
Hint:-
Using the formula
∫a2−x2=12xa2−x2+a22sin−1⁡(xa)+c∫x2−a2dx=x2x2−a2−a22log⁡|x+x2−a2|+c∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c

Given:-

∫3−x2dx
Solution:-
 Let, I=∫3−x2dx∴I=∫3−x2dx=∫(3)2−x2dx
As I match with the form
∫a2−x2=12xa2−x2+a22sin−1⁡(xa)+cI=x23−x2+32sin−1⁡(x3)+c

Indefinite Integrals Exercise 18.28 Question 17

Answer:-
x−12x2−2x−12log⁡|x−1+x2−2x|+c
Hint:-
∫a2−x2=12xa2−x2+a22sin−1⁡(xa)+c∫x2−a2dx=x2x2−a2−a22log⁡|x+x2−a2|+c∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c

Given:-

∫x2−2xdx
Solution:-
Let, I=∫x2−2xdx
We have
I=∫x2−2xdxI=∫x2−2x+12−12dxI=∫(x−1)2−(1)2dx

As I match with the form
∫x2−a2dx=x2x2−a2−a22log⁡|x+x2−a2|+cI=x−12(x−1)2−1−12log⁡|x−1+(x−1)2−1|+cI=x−12x2−2x−12log⁡|x−1+x2−2x|+c




Indefinite Integrals Exercise 18 point 28 Question 18

Answer:-
I=12(x−1)2x−x2+12sin−1⁡((x−1))+c
Hint:-
Using the formula
∫a2−x2=12xa2−x2+a22sin−1⁡(xa)+c∫x2−a2dx=x2x2−a2−a22log⁡|x+x2−a2|+c∫x2+a2dx=x2x2+a2+a22log⁡|x+x2+a2|+c

Given:-

∫2x−x2dx
Solution:-
 Let, I=∫2x−x2dx∴I=∫−(x2−2(1)x)dx
I=∫12−(x−a)2dxI=∫(1)2−(x−1)2dx
As I match with the form∫a2−x2=12xa2−x2+a22sin−1⁡(xa)+cI=x−12(1)2−(x−1)2+122sin−1⁡(x−11)+cI=12(x−1)2x−x2+12sin−1⁡(x−1)+c

Rd Sharma class 12 chapter 18 exercise 18.28 answers that are clear and reviewed. It further helps students with getting a handle on formulae and tending to methods. Rd Sharma Class 12th exercise 18.28 has around 18 inquiries, including its subparts, and it joins themes like: -

  • Questions Related to antiderivative

  • Joining of mathematical capacities

  • Combination of outstanding capacities

  • Random issues

  • Mathematical translation of endless vital

  • Correlation among separation and combination

  • Methods of combination

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  • Extraordinary yet arranged showing of the subjects.

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