e-Crash Course...Page | 1 CBSE Math Class 12th Preface to the First Edition This book is an effort...

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e-Crash Course

Math

Complete Course at a Glance for Class XII

[Prepared strictly according to the latest syllabus issued by the Central Board of

Secondary Education, New Delhi for 2020 Examination of Class XII]

By

Marksadda.com

(A Unit of Ultra Creations)

ISBN 978-81-943778-7-0

https://www.marksadda.com/https://www.marksadda.com/coursedetails/-Maths/1-6https://www.marksadda.com/


Preface to the First Edition

This book is an effort to provide basic concepts and explanations in a concise way.

The topics have been explained through the examples.

Some main features of this book are:

1. The language used in this particular book is simple, lucid and easily

understandable by the students.

2. Essential graphs and diagrams have been correctly drawn and labeled well.

3. The concepts and explanations are provided in a concise manner.

Improvement is a consistent process to make the things better. Ultra Creations will

add more values to this particular work regarding betterment and upgradation of

content and quality in further editions. Suggestions and feedbacks are always

welcomed from teachers and students on [email protected]



Copyright © 2019 Marks Adda

All rights reserved. No part of this publication may be reproduced, distributed or transmitted in any

for or by any means, including photocopying, recording, uploading, scanning, sharing on social

media or other electronic or mechanical methods, without the written permission of the publisher.

For permission request write to [email protected]

Disclaimer:

The information provided in this book is to help the students preparing for the examination. All

efforts have been done to terminate errors in the content provided in this book. Neither the

publisher nor the author shall be responsible for any errors, omissions or damages arising out of

this information. Neither the publisher nor the author do not take any guarantee for any success in

the examination.

Jurisdiction:

The jurisdiction area for any legal dispute will be Nainital, Uttarakhand, India only.

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CONTENTS

CHAPTER 1 – RELATIONS AND FUNCTION .…………….……………..…5

CHAPTER 2 – INVERSE TRIGNOMETRIC FUNCTIONS .…….…..……17

CHAPTER 3 – ALGEBRA OF MATRICES………………….………………...29

CHAPTER 4 – DETERMINANTS……..……………….…….…………..….….41

CHAPTER 5 – ADJOINT AND INVERSE OF A MATRIX ….….…….….52

CHAPTER 6 – CONTINUTY AND DIFFERENTIABILITY ……………….66

CHAPTER 7 – APPLICATIONS OF DERIVATIVES ………….……….….82

CHAPTER 8 – INTEGRALS ………………………….….…………..…….…….93

CHAPTER 9 – APPLICATION OF INTEGRALS …………………….…… 112

CHAPTER 10– DIFFERENTIAL EQUATIONS ………..………….………120

CHAPTER 11 – VECTOR ALGEBRA……..………………………………….132

CHAPTER 12 –THREE DIMENSIONAL GEOMETRY ………….……..149

CHAPTER 13 – LINEAR PROGRAMMING…………….…………...……159

CHAPTER 14 – PROBABILITY..………………. ………………..…….….…172



1. STANDARD INTEGRALS:

i. ∫ 𝐱𝐧 dx=𝒙𝒏+𝟏

𝒏+𝟏, n≠ −𝟏

ii. ∫𝟏

𝐱 dx=𝐥𝐨𝐠|𝒙| +𝒄

iii. ∫ 𝒌 𝒅𝒙 =kx+𝒄

iv. ∫ 𝐬𝐢𝐧 𝒙 dx= − 𝐜𝐨𝐬 𝒙 + 𝒄

v. ∫ 𝐜𝐨𝐬 𝒙 dx= 𝐬𝐢𝐧 𝒙 +c

vi. ∫ 𝒕𝒂𝒏 𝒙 𝒅𝒙 = − 𝐥𝐨𝐠|𝐜𝐨𝐬 𝒙 | +c

vii. ∫ 𝐜𝐨𝐭 𝒙 dx =𝐥𝐨𝐠|𝒔𝒊𝒏𝒙|

viii. ∫ 𝒔𝒆𝒄𝒙 dx =𝐥𝐨𝐠|𝒔𝒆𝒄𝒙 + 𝒕𝒂𝒏𝒙| +C

ix. ∫ 𝒄𝒐𝒔𝒆𝒄𝒙 𝒅𝒙 = 𝐥𝐨𝐠|𝒄𝒐𝒔𝒆𝒄𝒙 − 𝒄𝒐𝒕𝒙| + 𝒄

x. ∫ 𝒔𝒆𝒄𝟐x dx = 𝐭𝐚𝐧 𝒙 +𝒄

xi. ∫ 𝒄𝒐𝒔𝒆𝒄𝟐x dx = − 𝐜𝐨𝐭 𝒙 +𝒄

xii. ∫ 𝐬𝐞𝐜 𝒙 𝐭𝐚𝐧 𝒙 dx = 𝒔𝒆𝒄𝒙+c

xiii. ∫ 𝐜𝐨𝐬𝐞𝐜 𝒙 dx= − 𝐜𝐨𝐬𝐞𝐜 𝒙 +c

xiv. ∫ 𝒆𝒙 dx= 𝒆𝒙 +c

xv. ∫ 𝒂𝒙 dx=𝒂𝒙

𝐥𝐨𝐠 𝒂 +c

xvi. ∫𝟏

√𝒂𝟐−𝒙𝟐 dx=𝐬𝐢𝐧−𝟏(

𝒙

𝒂) + 𝒄

xvii. ∫𝟏

𝒂𝟐 +𝒙𝟐 dx=

𝟏

𝒂 𝐭𝐚𝐧−𝟏(

𝒙

𝒂) +c

8 INTEGRALS



xviii. ∫𝟏

𝒙√𝒙𝟐−𝒂𝟐 dx=

𝟏

𝒂𝐬𝐞𝐜−𝟏(

𝒙

𝒂)+c

xix. ∫𝟏

𝒙𝟐−𝒂𝟐 dx=

𝟏

𝟐𝒂 𝐥𝐨𝐠 |

𝒙−𝒂

𝒙+𝒂|+c

xx. ∫𝟏

𝒂𝟐−𝒙𝟐 dx=

𝟏

𝟐𝒂 𝐥𝐨𝐠 |

𝒂+𝒙

𝒂−𝒙| +c

xxi. ∫𝟏

√𝒂𝟐+𝒙𝟐 dx=𝐥𝐨𝐠|𝒙 + √𝒙𝟐 + 𝒂𝟐| +c

xxii. ∫𝟏

√𝒙𝟐−𝒂𝟐 dx=𝐥𝐨𝐠|𝒙 + √𝒙𝟐 − 𝒂𝟐| +c

xxiii. ∫ √𝒂𝟐 − 𝒙𝟐 dx= 𝟏

𝟐 x√𝒂𝟐 − 𝒙𝟐 +

𝟏

𝟐 𝒂𝟐 𝐬𝐢𝐧−𝟏

𝒙

𝒂 +c

xxiv. ∫ √𝒙𝟐 − 𝒂𝟐dx= 𝟏

𝟐x√𝒙𝟐 − 𝒂𝟐 −

𝟏

𝟐 𝒂𝟐 𝐥𝐨𝐠|𝒙 + √𝒙𝟐 − 𝒂𝟐|+c

xxv. ∫ √𝒂𝟐 + 𝒙𝟐dx=𝟏

𝟐x√𝒙𝟐 + 𝒂𝟐+

𝟏

𝟐𝒂𝟐 𝐥𝐨𝐠|𝒙 + √𝒙𝟐 + 𝒂𝟐|+c

2. Method of integrate special type:

Type1. ∫𝒅𝒙

𝒂+𝒃 𝐜𝐨𝐬 𝒙 , ∫

𝒅𝒙

𝒂+𝒃 𝐬𝐢𝐧 𝒙 ,∫

𝒅𝒙

𝒂 𝐜𝐨𝐬 𝒙+𝒃 𝐬𝐢𝐧 𝒙 , ∫

𝒅𝒙

𝒂 𝐬𝐢𝐧 𝒙+𝒃 𝐜𝐨𝐬 𝒙+𝒄

Step1: put sin 𝑥=1−𝑡𝑎𝑛2

𝑥

2

1+𝑡𝑎𝑛2 𝑥

2

and cos 𝑥=2 tan

𝑥

2

1+𝑡𝑎𝑛2 𝑥

2

Step 2:put tan𝑥

2 =t and dx=

2𝑡

1+𝑡2

Type2. ∫𝒅𝒙

𝒂+𝒃𝒔𝒊𝒏𝟐𝒙 , ∫

𝒅𝒙

𝒂+𝒃𝒄𝒐𝒔𝟐𝒙 ,∫

𝒅𝒙

𝒂𝒔𝒊𝒏𝟐𝒙+𝒃𝒄𝒐𝒔𝟐𝒙,

∫𝒅𝒙

𝒂+𝒃𝒔𝒊𝒏𝟐𝒙+𝒄𝒄𝒐𝒔𝟐𝒙 ,∫

𝒅𝒙

(𝒂 𝐬𝐢𝐧 𝒙+𝒃 𝐜𝐨𝐬 𝒙)𝟐

Step1: Divide numerator and denominator by 𝑐𝑜𝑠2𝑥

Step 2: Put tan 𝑥=t and 𝑠𝑒𝑐2𝑥dx= dt.



Type3. ∫𝟏

𝒂𝒙𝟐+𝒃𝒙+𝒄dx, ∫

𝟏

√𝒂𝒙𝟐+𝒃𝒙+𝒄dx,∫ √𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 dx

Step 1: Make coeff. Of 𝑥2 𝑎𝑠 𝑢𝑛𝑖𝑡𝑦.

Step 2: Make perfect square by adding and subtracting (𝑐𝑜𝑒𝑓𝑓.𝑜𝑓 𝑥

2)2.

Type4. ∫𝟏

𝒙(𝒙𝒏+𝒌) dx

Step 1: Multiply numerator and denominator by𝑥𝑛−1.

Step 2: put 𝑥𝑛+K =t so that

n𝑥𝑛−1dx=dt

Type5. ∫𝒇′(𝒙)

𝒇(𝒙) dx or ∫(𝒇(𝒙))

𝒏𝒇′(𝒙)𝒅𝒙

Step 1: put f(x)=t so that 𝑓′(x)dx=dt

Step 2: Apply ∫1

𝑡 dt =log|𝑡|+c or

∫ 𝑡𝑛𝑑𝑡 = 𝑡𝑛+1

𝑛+1 +c

Type6.∫𝒑𝒙+𝒒

𝒂𝒙𝟐+𝒃𝒙+𝒄dx or∫

𝒑𝒙+𝒒

√𝒂𝒙𝟐+𝒃𝒙+𝒄dx

Step 1: Put px+q=A𝑑(𝑎𝑥2+𝑏𝑥+𝑐)

𝑑𝑥 +c

Step 2: Equate coeff. Of x and constant term to get A and B.

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Type7.∫𝑥2±1

𝑥4±𝑘𝑥2+1 dx

Step 1: Divide numerator and denominator by 𝑥2.

Step 2: Make a perfect square in the denominator.

Step 3: Put x ±1

𝑥= t so that

(1 ∓1

𝑥2) dx=dt

Type8. ∫𝑎 sin 𝑥+𝑏 cos 𝑥

𝑐 sin 𝑥+𝑑 cos 𝑥 dx,

Where c≠0, d≠ 0 and at least one of a and b is non-zero.

Step 1: Put asin 𝑥+bcos 𝑥=A𝑑(𝑐 sin 𝑥+𝑑 cos 𝑥)

𝑑𝑥 + B(csin 𝑥 + 𝑑 cos 𝑥)

Step 2: Apply formula

∫𝑓′(𝑥)

𝑓(𝑥) dx =log|𝑓(𝑥)|+c

Type9. ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥

Step 1:choose first function and second function according to ILATE

I→Inverse trigonometric function



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L→Logarithmic function

A→Algebraic function

T→Trignometric function

E→Exponential function

Step 2:Let f(x) is first function and g(x) is second function. Write as

f(x)∫ 𝑔(𝑥)dx-∫(𝑓′(𝑥).∫ 𝑔(𝑥)𝑑𝑥)𝑑𝑥

Type10.∫𝑥2

(𝑥2+𝑎2)(𝑥2+𝑏2)dx

Step 1:Let 𝑥2=y to get 𝑦

(𝑦+𝑎2)(𝑦+𝑏2)

Step 2:Resolve it into partial fraction.

Step 3: Replace y by 𝑥2𝑎𝑛𝑑 integrate.

Type11. ∫𝑑𝑥

𝑃√𝑄

a) If ‘P’ is linear or quadratic and ‘Q’ is linear then put √𝑄=t

b) If ‘P’ is liner and ‘Q’ is quadratic then put P=1

𝑡

c) If ‘P’ and ‘Q’ are both pure quadratic then put x=1

𝑡 and then √𝑄 =z

Note: To integrate ∫𝒙 𝒅𝒙

𝑷 √𝑸



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If ‘P’ and ‘Q’ are both pure quadratic then put √𝑄=t

Type 12: ∫ 𝑒𝑥[f(x)+𝑓′(𝑥)]dx

Step 1: Write it as ∫ 𝑒𝑥 𝑓(𝑥)𝑑𝑥 + ∫ 𝑒𝑥𝑓′(𝑥)𝑑𝑥

Step 2: Apply integration by parts in ∫ 𝑒𝑥𝑓(𝑥)𝑑𝑥

Type13. Integration using partial fraction:

INTEGRAND FORM OF PARTIAL FRACTION

a) 𝑝𝑥+𝑞

(𝑥−𝑎)(𝑥−𝑏)

𝐴

𝑥−𝑎 +

𝐵

𝑥−𝑏

b) 𝑝𝑥2+𝑞𝑥+𝑟

(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)

𝐴

𝑥 − 𝑎+

𝐵

𝑥 − 𝑏+

𝐶

𝑥 − 𝑐

c) 𝑝𝑥+𝑞

(𝑥−𝑎)2

𝐴

𝑥−𝑎 +

𝐵

(𝑥−𝑎)2

d) 𝑝𝑥2+𝑞𝑥+𝑟

(𝑥−𝑎) (𝑥−𝑏)2

𝐴

𝑥 − 𝑎+

𝐵

𝑥 − 𝑏+

𝐶

(𝑥 − 𝑏)2

e) 𝑝𝑥2+𝑞𝑥+𝑟

(𝑥−𝑎) (𝑥2+𝑏𝑥+𝑐)

𝐴

𝑥−𝑎+

𝐵𝑥+𝐶

𝑥2+𝑏𝑥+𝑐

Type14.

a) ∫ 𝑠𝑖𝑛𝑚𝑥 𝑐𝑜𝑠𝑛𝑥 𝑑𝑥

i. If ‘m’ is positive odd integer then put cos 𝑥=t

ii. If ‘n’ is positive odd integer then put sin 𝑥=t

iii. If both m, n are positive even integer then apply

𝑠𝑖𝑛2𝑥 =1−cos 2𝑥

2 and 𝑐𝑜𝑠2𝑥 =

1+cos 2𝑥

2

iv. If m+n is negative even integer then put tan 𝑥 = t

b) ∫ 𝑠𝑖𝑛𝑛𝑥 dx; If ‘n’ is positive odd integer then put cos 𝑥 = t



c) ∫ 𝑐𝑜𝑠𝑛 𝑥𝑑𝑥; If ‘n’ is positive odd integer then put sin 𝑥 =t

d) ∫ 𝑠𝑒𝑐𝑛𝑥 𝑑𝑥; If ‘n’ is positive even integer then put tan 𝑥 =t

e) ∫ 𝑐𝑜𝑠𝑒𝑐𝑛𝑥 𝑑𝑥; If ‘n’ is positive even integer then put cot 𝑥 =t

3. Properties of Definite integrals:

i. ∫ 𝒇(𝒙)𝒃

𝒂 𝒅𝒙 = − ∫ 𝒇(𝒙)𝒅𝒙

𝒃

𝒂

ii. ∫ 𝒇(𝒙)𝒃

𝒂𝒅𝒙 = ∫ 𝒇(𝑿)𝒅𝒙

𝒄

𝒂+∫ 𝒇(𝒙)

𝒃

𝒄 𝒅𝒙, for a< c


IMPORTANT QUESTIONS

Find the integral of ∫𝑠𝑒𝑐2𝑥

𝑐𝑜 𝑠𝑒𝑐2𝑥𝑑𝑥

Solution

∫𝑠𝑒𝑐2𝑥

𝑐𝑜 𝑠𝑒𝑐2𝑥𝑑𝑥

1

𝑐𝑜𝑠2𝑥

= ∫1

𝑠𝑖𝑛2𝑥 𝑑𝑥

= ∫𝑠𝑖𝑛2𝑥

𝑐𝑜𝑠2𝑥𝑑𝑥

= ∫ 𝑡𝑎𝑛2𝑥𝑑𝑥

= ∫(𝑠𝑒𝑐2𝑥 − 1)𝑑𝑥

= ∫ 𝑠𝑒𝑐2𝑥𝑑𝑥 − ∫ 1𝑑𝑥

= 𝑡𝑎𝑛 𝑥 − 𝑥 + 𝑐

Integrate the function 2𝑥

1+𝑥2

Solution

= ∫2𝑥

1+𝑥2 𝑑𝑥

𝑝𝑢𝑡 1 + 𝑥2 = 𝑡 ⟹ 2𝑥𝑑𝑥 = 𝑑𝑡

⇒ ∫2𝑥

1+𝑥2 𝑑𝑥 = ∫ 1

𝑡𝑑𝑡

= 𝑙𝑜𝑔|𝑡| + 𝑐

= log(1 + 𝑥2) + 𝑐



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Integeate the function sin(ax+b)cos(ax+b)

Solution

sin(ax+b)cos(ax+b) = 2 sin(𝑎𝑥+𝑏)cos (𝑎𝑥+𝑏)

2=

𝑠𝑖𝑛2(𝑎𝑥+𝑏)

2

put 2(ax+b) = t

⇒2adx = dt

⇒∫𝑠𝑖𝑛2(𝑎𝑥+𝑏)

2 𝑑𝑥 =

1

2∫

𝑠𝑖𝑛𝑡𝑑𝑡

2𝑎

=1

4𝑎[− cos 𝑡] + 𝑐

=−1

4𝑎𝑐𝑜𝑠2(𝑎𝑥 + 𝑏) + 𝑐

Integrate the function 𝑒2𝑥− 𝑒−2𝑥

𝑒2𝑥+ 𝑒−2𝑥

Solution

Let 𝑒2𝑥 + 𝑒−2𝑥 = 𝑡

∴ (2𝑒2𝑥 − 2𝑒−2𝑥)𝑑𝑥 = 𝑑𝑡

⇒ 2(𝑒2𝑥 − 2𝑒−2𝑥)𝑑𝑥 = 𝑑𝑡

⇒ ∫ (𝑒2𝑥− 𝑒−2𝑥

𝑒2𝑥+ 𝑒−2𝑥) = 𝑑𝑥 ∫

𝑑𝑡

2𝑡

= 1

2∫

1

2𝑑𝑡

=1

2𝑙𝑜𝑔|𝑡| + 𝑐

=1

2𝑙𝑜𝑔|𝑒2𝑥 + 𝑒−2𝑥| + 𝑐



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Find the integral of the function 𝑠𝑖𝑛4𝑥

Solution

𝑠𝑖𝑛4𝑥 = 𝑠𝑖𝑛2𝑥 𝑠𝑖𝑛2𝑥

= (1−𝑐𝑜𝑠2𝑥

2) (

1−𝑐𝑜𝑠2𝑥

2)

=1

4(1 − 𝑐𝑜𝑠2𝑥)2

=1

4[1 + 𝑐𝑜𝑠22𝑥 − 2𝑐𝑜𝑠2𝑥]

=1

4[1 + (

1+𝑐𝑜𝑠4𝑥

2) − 2𝑐𝑜𝑠2𝑥]

=1

4[1 +

1

2+

1

2𝑐𝑜𝑠4𝑥 − 2𝑐𝑜𝑠2𝑥]

= 1

4[

3

2+

1

2𝑐𝑜𝑠4𝑥 − 2𝑐𝑜𝑠2𝑥]

∴ ∫ 𝑠𝑖𝑛4𝑥𝑑𝑥 =1

4∫ [

3

2+

1

2𝑐𝑜𝑠4𝑥 − 2𝑐𝑜𝑠2𝑥]dx

=1

4[

3

2𝑥+

1

2(

𝑠𝑖𝑛4𝑥

4) −

2𝑠𝑖𝑛2𝑥

2] + 𝑐

=1

8[3𝑥 +

𝑠𝑖𝑛4𝑥

4− 2𝑠𝑖𝑛2𝑥] + 𝑐

=3𝑥

8−

1

4𝑠𝑖𝑛2𝑥 +

1

32𝑠𝑖𝑛4𝑥 + 𝑐

Integrate the retional function 3𝑥+5

𝑥3=𝑥2−𝑥+1

Solution

3𝑥+5

𝑥3−𝑥2−𝑥+1=

3𝑥+5

(𝑥−1)2(𝑥+1)

Let 3𝑥+5

(𝑥−1)2(𝑥+1)=

𝐴

(𝑥−1)+

𝐵

(𝑥−1)2+

𝑐

(𝑥+1)



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⇒ 3𝑥 + 5 = 𝐴(𝑥 − 1)(𝑥 + 1) + 𝐵(𝑥 + 1) + 𝐶(𝑥 − 1)2

⇒3𝑥 + 5 = 𝐴(𝑥2 − 1) + 𝐵(𝑋 + 1) + 𝐶(𝑥2 + 1 − 2𝑥) … … … . . (1)

Substituting x=1 in equation (1). We obtain

B=4

Equating the coefficients of 𝑥2 and x , we obtain

A + C = 0

B – 2C = 3

On solving, we obtain

𝐴 = −1

2 𝑎𝑛𝑑 𝐶 =

1

2

∴ 3𝑥+5

(𝑥−1)2(𝑥+1)=

−1

2(𝑥−1)+

4

(𝑥−1)2+

1

2(𝑥+1)

⇒ ∫3𝑥 + 5

(𝑥 − 1)2(𝑥 + 1)𝑑𝑥 = −

1

2∫

1

𝑥 − 1𝑑𝑥 + 4 ∫

1

(𝑥 − 1)2𝑑𝑥 +

1

2∫

1

(𝑥 + 1)𝑑𝑥

= −1

2𝑙𝑜𝑔|𝑥 − 1| + 4 (

−1

𝑥−1) +

1

2𝑙𝑜𝑔|𝑥 + 1| + 𝐶

=1

2𝑙𝑜𝑔 |

𝑥+1

𝑥−1| −

4

(𝑥−1+ 𝐶

Integrate the function 1

√8+3𝑥−𝑥2

Solution

8 + 3𝑥 − 𝑥2 = 8 − (𝑥2 − 3𝑥 +9

4−

9

4)

=41

4− (𝑥 −

3

2)2

⇒∫1

√8+3𝑥−𝑥2𝑑𝑥 = ∫

1

√41

4−(𝑥−

3

2)2 𝑑𝑥

Let 𝑥 −3

2= 𝑡



⇒∫1

√41

4−(𝑥−

3

2)2 𝑑𝑥

∫1

√(√41

2)2−𝑡2 𝑑𝑡

= 𝑠𝑖𝑛 − 1 (𝑡

√41

2

) + 𝐶

𝑥 −3

2

= 𝑠𝑖𝑛 − 1 (√41

2) + 𝐶

= 𝑠𝑖𝑛 − 1 (2𝑥−3

√41) + 𝐶

By using the properties of definite intregals, evaluate the intergral ∫ |𝑥 + 2|𝑑𝑥5

−5

Solution

Let 𝐼 = ∫ |𝑥 + 2|𝑑𝑥5

−5

It can be see that (x + 2)≤ 0 on [- 5, -2] and ≥ 0 on [- 2,5]

∴ 𝐼 = ∫−2

−5− (𝑥 + 2)𝑑𝑥 + ∫

−2

−5− (𝑥 + 2)𝑑𝑥

= − [𝑥2

2+ 2𝑥] −2

−5 +[

𝑥2

2+ 2𝑥] −2

−5

= − [(−2)2

2+ 2(−2) −

(−5)2

2− 2(−5)] + [

(5)2

2+ 2(5) −

(−2)2

2− 2(−2)]

= − [−2 − 4 −25

2+ 10]+[

25

2+ 10 − 2 + 4]

= −2 + 4 +25

2− 10 +

25

2+ 10 − 2 + 4 = 29



PREVIOUS YEAR QUESTION PAPER

Evaluate ∫ (2𝑥2 + 5𝑥)𝑑𝑥 3

1as a limit of a sum.

[CBSE 2010]

Prove that ∫ (√tan 𝑥 + √cot 𝑥)) 𝑑𝑥 = √2.𝜋/4

0 𝜋

2

[CBSE 2010]

Evaluate : ∫2

(1−𝑥)(1+𝑥2)𝑑𝑥

[CBSE 2010]

Evaluate : ∫ sin 𝑥 sin 2𝑥 sin 3𝑥 𝑑𝑥

[CBSE 2010]

Evaluate ∫(1 − 𝑥)√𝑥 𝑑𝑥.

[CBSE 2010]

Evaluate : ∫1

𝑥𝑑𝑥

3

2

[CBSE 2010]

Evaluate : ∫sin 𝑥+cos 𝑥

9+16 sin 2𝑥𝑑𝑥

𝜇

40

[CBSE 2011]



Evaluate ∫ (𝑥2 − 𝑥)𝑑𝑥4

1 as a limit of sums.

[CBSE 2011]

Evaluate : ∫𝑑𝑥

1+𝑥2√3

11

[CBSE 2011]

Evaluate : ∫(1+log 𝑥)

𝑥

2𝑑𝑥

[CBSE 2011]

Evaluate: ∫ (2𝑥2 + 5𝑥)𝑑𝑥3

1 as a limit of a sum.

[CBSE 2011]

Evaluate: ∫3𝑥−1

(𝑥+2)2𝑑𝑥

[CBSE 2012]

Prove that ∫ (√tan 𝑥 + √cot 𝑥) 𝑑𝑥 = √2.𝜋/4

0 𝜋

2

[CBSE 2012]

Evaluate: ∫2

(1−𝑥)(1+𝑥2)𝑑𝑥

[CBSE 2012]

Evaluate : ∫𝑐𝑜𝑠2𝑥

𝑐𝑜𝑠2𝑥+4𝑠𝑖𝑛2𝑥𝑑𝑥

𝜋/4

0

[CBSE 2012]

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Evaluate : ∫ (𝑥2 + 𝑥)𝑑𝑥3

1 as a limit of a sum.

[CBSE 2012]

Evaluate : ∫ 𝑥2

𝑡𝑎𝑛−1𝑥 𝑑𝑥

[CBSE 2012]

Evaluate: ∫ sin 𝑥 sin 2𝑥𝑠𝑖𝑛 3𝑥 𝑑𝑥

[CBSE 2012]

If ∫ 3𝑥2 𝑑𝑥 = 8,𝑎

0 write the value of ‘a’

[CBSE 2012]

If ∫ (𝑥−1

𝑥2) 𝑒𝑥 𝑑𝑥 = 𝑓(𝑥)𝑒2 + 𝑐, then write the value of 𝑓(𝑥).

[CBSE 2012]

Evaluate ∫(1 − 𝑥)√𝑥 𝑑𝑥

[CBSE 2012]

Evaluate : ∫1

𝑥𝑑𝑥

3

1

[CBSE 2012]

Evaluate : ∫ (|𝑥| + |𝑥 − 2| + |𝑥 − 4|)𝑑𝑥4

0

[CBSE 2013]

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Evaluate: ∫𝑥2

(𝑥2+4)(𝑥2+9)𝑑𝑥

[CBSE 2013]

Evaluate : ∫5𝑥−2

1+2𝑥+3𝑥2𝑑𝑥

[CBSE 2013]

Evaluate : ∫sin (𝑥−𝑎)

sin (𝑥+𝑎)𝑑𝑥

[CBSE 2013]

Evaluate: ∫𝑑𝑥

1+√cot 𝑥

𝜋/3

𝜋/6

[CBSE 2014]

Evaluate : ∫(𝑥 − 3)√𝑥2 + 3𝑥 − 18 dx

[CBSE 2014]

Evaluate : ∫𝑠𝑖𝑛6𝑥+𝑐𝑜𝑠6𝑥

𝑠𝑖𝑛2𝑥.𝑐𝑜𝑠2𝑑𝑥

[CBSE 2014]

Evaluate : ∫𝑑𝑥

9+𝑥23

0

[CBSE 2014]



Write the antiderivative of ( √𝑥3

+1

√𝑥).

[CBSE 2014]

Find ∫𝑠𝑖𝑛−1𝑥

(1−𝑥2)3/2𝑑𝑥

1

√2

0

[CBSE 2015]

Find ∫𝑥

(𝑥2+1)(𝑥−1)𝑑𝑥.

[CBSE 2015]

Evaluate : ∫ (1 + tan 𝑥) 𝑑𝑥𝜋/4

0

[CBSE 2015]

Find : ∫𝑥 𝑠𝑖𝑛−1𝑥

√1−𝑥2𝑑𝑥

[CBSE 2016]

Find ∫(𝑥2+1)(𝑥2+4)

(𝑥2+3)(𝑥2−5)𝑑𝑥

[CBSE 2016]

Find ∫(2𝑥 + 5)√10 − 4𝑥 − 3𝑥2𝑑𝑥

[CBSE 2016]



Find ∫3𝑥

3𝑥−1𝑑𝑥

[CBSE 2017]

Find ∫𝑥

√32−𝑥2𝑑𝑥

[CBSE 2017]

Evaluate : ∫𝑥 sin 𝑥

1+𝑐𝑜𝑠2𝑥𝑑𝑥

𝜋

0

[CBSE 2017]

Find ∫𝑥2+𝑥+1

(𝑥+1)2(𝑥−2)𝑑𝑥

[CBSE 2017]

Find ∫(𝑥 − 3)√3 − 2𝑥 − 𝑥2 𝑑𝑥

[CBSE 2017]

Evaluate : ∫cos 2𝑥+2𝑠𝑖𝑛2 𝑥

𝑐𝑜𝑠2𝑥𝑑𝑥

[CBSE 2018]

Find ∫2 cos 𝑥

(1−sin 𝑥)(1+𝑠𝑖𝑛2 𝑥)𝑑𝑥

[CBSE 2018]



Evaluate : ∫ sin 𝑥+cos 𝑥

16+9 sin 2𝑥𝑑𝑥

𝜋/4

0

[CBSE 2018]

Evaluate: ∫ (𝑥2 + 3𝑥 + 𝑒2) 𝑑𝑥,3

1 as the limit of the sum.

[CBSE 2018]



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