Karnataka 2nd PUC Mathematics Integrals MCQ with Answers

GOVERNMENT OF KARNATAKA
DEPARTMENT OF SCHOOL EDUCATION (PRE-UNIVERSITY)
18TH CROSS, MALLESHWARAM, BENGALURU – 560 012
CHAPTERWISE MULTIPLE CHOICE QUESTIONS FOR COMPETATIVE EXAM
SUBJECT: MATHEMATICS – II PUC
NAME OF THE CHAPTER: INTEGRALS

INDEFINITE INTEGRAL
Definition: Let 𝑓(𝑥) be a function of 𝑥.
𝑑
If there exists a function F(𝑥) such that 𝑑𝑥 F(𝑥) = 𝑓(𝑥) then F(𝑥) is called integral of

𝑓(𝑥) with respect x.
In symbols we write ∫ 𝑓(𝑥)𝑑𝑥 = F(𝑥)
𝑑
Note: 𝑑𝑥 [F(𝑥) + 𝐶] = 𝑓(𝑥) + 0 ∴ ∫ 𝑓(𝑥)𝑑𝑥 = F(𝑥) + 𝐶

Integral of some standard functions
𝑥 𝑛+1
1. ∫ 𝑥 𝑛 𝑑𝑥 = 𝑛+1 + 𝐶 , (𝑛 ≠ −1)

2. ∫ 𝑑𝑥 = 𝑥 + 𝐶
1
3. ∫ 𝑥 𝑑𝑥 = log|𝑥| + 𝐶

4. ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶
𝑎𝑥
5. ∫ 𝑎 𝑥 𝑑𝑥 = log𝑎 + 𝐶, 𝑎 > 0, 𝑎 ≠ 1

6. ∫ 𝑠𝑖𝑛𝑥 𝑑𝑥 = −𝑐𝑜𝑠𝑥 + 𝐶
7. ∫ 𝑐𝑜𝑠𝑥 𝑑𝑥 = 𝑠𝑖𝑛𝑥 + 𝐶
8. ∫ 𝑡𝑎𝑛𝑥 𝑑𝑥 = log|𝑠𝑒𝑐𝑥| + 𝐶 or − log|𝑐𝑜𝑠𝑥| + 𝐶
9. ∫ 𝑐𝑜𝑡𝑥 𝑑𝑥 = log|𝑠𝑖𝑛𝑥| + 𝐶 or − log|𝑐𝑜𝑠𝑒𝑐𝑥| + 𝐶
𝑥 𝜋
10. ∫ 𝑠𝑒𝑐𝑥 𝑑𝑥 = log|𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥| + 𝐶 = log |𝑡𝑎𝑛 (2 + 4 )| + 𝐶
𝑥
11. ∫ 𝑐𝑜𝑠𝑒𝑐𝑥 𝑑𝑥 = log|𝑐𝑜𝑠𝑒𝑐𝑥 − 𝑐𝑜𝑡𝑥| + 𝐶 = log |𝑡𝑎𝑛 2| + 𝐶

12. ∫ 𝑠𝑒𝑐 2 𝑥 𝑑𝑥 = 𝑡𝑎𝑛𝑥 + 𝐶
13. ∫ 𝑐𝑜𝑠𝑒𝑐 2 𝑥 𝑑𝑥 = − 𝑐𝑜𝑡𝑥 + 𝐶
14. ∫ 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 𝑑𝑥 = 𝑠𝑒𝑐𝑥 + 𝐶
15. ∫ 𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥 𝑑𝑥 = −𝑐𝑜𝑠𝑒𝑐𝑥 + 𝐶

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1
16. ∫ √1−𝑥 2 𝑑𝑥 = 𝑠𝑖𝑛−1 𝑥 + 𝐶 or −𝑐𝑜𝑠 −1 𝑥 + 𝐶
1
17. ∫ 1+𝑥 2 𝑑𝑥 = 𝑡𝑎𝑛−1 𝑥 + 𝐶 or − 𝑐𝑜𝑡 −1 𝑥 + 𝐶
1 1 𝑥−𝑎
18. ∫ 𝑥 2 −𝑎2 𝑑𝑥 = 2𝑎 log |𝑥+𝑎| + 𝐶
1 1 𝑎+𝑥
19. ∫ 𝑎2 −𝑥 2 𝑑𝑥 = 2𝑎 log |𝑎−𝑥| + 𝐶
1 1 𝑥
20. ∫ 𝑎2 +𝑥 2 𝑑𝑥 = 𝑎 𝑡𝑎𝑛−1 𝑎 + 𝐶
1
21. ∫ √𝑥 2 2 𝑑𝑥 = log|𝑥 + √𝑥 2 − 𝑎2 | + 𝐶
−𝑎
1 𝑥
22. ∫ √𝑎2 2 𝑑𝑥 = 𝑠𝑖𝑛−1 (𝑎) + 𝐶
−𝑥
1
23. ∫ √𝑥 2 2 𝑑𝑥 = log|𝑥 + √𝑥 2 + 𝑎2 | + 𝐶
+𝑎
𝑥 𝑎2
24. ∫ √𝑥 2 − 𝑎2 𝑑𝑥 = 2 √𝑥 2 − 𝑎2 − 2 log|𝑥 + √𝑥 2 − 𝑎2 | + 𝐶
𝑥 𝑎2
25. ∫ √𝑥 2 + 𝑎2 𝑑𝑥 = 2 √𝑥 2 + 𝑎2 + 2 log|𝑥 + √𝑥 2 + 𝑎2 | + 𝐶
𝑥 𝑎2 𝑥
26. ∫ √𝑎2 − 𝑥 2 𝑑𝑥 = 2 √𝑎2 − 𝑥 2 + 2 sin−1 (𝑎) + 𝐶

Note:
𝑥|𝑥|
 ∫|𝑥|𝑑𝑥 = +𝐶
2

 If ‘n’ is a positive integer, then
𝑥|𝑥 𝑛 | 𝑥 2 |𝑥 𝑛 |
∫|𝑥 𝑛 |𝑑𝑥 = 𝑛+1 + 𝐶 and ∫ 𝑥|𝑥 𝑛 |𝑑𝑥 = 𝑛+2 + 𝐶
Some properties of Indefinite Integrals
 ∫ 𝑓 ′ (𝑥)𝑑𝑥 = 𝑓(𝑥) + 𝐶
 ∫[𝑓(𝑥) ± g(𝑥)] = ∫ 𝑓(𝑥) 𝑑𝑥 ± ∫ g(𝑥) 𝑑𝑥
 ∫ 𝑘𝑓(𝑥) 𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥) 𝑑𝑥, where 𝑘 ∈ 𝑅
Integration By Substitution
Let 𝐼 = ∫ 𝑓(𝑥)𝑑𝑥
𝑑𝑥
Put 𝑥 = g(𝑡) then 𝑑𝑡 = g ′ (𝑡) or 𝑑𝑥 = g ′ (𝑡)𝑑𝑡

Thus 𝐼 = ∫ 𝑓(g(𝑡))g ′ (𝑡)𝑑𝑡
Some Standard Results
𝑓(𝑎𝑥+𝑏)
 ∫ 𝑓 ′ (𝑎𝑥 + 𝑏)𝑑𝑥 = 𝑎
+𝐶

 ∫ 𝑓 ′ {g(𝑥)} g ′ (𝑥)𝑑𝑥 = 𝑓{𝑔(𝑥)} + 𝐶
[𝑓(𝑥)]𝑛+1
 ∫[𝑓(𝑥)]𝑛 𝑓 ′ (𝑥)𝑑𝑥 = 𝑛+1
+ 𝐶 , for 𝑛 ≠ −1
𝑓 ′ (𝑥)
 ∫ 𝑓(𝑥) 𝑑𝑥 = log|𝑓(𝑥)| + 𝐶

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𝑓 ′ (𝑥)
 ∫ 𝑑𝑥 = 2√𝑓(𝑥) + 𝐶
√𝑓(𝑥)

 ∫ 𝑒 𝑓(𝑥) 𝑓 ′ (𝑥) 𝑑𝑥 = 𝑒 𝑓(𝑥) + 𝐶
𝑎𝑓(𝑥)
 ∫ 𝑎 𝑓(𝑥) 𝑓 ′ (𝑥) 𝑑𝑥 = log𝑎 + 𝐶 , 𝑎 > 0, 𝑎 ≠ 1

Some Special Substitutions
Functions Substitution

√𝑎 2 − 𝑥 2 𝑥 = 𝑎𝑠𝑖𝑛𝜃 𝑜𝑟 𝑥 = 𝑎𝑐𝑜𝑠𝜃
√𝑎2 +𝑥 2 or 𝑎2 +𝑥 2 𝑥 = 𝑎𝑡𝑎𝑛𝜃 𝑜𝑟 𝑥 = 𝑎𝑐𝑜𝑡𝜃
√𝑥 2 − 𝑎 2 𝑥 = 𝑎𝑠𝑒𝑐𝜃 𝑜𝑟 𝑥 = 𝑎𝑐𝑜𝑠𝑒𝑐𝜃

𝑎−𝑥 𝑎+𝑥
√ or √ 𝑥 = 𝑎𝑐𝑜𝑠2𝜃
𝑎+𝑥 𝑎−𝑥
𝑥 𝑎−𝑥
√ or √ 𝑥 = 𝑎𝑠𝑖𝑛2 𝜃 𝑜𝑟 𝑥 = 𝑎𝑐𝑜𝑠 2 𝜃
𝑎−𝑥 𝑥
𝑥 𝑎+𝑥
√ or √ 𝑥 = 𝑎𝑡𝑎𝑛2 𝜃 𝑜𝑟 𝑥 = 𝑎𝑐𝑜𝑡 2 𝜃
𝑎+𝑥 𝑥
𝑥−𝑎
√ or √(𝑥 − 𝑎)(𝑥 − 𝑏) 𝑥 = 𝑎𝑐𝑜𝑠 2 𝜃 + 𝑏𝑠𝑖𝑛2 𝜃
𝑏−𝑥

Integration Using Trigonometric Identities
Type I :
o ∫ 𝑠𝑖𝑛𝑛 𝑥 𝑑𝑥 , put 𝑐𝑜𝑠𝑥 = 𝑡 if 𝑛 is odd positive integer.
o ∫ 𝑐𝑜𝑠 𝑛 𝑥 𝑑𝑥 , put 𝑠𝑖𝑛𝑥 = 𝑡 if 𝑛 is odd positive integer.
o ∫ 𝑠𝑒𝑐 𝑛 𝑥 𝑑𝑥 , put 𝑡𝑎𝑛𝑥 = 𝑡 if 𝑛 is even positive integer.
o ∫ 𝑐𝑜𝑠𝑒𝑐 𝑛 𝑥 𝑑𝑥 , put 𝑐𝑜𝑡𝑥 = 𝑡 if 𝑛 is even positive integer.

Type II :
∫ 𝑠𝑖𝑛𝑎𝑥 ∙ 𝑐𝑜𝑠𝑏𝑥 𝑑𝑥 or ∫ 𝑠𝑖𝑛𝑎𝑥 ∙ 𝑠𝑖𝑛𝑏𝑥 𝑑𝑥 or ∫ 𝑐𝑜𝑠𝑎𝑥 ∙ 𝑐𝑜𝑠𝑏𝑥 𝑑𝑥
 First multiply and divide the integrand by 2.
 Apply product to sum or difference trigonometric formulae.
Type III :
∫ 𝑠𝑖𝑛𝑚 𝑥 𝑐𝑜𝑠 𝑛 𝑥 𝑑𝑥 , where 𝑚 and 𝑛 are integers.
 Put 𝑐𝑜𝑠𝑥 = 𝑡 if 𝑚 is odd or put 𝑠𝑖𝑛𝑥 = 𝑡 if 𝑛 is odd.
 If 𝑚 + 𝑛 is an even negative integers, apply the substitution 𝑡𝑎𝑛𝑥 = 𝑡.
 If 𝑚 and 𝑛 are non negative even integers, use
1−𝑐𝑜𝑠2𝑥 1+𝑐𝑜𝑠2𝑥
o 𝑠𝑖𝑛2 𝑥 = and 𝑐𝑜𝑠 2 𝑥 =
2 2
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Integration By Partial Fractions
Let 𝑓(𝑥) and g(𝑥) be polynomials of degree 𝑚 and 𝑛 respectively.
𝑓(𝑥) 𝑅(𝑥)
 If ≥ 𝑛 , then g(𝑥) = ℎ(𝑥) + g(𝑥) where deg(𝑅(𝑥)) < deg(g(𝑥)).
𝑅(𝑥)
 Now resolve g(𝑥) into partial fractions.

Case I: If g(𝑥) = (𝑥 − 𝑎1 )(𝑥 − 𝑎2 )(𝑥 − 𝑎3 ) … … (𝑥 − 𝑎𝑛 ).
𝑹(𝒙) 𝑨 𝑨 𝑨 𝑨
Then, = (𝒙−𝒂𝟏 ) + (𝒙−𝒂𝟐 ) + (𝒙−𝒂𝟑 ) + … … + (𝒙−𝒂𝒏 ) .
𝐠(𝒙) 𝟏 𝟐 𝟑 𝒏

Case II: If g(𝑥) = (𝑥 − 𝑎)𝑘 (𝑥 − 𝑎1 )(𝑥 − 𝑎2 )(𝑥 − 𝑎3 ) … … (𝑥 − 𝑎𝑟 ).
𝑹(𝒙) 𝑨 𝑨 𝑨 𝑩 𝑩 𝑩
Then, 𝟏
= (𝒙−𝒂) 𝟐
+ (𝒙−𝒂)𝟐
𝒌
+ ⋯ + (𝒙−𝒂)𝒌
+ (𝒙−𝒂𝟏 ) + (𝒙−𝒂𝟐 ) + ⋯ + (𝒙−𝒂𝒓 ) .
𝐠(𝒙) 𝟏 𝟐 𝒓

Case III: If g(𝑥) = (𝑥 − 𝑎1 )(𝑥 − 𝑎2 )(𝑥 − 𝑎3 ) … … (𝑎𝑥 + 𝑏𝑥 + 𝑐).
2

𝑹(𝒙) 𝑨 𝑨 𝑨 𝑩𝒙+𝑪
Then, = (𝒙−𝒂𝟏 ) + (𝒙−𝒂𝟐 ) + (𝒙−𝒂𝟑 ) + … … + (𝒂𝒙𝟐 +𝒃𝒙+𝒄) .
𝐠(𝒙) 𝟏 𝟐 𝟑

Find the values of constant in partial fractions and solve.
Some Standard Results
𝒇(𝒙) 𝒇(𝒂) 𝒇(𝒃)
 ∫ (𝒙−𝒂)(𝒙−𝒃) 𝒅𝒙 = 𝑨 𝐥𝐨𝐠|𝒙 − 𝒂| + 𝑩 𝐥𝐨𝐠|𝒙 − 𝒃| + 𝑪, where 𝑨 = (𝒂−𝒃) , 𝑩 = (𝒃−𝒂)
𝒇(𝒙)
 ∫ (𝒙−𝒂)(𝒙−𝒃)(𝒙−𝒄) 𝒅𝒙 = 𝑨 𝐥𝐨𝐠|𝒙 − 𝒂| + 𝑩 𝐥𝐨𝐠|𝒙 − 𝒃| + 𝑪 𝐥𝐨𝐠|𝒙 − 𝒃| + 𝑲
𝒇(𝒂) 𝒇(𝒃) 𝒇(𝒄)
where 𝑨 = (𝒂−𝒃)(𝒂−𝒄) , 𝑩 = (𝒃−𝒂)(𝒃−𝒄) , 𝑪 = (𝒄−𝒂)(𝒄−𝒃)

Integration By Parts
If 𝑓 and g are two functions of 𝑥, then
𝐝
∫ 𝐟(𝐱)𝐠(𝐱)𝐝𝐱 = 𝐟(𝐱) ∫ 𝐠(𝐱)𝐝𝐱 − ∫ { 𝐟(𝐱) ∫ 𝐠(𝐱)𝐝𝐱} 𝐝𝐱
𝐝𝐱
If 𝑓(𝑥) = first function(that can be differentiated) and
g(𝑥) = second function(that can be integrated), then the preference of
this order can be decided by the word “ILATE”, where
I stands for Inverse Function
L stands for Logarithmic Function
A stands for Algebraic Function
T stands for Trigonometric Function
E stands for Exponential Function.
Generalised integration by parts
∫ 𝑢 𝑣 𝑑𝑥 = 𝑢𝑣1 − 𝑢′ 𝑣2 + 𝑢′′ 𝑣3 − 𝑢′′′ 𝑣4 + … … … …….
where 𝑢′ , 𝑢′′ , 𝑢′′′ , … … representing successive differentiation of 𝑢 and
𝑣1 , 𝑣2 , 𝑣3 , 𝑣4 ,….. representing successive integration of 𝑣.

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Some Standard Results
1. ∫ 𝑒 𝑥 [𝑓(𝑥) + 𝑓 ′ (𝑥)]𝑑𝑥 = 𝑒 𝑥 𝑓(𝑥) + 𝐶
2. ∫ 𝑒 𝑘𝑥 [𝑘𝑓(𝑥) + 𝑓 ′ (𝑥)]𝑑𝑥 = 𝑒 𝑘𝑥 𝑓(𝑥) + 𝐶
𝑓 ′ (𝑥) 𝑒 𝑘𝑥 𝑓(𝑥)
3. ∫ 𝑒 𝑘𝑥 [𝑓(𝑥) + ] 𝑑𝑥 = +𝐶
𝑘 𝑘

4. ∫ 𝑒 𝑥 {[𝑓(𝑥) + g(𝑥)] + [𝑓 ′ (𝑥) + g ′ (𝑥)]}𝑑𝑥 = 𝑒 𝑥 [𝑓(𝑥) + g(𝑥)] + 𝐶
5. ∫[𝑓(𝑥) + 𝑥𝑓 ′ (𝑥)] 𝑑𝑥 = 𝑥𝑓(𝑥) + 𝐶
6. ∫[𝑓(𝑥)g ′ (𝑥) + 𝑓 ′ (𝑥)g(𝑥)]𝑑𝑥 = ∫[𝑓(𝑥)g(𝑥)]′ 𝑑𝑥 = 𝑓(𝑥)g(𝑥) + 𝐶
𝑒 𝑎𝑥 𝑒 𝑎𝑥 𝑏
7. ∫ 𝑒 𝑎𝑥 𝑠𝑖𝑛𝑏𝑥𝑑𝑥 = 𝑎2 +𝑏2 [𝑎𝑠𝑖𝑛𝑏𝑥 − 𝑏𝑐𝑜𝑠𝑏𝑥] + 𝐶 = √𝑎2 +𝑏2 𝑠𝑖𝑛 [𝑏𝑥 − tan−1 (𝑎)] + 𝐶
𝑒 𝑎𝑥 𝑒 𝑎𝑥 𝑏
8. ∫ 𝑒 𝑎𝑥 𝑐𝑜𝑠𝑏𝑥𝑑𝑥 = 𝑎2 +𝑏2 [𝑎𝑐𝑜𝑠𝑏𝑥 + 𝑏𝑠𝑖𝑛𝑏𝑥] + 𝐶 = √𝑎2 +𝑏2 𝑐𝑜𝑠 [𝑏𝑥 − tan−1 (𝑎)] + 𝐶
𝑒 𝑎𝑥
9. ∫ 𝑒 𝑎𝑥 𝑠𝑖𝑛(𝑏𝑥 + 𝑐)𝑑𝑥 = 𝑎2 +𝑏2 [𝑎𝑠𝑖𝑛(𝑏𝑥 + 𝑐) − 𝑏𝑐𝑜𝑠(𝑏𝑥 + 𝑐)] + 𝐾
𝑒 𝑎𝑥 𝑏
= 𝑠𝑖𝑛 [(𝑏𝑥 + 𝑐) − tan−1 ( )] + 𝐾
√𝑎2 + 𝑏 2 𝑎
𝑒 𝑎𝑥
10. ∫ 𝑒 𝑎𝑥 𝑐𝑜𝑠(𝑏𝑥 + 𝑐)𝑑𝑥 = 𝑎2 +𝑏2 [𝑎𝑐𝑜𝑠(𝑏𝑥 + 𝑐) + 𝑏𝑠𝑖𝑛(𝑏𝑥 + 𝑐)] + 𝐶
𝑒 𝑎𝑥 𝑏
= √𝑎2 𝑐𝑜𝑠 [(𝑏𝑥 + 𝑐) − tan−1 (𝑎)] + 𝐶
+𝑏2

DEFINITE INTEGRAL
𝑏
Definition: If ∫ 𝑓(𝑥)𝑑𝑥 = F(𝑥) + 𝐶 then ∫𝑎 𝑓(𝑥)𝑑𝑥 = |F(𝑥)| 𝑎𝑏 = F(𝑏) − F(𝑎).
The definite integral has a unique value.
Fundamental Theorem Of Integral Calculus
 Let 𝑓 be a continuous function on the closed interval [𝑎, 𝑏] and let 𝐴 (𝑥) be the area
function. Then 𝐴′(𝑥) = 𝑓 (𝑥), for all 𝑥 ∈ [𝑎, 𝑏].
 Let 𝑓 be continuous function defined on the closed interval [𝑎, 𝑏] and 𝐹 be an anti
𝑏
derivative of 𝑓. Then ∫𝑎 𝑓(𝑥)𝑑𝑥 = |F(𝑥)| 𝑎𝑏 = F(𝑏) − F(𝑎).
Evaluation Of Definite Integrals By Substitution
𝑏
Consider the integral ∫𝑎 𝑓(g(𝑥))g ′ (𝑥)𝑑𝑥
Put g(𝑥) = 𝑡, then g ′ (𝑥)𝑑𝑥 = 𝑑𝑡
Also, when 𝑥 = 𝑎, 𝑡 = g(𝑎) and 𝑥 = 𝑏, 𝑡 = g(𝑏)
g(𝑏)
Then, we get ∫g(𝑎) 𝑓(𝑡) 𝑑𝑡

2026-27 MATHEMATICS CET MATERIAL Page 5 of 20

Properties Of Definite Integrals
𝑏 𝑎
1. ∫𝑎 𝑓(𝑥)𝑑𝑥 = − ∫𝑏 𝑓(𝑥)𝑑𝑥
𝑎
2. ∫𝑎 𝑓(𝑥)𝑑𝑥 = 0
𝑏 𝑏 𝑏
3. ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑦)𝑑𝑦 = ∫𝑎 𝑓(𝑡)𝑑𝑡
(Definite integral is independent of variables of integration)
𝑏 𝑐 𝑏
4. If 𝑎 < 𝑐 < 𝑏 then ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑥)𝑑𝑥 + ∫𝑐 𝑓(𝑥)𝑑𝑥
𝑏 𝑏
5. ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑎 + 𝑏 − 𝑥)𝑑𝑥
𝑎 𝑎
6. ∫0 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥)𝑑𝑥
2𝑎 𝑎 𝑎
7. ∫0 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑥)𝑑𝑥 + ∫0 𝑓(2𝑎 − 𝑥)𝑑𝑥
𝑎 𝑎⁄2 𝑎⁄2
8. ∫0 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑥)𝑑𝑥 + ∫0 𝑓(𝑎 − 𝑥)𝑑𝑥
𝑎
2𝑎 2 ∫0 𝑓(𝑥)𝑑𝑥 𝑖𝑓 𝑓(2𝑎 − 𝑥) = 𝑓(𝑥)
9. ∫0 𝑓(𝑥)𝑑𝑥 = {
0 𝑖𝑓 𝑓(2𝑎 − 𝑥) = −𝑓(𝑥)
𝑎
𝑎 2 ∫0 𝑓(𝑥)𝑑𝑥 𝑖𝑓 𝑓 𝑖𝑠 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖. 𝑒. 𝑓(−𝑥) = 𝑓(𝑥)
10. ∫–𝑎 𝑓(𝑥)𝑑𝑥 = {
0 𝑖𝑓 𝑓 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖. 𝑒. 𝑓(−𝑥) = −𝑓(𝑥)
𝑎 𝑎
11. ∫−𝑎 𝑓(𝑥)𝑑𝑥 = ∫0 [𝑓(𝑥) + 𝑓(−𝑥)]𝑑𝑥
𝑏
12. ∫𝑎 𝑓(𝑥)𝑑𝑥 = 0 if 𝑓(𝑎 + 𝑥) = −𝑓(𝑏 − 𝑥)
𝑏 𝑎+𝑏 𝑏
13. ∫𝑎 𝑥𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑥)𝑑𝑥 if 𝑓(𝑎 + 𝑏 − 𝑥) = 𝑓(𝑥)
2
𝑎 𝑎 𝑎 𝑎⁄2
14. ∫0 𝑥𝑓(𝑥)𝑑𝑥 = 2 ∫0 𝑓(𝑥)𝑑𝑥 = 𝑎 ∫0 𝑓(𝑥)𝑑𝑥 , if 𝑓(𝑎 − 𝑥) = 𝑓(𝑥)
𝑎+𝑏
𝑏
15. ∫𝑎 𝑓(𝑥)𝑑𝑥 = 2 ∫𝑎 2 𝑓(𝑥)𝑑𝑥 , if 𝑓(𝑎 + 𝑥) = 𝑓(𝑏 − 𝑥)
𝑏 𝑏
16. |∫𝑎 𝑓(𝑥)𝑑𝑥| ≤ ∫𝑎 |𝑓(𝑥)|𝑑𝑥

17. If f(x) is a periodic function with period T then
𝑛𝑇 𝑇
∫ 𝑓(𝑥)𝑑𝑥 = 𝑛 ∫ 𝑓(𝑥)𝑑𝑥 , 𝑛∈𝑁
0 0

18. If 𝑓: 𝑅 → 𝑅 is a continuous periodic function with period T and 𝑎 ∈ 𝑅 and 𝑛 is a positive
integer then
𝑎+𝑛𝑇 𝑎+𝑇 𝑇
∫ 𝑓(𝑥)𝑑𝑥 = 𝑛 ∫ 𝑓(𝑥)𝑑𝑥 = 𝑛 ∫ 𝑓(𝑥)𝑑𝑥, 𝑛 ∈ 𝑁
𝑎 𝑎 0

19. If f(x) is a periodic function with period T and 𝑎 ∈ 𝑅 + , then
𝑎+𝑛𝑇 𝑎
∫𝑛𝑇 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑥)𝑑𝑥, 𝑛 ∈ 𝑁

2026-27 MATHEMATICS CET MATERIAL Page 6 of 20

20. If f(x) is a periodic function with period T then
𝑏+𝑇 𝑏 𝑏+𝑛𝑇 𝑏
(i) ∫𝑎+𝑇 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑥)𝑑𝑥 (𝑜𝑟) ∫𝑎+𝑛𝑇 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑥)𝑑𝑥 , 𝑛 ∈ 𝑍
𝑛𝑇 𝑇
(ii) ∫𝑚𝑇 𝑓(𝑥)𝑑𝑥 = (𝑛 − 𝑚) ∫0 𝑓(𝑥)𝑑𝑥 , 𝑚, 𝑛 ∈ 𝑍
𝑑 𝜙(𝑥)
21. 𝑑𝑥 (∫𝜑(𝑥) 𝑓(𝑡)𝑑𝑡) = 𝑓(𝜙(𝑥))∅′ (𝑥) − 𝑓(𝜑(𝑥))𝜑 ′ (𝑥)
𝑏
22. If 𝑓(𝑥) ≥ 0 in [𝑎, 𝑏] then ∫𝑎 𝑓(𝑥)𝑑𝑥 ≥ 0
𝑏 𝑏
23. If 𝑓(𝑥) ≤ 𝑔(𝑥) on [𝑎, 𝑏] then ∫𝑎 𝑓(𝑥)𝑑𝑥 ≤ ∫𝑎 𝑔(𝑥)𝑑𝑥
24. If m, M are smallest and greatest values of a function 𝑓(𝑥) defined
𝑏
on [𝑎, 𝑏] then 𝑚(𝑏 − 𝑎) ≤ ∫𝑎 𝑓(𝑥)𝑑𝑥 ≤ 𝑀(𝑏 − 𝑎)
𝜋 𝜋
𝑛−1
25. If 𝐼𝑛 = ∫02 𝑠𝑖𝑛𝑛 𝑥𝑑𝑥 = ∫02 𝑐𝑜𝑠 𝑛 𝑥𝑑𝑥 , then 𝐼𝑛 = 𝐼𝑛−2 where 𝑛 ∈ 𝑁
𝑛
𝑛−1 𝑛−3 𝑛−5 1 𝜋
i. 𝐼𝑛 = ( 𝑛 ) (𝑛−2) (𝑛−4) ⋯ ⋯ 2 ∙ 2 ; 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
𝑛−1 𝑛−3 𝑛−5 2
ii. 𝐼𝑛 = ( 𝑛 ) (𝑛−2) (𝑛−4) ⋯ ⋯ 3 ; 𝑛 𝑖𝑠 𝑜𝑑𝑑.
𝜋
1
26. If 𝐼𝑛 = ∫04 𝑡𝑎𝑛𝑛 𝑥𝑑𝑥 , then 𝐼𝑛 = 𝑛−1 − 𝐼𝑛−2 and hence
1 1 1 1
i. 𝐼𝑛 = 𝑛−1 − 𝑛−3 + 𝑛−5 − 𝑛−7 + ⋯ ⋯ 𝐼0 𝑜𝑟 𝐼1
𝜋 1
ii. according as n is even or odd where 𝐼0 = 4 𝑎𝑛𝑑 𝐼1 = 2 𝑙𝑛2 .
𝜋
1
27. ∫04 (𝑡𝑎𝑛𝑛 𝑥 + 𝑡𝑎𝑛𝑛−2 𝑥)𝑑𝑥 = 𝑛−1
𝜋
1
28. ∫𝜋2 (𝑐𝑜𝑡 𝑛 𝑥 + 𝑐𝑜𝑡 𝑛−2 𝑥)𝑑𝑥 = 𝑛−1
4
𝜋 𝑛−2
(√2) 𝑛−2
29. If 𝐼𝑛 = ∫0 𝑠𝑒𝑐 𝑥𝑑𝑥 , then 𝐼𝑛 =
4 𝑛
+ 𝑛−1 𝐼𝑛−2
𝑛−1
𝜋
[(𝑚−1)(𝑚−3)(𝑚−5)…..][(𝑛−1)(𝑛−3)(𝑛−5)…..]
30. ∫02 𝑠𝑖𝑛𝑚 𝑥 ∙ 𝑐𝑜𝑠 𝑛 𝑥𝑑𝑥 = (𝑚+𝑛)(𝑚+𝑛−2)(𝑚+𝑛−4)(𝑚+𝑛−6)…….
∙𝑘
𝜋
If m and n are even then 𝑘 = 2 , otherwise 𝑘 = 1.
𝜋
𝜋
31. ∫0 𝑠𝑖𝑛𝑛 𝑥𝑑𝑥 = 2 ∫02 𝑠𝑖𝑛𝑛 𝑥𝑑𝑥 for all positive integral values of n.
Some Standard Results
𝑏 𝑥−𝑎 𝜋
A. ∫𝑎 √𝑏−𝑥 𝑑𝑥 = 2 (𝑏 − 𝑎)

𝑏 𝜋 𝑏 1
B. (i) ∫𝑎 √(𝑥 − 𝑎)(𝑏 − 𝑥) 𝑑𝑥 = 8 (𝑏 − 𝑎)2 (ii) ∫𝑎 𝑑𝑥 = 𝜋
√(𝑥−𝑎)(𝑏−𝑥)

𝑎 𝑎+𝑥 𝜋+2 𝑎 𝑎−𝑥 𝜋−2
C. (i) ∫0 √𝑎−𝑥 𝑑𝑥 = 𝑎 ( 2 ) (ii) ∫0 √𝑎+𝑥 𝑑𝑥 = 𝑎 ( 2 )
𝜋 𝜋 𝜋
𝑓(𝑠𝑖𝑛𝑥) 𝑓(𝑡𝑎𝑛𝑥) 𝑓(𝑠𝑒𝑐𝑥) 𝜋
D. ∫02 𝑓(𝑠𝑖𝑛𝑥)+𝑓(𝑐𝑜𝑠𝑥) 𝑑𝑥 = ∫02 𝑓(𝑡𝑎𝑛𝑥)+𝑓(𝑐𝑜𝑡𝑥) 𝑑𝑥 = ∫02 𝑓(𝑠𝑒𝑐𝑥)+𝑓(𝑐𝑜𝑠𝑒𝑐𝑥) 𝑑𝑥 = 4

2026-27 MATHEMATICS CET MATERIAL Page 7 of 20

𝜋 𝜋 𝜋
𝑎𝑠𝑖𝑛𝑥+𝑏𝑐𝑜𝑠𝑥 𝑎𝑡𝑎𝑛𝑥+𝑏𝑐𝑜𝑡𝑥 𝑎𝑠𝑒𝑐𝑥+𝑏𝑐𝑜𝑠𝑒𝑐𝑥 𝜋
E. ∫02 𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥 𝑑𝑥 = ∫02 𝑡𝑎𝑛𝑥+𝑐𝑜𝑡𝑥 𝑑𝑥 = ∫02 𝑠𝑒𝑐𝑥+𝑐𝑜𝑠𝑒𝑐𝑥 𝑑𝑥 = (𝑎 + 𝑏) 4
𝜋 𝜋
𝜋
F. ∫0 𝑙𝑛(𝑠𝑖𝑛𝑥)𝑑𝑥 = ∫0 𝑙𝑛(𝑐𝑜𝑠𝑥)𝑑𝑥 = − 2 𝑙𝑛2
2 2

𝜋 𝜋
G. ∫02 𝑙𝑛(𝑡𝑎𝑛𝑥)𝑑𝑥 = ∫02 𝑙𝑛(𝑐𝑜𝑡𝑥)𝑑𝑥 = 0
𝜋
𝑑𝑥 𝜋 𝜋 𝑑𝑥 𝜋
H. (i) ∫02 𝑎2 𝑐𝑜𝑠2 𝑥+𝑏2𝑠𝑖𝑛2 𝑥 = 2𝑎𝑏 (ii) ∫0 𝑎2 𝑐𝑜𝑠2 𝑥+𝑏2𝑠𝑖𝑛2 𝑥 = 𝑎𝑏
∞ 1 𝑛
I. ∫0 𝑛 𝑑𝑥 = 𝑛2 −1
(𝑥+√𝑥 2 +1)

𝑛 𝑛(𝑛−1) 𝑛2 𝑛(𝑛−1)(4𝑛+1)
J. (i) ∫0 [𝑥]𝑑𝑥 = ,∀ 𝑛 ∈ 𝑁 (ii) ∫0 [√𝑥]𝑑𝑥 = ,∀ 𝑛 ∈ 𝑁
2 6
𝑏 𝑓(𝑎+𝑏−𝑥) 𝑏 𝑓(𝑥) (𝑏−𝑎)
K. ∫𝑎 𝑓(𝑥)+𝑓(𝑎+𝑏−𝑥) = ∫𝑎 𝑓(𝑥)+𝑓(𝑎+𝑏−𝑥) = 2
1 1 1
L. ∫0 𝑥(1 − 𝑥)𝑛 𝑑𝑥 = ∫0 (1 − 𝑥)𝑥 𝑛 𝑑𝑥 = (𝑛+1)(𝑛+2)

INDEFINITE INTEGRAL
Integration As An Inverse Process Of Differentiation
𝒙𝟑 +𝒙𝟐 +𝟏
1. ∫ 𝒅𝒙 = (Easy)
𝒙+𝟏
𝑥2 𝑥2
(A) 2 + log(𝑥) + 𝐶 (B) 2 + log(𝑥 + 1) + 𝐶
𝑥3 𝑥3
(C) 3 + log(𝑥) + 𝐶 (D) 3 + log(𝑥 + 1) + 𝐶
𝒙𝟑
2. ∫ 𝒙+𝟏 𝒅𝒙 is equal to (Average)
𝑥2 𝑥3 𝑥2 𝑥3
(A) 𝑥 − 2 + 3 − log|1 + 𝑥| + 𝐶 (B) 𝑥 − 2 − 3 − log|1 + 𝑥| + 𝐶
𝑥2 𝑥3 𝑥2 𝑥3
(C) 𝑥 + 2 − 3 − log|1 − 𝑥| + 𝐶 (D) 𝑥 + 2 + 3 − log|1 − 𝑥| + 𝐶
𝟏−𝒙𝟒
3. ∫ 𝟏−𝒙 𝒅𝒙 is equal to (Easy)
𝑥2 𝑥3 𝑥4 𝑥2 𝑥3 𝑥4
(A) 𝑥 − 2 − 3 − 4 + 𝐶 (B) 𝑥 + 2 + 3 + 4 + 𝐶
𝑥2 𝑥3 𝑥4 𝑥2 𝑥3 𝑥4
(C) 2𝑥 + 2 + 3 + 4 + 𝐶 (D) 𝑥 2 + 2 + 3 + 4 + 𝐶
𝟏+𝒙𝟒
4. The value of ∫ 𝟏+𝒙𝟔 𝒅𝒙 is equal to [KCET 2020] (Difficult)
1
(A) tan−1 𝑥 + 3 tan−1 𝑥 2 + 𝐶 (B) tan−1 𝑥 + tan−1 𝑥 3 + 𝐶
1 1
(C) tan−1 𝑥 + 3 tan−1 𝑥 3 + 𝐶 (D) tan−1 𝑥 − 3 tan−1 𝑥 3 + 𝐶
(𝒙𝟑 +𝟑𝒙𝟐 +𝟑𝒙+𝟏)𝒅𝒙
5. ∫ (𝒙+𝟏)𝟓
= (Easy)
1 1
(A) tan−1 𝑥 + 𝐶 (B) log(𝑥 + 1) + 𝐶 (C) 5 log(𝑥 + 1) + 𝐶 (D) − 𝑥+1 + 𝐶

2026-27 MATHEMATICS CET MATERIAL Page 8 of 20

6. ∫ 𝒆𝒍𝒐𝒈𝒕𝒂𝒏𝒙 𝒅𝒙 = (Easy)
(A) 𝑒 𝑡𝑎𝑛𝑥 + 𝐶 (B) 𝑡𝑎𝑛𝑥 + 𝐶 (C) log(𝑠𝑒𝑐𝑥) + 𝐶 (D) log(𝑡𝑎𝑛𝑥) + 𝐶
7. ∫ 𝒆−𝒙 𝐥𝐨𝐠 𝟐 𝟐𝒙 = [KCET 2026] (Average)
1 𝑥2
(A) log 𝑥 + 𝐶 (B) 𝑥 + 𝐶 (C) 𝑥 + C (D) 3 + 𝐶
𝒆𝟔𝐥𝐨𝐠𝒙 −𝒆𝟓𝐥𝐨𝐠𝒙
8. The value of ∫ 𝒆𝟒𝐥𝐨𝐠𝒙 −𝒆𝟑𝐥𝐨𝐠𝒙 𝒅𝒙 is equal to [KCET 2016] (Average)
𝑥3 3 1
(A) 0 (B) 3 + 𝐶 (C) 𝑥 3 + 𝐶 (D) 𝑥 + C

9. ∫(𝒆𝒙𝐥𝐨𝐠𝒂 + 𝒆𝒂𝐥𝐨𝐠𝒙 + 𝒆𝒂𝐥𝐨𝐠𝒂 )𝒅𝒙 is equal to (Average)
𝑎𝑥 𝑥𝑎 𝑎𝑥 𝑥𝑎
(A) log 𝑎 + 𝑎+1 + 𝑎𝑎 𝑥 + 𝐶 (B) log 𝑎 + 𝑎−1 + 𝑎𝑥 𝑎 + 𝐶
𝑎𝑥 𝑥𝑎 𝑎𝑥 𝑥𝑎
(C) log 𝑎 + 𝑎+1 + 𝑎𝑥 𝑎 + 𝐶 (D) log 𝑥 + 𝑎+1 + 𝑎𝑎 𝑥 + 𝐶

𝟖𝟏+𝒙 +𝟒𝟏−𝒙
10. ∫ 𝒅𝒙 = (Difficult)
𝟐𝒙
22𝑥+2 22−3𝑥 22𝑥+2 22−3𝑥
(A) + 3log 2 + 𝐶 (B) − 3log 2 + 𝐶
log 2 log 2

22𝑥+2 22−3𝑥 22𝑥+2 22−3𝑥
(C) − log 2 + 3log 2 + 𝐶 (D) − log 2 − 3log 2 + 𝐶
𝟏
11. ∫ 𝒇(𝒙)𝒔𝒊𝒏𝒙 𝒄𝒐𝒔𝒙 𝒅𝒙 = 𝟐(𝒃𝟐 −𝒂𝟐 ) 𝐥𝐨𝐠(𝒇(𝒙)) + 𝑪, where C is constant of

integration, then the value of 𝒇(𝒙) = (Difficult)
2 2 2 2
(A) 𝑎𝑏𝑠𝑖𝑛2𝑥 + 𝐶 (B) (𝑏2 −𝑎2)𝑠𝑖𝑛2𝑥 + 𝐶 (C) 𝑎𝑏𝑐𝑜𝑠2𝑥 + 𝐶 (D) (𝑏2 −𝑎2)𝑐𝑜𝑠2𝑥 + 𝐶

√𝟏−𝒙𝟐 +√𝟏+𝒙𝟐
12. ∫ 𝒅𝒙 = (Average)
√𝟏−𝒙𝟒

(A) log(𝑥 + √1 + 𝑥 2 ) + 𝑠𝑖𝑛−1 𝑥 + 𝐶 (B) log(𝑥 + √1 + 𝑥 2 ) − 𝑠𝑖𝑛−1 𝑥 + 𝐶
(C) 𝑐𝑜𝑠 −1 𝑥 − 𝑠𝑖𝑛−1 𝑥 + 𝐶 (D) 𝑡𝑎𝑛−1 𝑥 − 𝑠𝑖𝑛−1 𝑥 + 𝐶
Integration By Substitution
13. ∫ 𝒆𝟑𝐥𝐨𝐠𝒙 (𝒙𝟒 + 𝟏)−𝟏 𝒅𝒙 = (Easy)
(A) log(𝑥 4 + 1) + 𝐶 (B) 3log(𝑥 4 + 1) + 𝐶
1
(C) −log(𝑥 4 + 1) + 𝐶 (D) 4 log(𝑥 4 + 1) + 𝐶
𝒙𝟗
14. ∫ (𝟒𝒙𝟐 +𝟏)𝟔 𝒅𝒙 is equal to (Average)

1 1 −5 1 1 −5
(A) 5𝑥 (4𝑥 + 𝑥 2 ) +𝐶 (B) 5 (4 + 𝑥 2 ) +𝐶

1 1 −5 1 1 −5
(C) 10𝑥 (4 + 𝑥 2 ) +𝐶 (D) 10 (4 + 𝑥 2 ) +𝐶
𝟏
15. ∫ 𝟑 𝒅𝒙 = [KCET 2015, 2025] (Difficult)
𝒙𝟐 (𝒙𝟒 +𝟏) ⁄𝟒
1 3 1 1
–(1+𝑥 4 ) ⁄4 –(1+𝑥 4 ) ⁄4 –(1+𝑥 4 ) ⁄4 –(1+𝑥 4 ) ⁄4
(A) +𝐶 (B) +𝐶 (C) +𝐶 (D) +𝐶
𝑥2 𝑥 𝑥 2𝑥
2026-27 MATHEMATICS CET MATERIAL Page 9 of 20

𝟏
16. ∫ 𝒙+𝒙 𝒙 𝒅𝒙 = [KCET 2019] (Easy)
√ √
1
(A) tan−1 √𝑥 + 𝐶 (B) 2 log(√𝑥 + 1) + 𝐶 (C) 2tan−1 √𝑥 + 𝐶 (D) 2 tan−1 √𝑥 + 𝐶
√𝒙
17. If ∫ 𝒙(𝒙+𝟏) 𝒅𝒙 = 𝒌 𝐭𝐚𝐧−𝟏 𝒎 , then (𝒌, 𝒎) is……… (Average)

(A) (2, 𝑥) (B) (1, 𝑥) (C) (1, √𝑥 ) (D) (2, √𝑥 )
𝒙𝟑 𝒅𝒙
18. If ∫ = 𝒂(𝟏 + 𝒙𝟐 )𝟑⁄𝟐 + 𝒃√𝟏 + 𝒙𝟐 + 𝑪, then (Difficult)
√𝟏+𝒙𝟐
1 1 1 1
(A) 𝑎 = 3 , 𝑏 = −1 (B) 𝑎 = − 3 , 𝑏 = 1 (C) 𝑎 = − 3 , 𝑏 = −1 (D) 𝑎 = 3 , 𝑏 = 1
𝒅𝒙
19. ∫ (𝟏+𝒆𝒙)(𝟏+𝒆−𝒙 ) = (Average)
1 1 1 1
(A) 𝑒 𝑥 + 𝐶 (B) (1+𝑒 𝑥 )2 + 𝐶 (C) (1+𝑒 𝑥 ) + 𝐶 (D) − (1+𝑒 𝑥 ) + 𝐶
𝟐𝒙 𝒙
20. The value of ∫ 𝟐𝟐 𝟐𝟐 𝟐𝒙 𝒅𝒙 = (Average)
1 2𝑥 1 𝑥 1 𝑥 1 2𝑥
(A) (log 2)3 22 +𝐶 (B) (log 2)3 22 + 𝐶 (C) (log 2)2 22 + 𝐶 (D) (log 2)4 22 +𝐶

21. The value of ∫ 𝟐𝟑𝒙 𝟑𝟐𝒙 𝟓𝒙 𝒅𝒙 = (Difficult)
23𝑥 32𝑥 5𝑥 23𝑥 32𝑥 5𝑥 23𝑥 32𝑥 5𝑥 23𝑥 32𝑥 5𝑥
(A) log(720) + 𝐶 (B) log(360) + 𝐶 (C) log(180) + 𝐶 (D) log(90) + 𝐶
𝒔𝒊𝒏𝟐𝒙
22. ∫ 𝐬𝐢𝐧𝟐 𝒙+𝟐 𝐜𝐨𝐬𝟐 𝒙 𝒅𝒙 = [KCET 2014] (Easy)

(A) log(1 + 𝑡𝑎𝑛2 𝑥) + 𝐶 (B) log(1 + 𝑐𝑜𝑠 2 𝑥) + 𝐶
(C) − log(1 + 𝑐𝑜𝑠 2 𝑥) + 𝐶 (D) −log(1 + 𝑠𝑖𝑛2 𝑥) + 𝐶
𝒔𝒊𝒏𝒏 𝒙
23. If ‘ 𝒏’ is a natural number, then ∫ 𝒄𝒐𝒔𝒏+𝟐 𝒙 𝒅𝒙 = [KCET 2026] (Easy)
𝑡𝑎𝑛𝑛−1 𝑥 𝑡𝑎𝑛𝑛 𝑥 𝑡𝑎𝑛𝑛+1 𝑥 𝑡𝑎𝑛𝑛+2 𝑥
(A) +𝐶 (B) +𝐶 (C) +𝐶 (D) +𝐶
𝑛−1 𝑛 𝑛+1 𝑛+2
𝒔𝒊𝒏𝒙𝒄𝒐𝒔𝒙𝒅𝒙
24. ∫ = (Average)
√𝟏−𝒔𝒊𝒏𝟒 𝒙
1
(A) tan−1(𝑠𝑖𝑛2 𝑥) + 𝐶 (B) 2 sin−1(𝑠𝑖𝑛2 𝑥) + 𝐶
1
(C) 2 cos −1(𝑠𝑖𝑛2 𝑥) + 𝐶 (D) tan−1(2𝑠𝑖𝑛𝑥) + 𝐶
𝐜𝐨𝐬𝐞𝐜𝒙
25. ∫ 𝒙 𝒅𝒙 = (Difficult)
𝒄𝒐𝒔𝟐 (𝟏+𝒍𝒐𝒈𝒕𝒂𝒏 )
𝟐

𝑥 𝑥
(A) 𝑡𝑎𝑛 (1 + log𝑡𝑎𝑛 2) + 𝐶 (B) 𝑠𝑒𝑐 2 (1 + log𝑡𝑎𝑛 2) + 𝐶
𝑥 𝑥
(C) −𝑡𝑎𝑛 (1 + log𝑡𝑎𝑛 2) + 𝐶 (D) 𝑠𝑖𝑛2 (1 + log𝑡𝑎𝑛 2) + 𝐶
𝒙𝟑 𝐬𝐢𝐧(𝐭𝐚𝐧−𝟏 (𝒙𝟒 ))
26. ∫ 𝒅𝒙 is equal to [KCET 2021] (Average)
𝟏+𝒙𝟖
𝑐𝑜𝑠(tan−1 (𝑥 4 )) 𝑐𝑜𝑠(tan−1 (𝑥 4 ))
(A) − +𝐶 (B) +𝐶
4 4
𝑐𝑜𝑠(tan−1 (𝑥 3 )) 𝑠𝑖𝑛(tan−1 (𝑥 4 ))
(C) − +𝐶 (D) +𝐶
3 4

2026-27 MATHEMATICS CET MATERIAL Page 10 of 20

𝒇(𝒙)
27. ∫ 𝒙𝒇(𝒙)𝒅𝒙 + = 𝟎, then 𝒇(𝒙) is equal to [KCET 2026] (Difficult)
𝟐
2 2
(A) 𝑒 −2𝑥 (B) 𝑒 2𝑥 (C) 𝑒 −𝑥 (D) 𝑒 𝑥

Integration Using Trigonometric Identities
𝐬𝐢𝐧𝟐 𝒙
28. ∫ 𝟏+𝒄𝒐𝒔𝒙 𝒅𝒙 is equal to [KCET 2015] (Easy)

(A) 𝑥 + 𝑠𝑖𝑛𝑥 + 𝐶 (B) 𝑐𝑜𝑠𝑥 + 𝐶 (C) 𝑥 − 𝑠𝑖𝑛𝑥 + 𝐶 (D) 𝑠𝑖𝑛𝑥 + 𝐶
𝒄𝒐𝒔𝟐𝒙−𝒄𝒐𝒔𝟐𝜽
29. ∫ 𝒄𝒐𝒔𝒙−𝒄𝒐𝒔𝜽 𝒅𝒙 is equal to [KCET 2017, 2022] (Average)

(A) 2(𝑠𝑖𝑛𝑥 + 𝑥𝑐𝑜𝑠𝜃) + 𝐶 (B) 2(𝑠𝑖𝑛𝑥 − 𝑥𝑐𝑜𝑠𝜃) + 𝐶
(C) 2(𝑠𝑖𝑛𝑥 + 2𝑥𝑐𝑜𝑠𝜃) + 𝐶 (D) 2(𝑠𝑖𝑛𝑥 − 2𝑥𝑐𝑜𝑠𝜃) + 𝐶
30. ∫ √𝒄𝒐𝒔𝒆𝒄𝒙 − 𝒔𝒊𝒏𝒙 𝒅𝒙 = [KCET 2023] (Easy)
√𝑠𝑖𝑛𝑥 2
(A) √𝑠𝑖𝑛𝑥 + 𝐶 (B) 2√𝑠𝑖𝑛𝑥 + 𝐶 (C) +𝐶 (D) +𝐶
2 √𝑠𝑖𝑛𝑥
𝒄𝒐𝒔(𝒙−𝒂)
31. ∫ 𝒅𝒙 = (Easy)
𝒔𝒊𝒏𝒙

(A) 𝑐𝑜𝑠𝑎 log|𝑠𝑖𝑛𝑥| − 𝑥𝑠𝑖𝑛𝑎 + 𝐶 (B) 𝑐𝑜𝑠𝑎 log|𝑠𝑖𝑛𝑥| + 𝑥𝑠𝑖𝑛𝑎 + 𝐶
(C) 𝑐𝑜𝑠𝑎 log|𝑠𝑖𝑛𝑥| − 𝑥𝑡𝑎𝑛𝑎 + 𝐶 (D) 𝑐𝑜𝑠𝑎 log|𝑠𝑖𝑛𝑥| + 𝑥𝑡𝑎𝑛𝑎 + 𝐶
𝒔𝒊𝒏𝒙
32. If ∫ 𝐬𝐢𝐧(𝒙+𝜶) 𝒅𝒙 = 𝑨𝒙 + 𝑩𝒍𝒐𝒈𝒔𝒊𝒏(𝒙 + 𝜶) + 𝑪,then the value of (𝑨, 𝑩)is (Average)

(A) (−𝑠𝑖𝑛𝛼, 𝑐𝑜𝑠𝛼) (B) (𝑐𝑜𝑠𝛼, 𝑠𝑖𝑛𝛼) (C) (𝑠𝑖𝑛𝛼 , 𝑐𝑜𝑠𝛼) (D) (𝑐𝑜𝑠𝛼 , −𝑠𝑖𝑛𝛼)
𝒅𝒙
33. ∫ 𝒔𝒊𝒏(𝒙−𝒂)𝒔𝒊𝒏(𝒙−𝒃) is equal to (Difficult)
𝑠𝑖𝑛(𝑥−𝑏) 𝑠𝑖𝑛(𝑥−𝑎)
(A) 𝑠𝑖𝑛(𝑏 − 𝑎)log |𝑠𝑖𝑛(𝑥−𝑎)| + 𝐶 (B) 𝑐𝑜𝑠𝑒𝑐(𝑏 − 𝑎)log |𝑠𝑖𝑛(𝑥−𝑏)| + 𝐶
𝑠𝑖𝑛(𝑥−𝑏) 𝑠𝑖𝑛(𝑥−𝑎)
(C) 𝑐𝑜𝑠𝑒𝑐(𝑏 − 𝑎)log |𝑠𝑖𝑛(𝑥−𝑎)| + 𝐶 (D) 𝑠𝑖𝑛(𝑏 − 𝑎)log |𝑠𝑖𝑛(𝑥−𝑏)| + 𝐶
𝟏
34. ∫ 𝒔𝒊𝒏𝒙𝒄𝒐𝒔𝟐 𝒙 𝒅𝒙 = (Average)
𝑥 𝑥
(A) log |𝑡𝑎𝑛 2| + 𝑠𝑒𝑐𝑥 + 𝐶 (B) log |𝑡𝑎𝑛 2| − 𝑠𝑒𝑐𝑥 + 𝐶
𝑥 𝑥
(C) log |𝑡𝑎𝑛 2| + 𝑠𝑖𝑛𝑥 + 𝐶 (D) log |𝑡𝑎𝑛 2| − 𝑠𝑖𝑛𝑥 + 𝐶
𝒆𝒙 (𝟏+𝒙)𝒅𝒙
35. The value of ∫ 𝐜𝐨𝐬𝟐 (𝒆𝒙 .𝒙) = [KCET 2016] (Average)

(A) −𝑐𝑜𝑡(𝑒𝑥 𝑥 ) + 𝐶 (B) 𝑡𝑎𝑛(𝑥𝑒 𝑥 ) + 𝐶 (C) 𝑡𝑎𝑛(𝑒 𝑥 ) + 𝐶 (D) cot(𝑒 𝑥 ) + 𝐶
𝒔𝒊𝒏𝟖 𝒙−𝒄𝒐𝒔𝟖 𝒙
36. ∫ 𝟏−𝟐𝒔𝒊𝒏𝟐 𝒙𝒄𝒐𝒔𝟐 𝒙 𝒅𝒙 = (Difficult)
1 1 1
(A) 2 𝑠𝑖𝑛2𝑥 + 𝐶 (B) − 2 𝑠𝑖𝑛𝑥 + 𝐶 (C) − 2 𝑠𝑖𝑛2𝑥 + 𝐶 (D) −𝑠𝑖𝑛2 𝑥 + 𝐶
𝟓𝒙
𝒔𝒊𝒏
37. ∫ 𝟐
𝒙 𝒅𝒙 = [KCET 2024] (Difficult)
𝒔𝒊𝒏
𝟐

(A) 𝑥 + 2𝑠𝑖𝑛𝑥 + 𝑠𝑖𝑛2𝑥 + 𝐶 (B) 𝑥 + 2𝑠𝑖𝑛𝑥 + 2𝑠𝑖𝑛2𝑥 + 𝐶
(C) 2𝑥 + 2𝑠𝑖𝑛𝑥 + 𝑠𝑖𝑛2𝑥 + 𝐶 (D) 2𝑥 + 𝑠𝑖𝑛𝑥 + 2𝑠𝑖𝑛2𝑥 + 𝐶
2026-27 MATHEMATICS CET MATERIAL Page 11 of 20

𝟓𝒕𝒂𝒏𝒙
38. If the integral ∫ 𝒕𝒂𝒏𝒙−𝟐 𝒅𝒙 = 𝒙 + 𝒂𝐥𝐨𝐠|𝒔𝒊𝒏𝒙 − 𝟐𝒄𝒐𝒔𝒙| + 𝑪, then 𝒂 is equal to (Difficult)

(A) −2 (B) −1 (C) 1 (D) 2
Integrals Of Some Particular Functions
𝒅𝒙
39. ∫ 𝟒𝒙𝟐 +𝟏 = (Easy)
1 𝑥 1 𝑥
(A) tan−1 2𝑥 + 𝐶 (B) 2 tan−1 2𝑥 + 𝐶 (C) 3 tan−1 2 + 𝐶 (D) 2 tan−1 2 + 𝐶
𝟐𝒅𝒙
40. The value of ∫ = (Easy)
√𝟏−𝟒𝒙𝟐

(A) sin−1 2𝑥 + 𝐶 (B) cos −1 2𝑥 + 𝐶 (C) tan−1 2𝑥 + 𝐶 (D) cot −1 2𝑥 + 𝐶
𝒅𝒙
41. The value of ∫ = [KCET 2018] (Average)
√𝟑−𝟔𝒙−𝟗𝒙𝟐

3𝑥+1 3𝑥+1
(A) sin−1 ( 2 ) + 𝐶 (B) sin−1 ( 6 ) + 𝐶
1 3𝑥+1 2𝑥+1
(C) 3 sin−1 ( 2 ) + 𝐶 (D) sin−1 ( 3 ) + 𝐶
𝒙𝟐 𝒅𝒙
42. The value of ∫ is equal to [KCET 2021] (Average)
√𝒙𝟔 +𝒂𝟔

(A) log|𝑥 3 − √𝑥 6 + 𝑎6 | + 𝐶 (B) log|𝑥 3 + √𝑥 6 + 𝑎6 | + 𝐶
1 1
(C) 3 log|𝑥 3 − √𝑥 6 + 𝑎6 | + 𝐶 (D) 3 log|𝑥 3 + √𝑥 6 + 𝑎6 | + 𝐶
𝟏
43. ∫ 𝟏+𝟑 𝒔𝒊𝒏𝟐 𝒙+𝟖 𝒄𝒐𝒔𝟐 𝒙 𝒅𝒙 = [KCET 2023] (Difficult)
1 2𝑡𝑎𝑛𝑥 2𝑡𝑎𝑛𝑥
(A) 6 𝑡𝑎𝑛−1 ( )+𝐶 (B) 6 𝑡𝑎𝑛−1 ( )+𝐶
3 3
1 2𝑡𝑎𝑛𝑥
(C) 6 𝑡𝑎𝑛−1(2𝑡𝑎𝑛𝑥) + 𝐶 (D) 𝑡𝑎𝑛−1 ( )+𝐶
3
𝒔𝒊𝒏 𝒙
44. ∫ 𝟑+𝟒 𝒄𝒐𝒔𝟐 𝒙 𝒅𝒙 = [KCET 2024] (Average)
1 𝑐𝑜𝑠𝑥 1 2𝑐𝑜𝑠𝑥
(A) 𝑡𝑎𝑛−1 ( 3 ) + 𝐶 (B) − 2√3 𝑡𝑎𝑛−1 ( )+𝐶
√3 √3
1 𝑐𝑜𝑠𝑥 1 2𝑐𝑜𝑠𝑥
(C) 2√3 𝑡𝑎𝑛−1 ( 3 ) + 𝐶 (D) − 𝑡𝑎𝑛−1 ( )+𝐶
√3 3
𝟏
45. ∫ 𝒙[𝟔(𝐥𝐨𝐠 𝒙)𝟐 +𝟕𝐥𝐨𝐠 𝒙+𝟐] 𝒅𝒙 = [KCET 2024] (Difficult)
2log𝑥+1 1 2log𝑥+1 1 3log𝑥+2 3log𝑥+2
(A)log |3log𝑥+2| + 𝐶 (B) 2 log |3log𝑥+2| + 𝐶 (C) 2 log |2log𝑥+1 | + 𝐶 (D)log |2log𝑥+1 | + 𝐶

Integration By Partial Fractions
𝒅𝒙
46. 𝐓𝐡𝐞 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 ∫ (𝒙+𝟏)(𝒙+𝟐) is [KCET 2025] (Easy)
𝑥−1 𝑥+2 𝑥+1 𝑥−1
(A) log |𝑥−2| + 𝐶 (B) log |𝑥+1| + 𝐶 (C) log |𝑥+2| + 𝐶 (D) log |𝑥+2| + 𝐶
𝟐𝒙−𝟏
47. ∫ (𝒙−𝟏)(𝒙+𝟐)(𝒙−𝟑) 𝒅𝒙 = 𝑨 𝐥𝐨𝐠|𝒙 − 𝟏| + 𝑩 𝐥𝐨𝐠|𝒙 + 𝟐| + 𝑪 𝐥𝐨𝐠|𝒙 − 𝟑| + 𝑲.

Then A, B, C are respectively. [KCET 2019] (Average)
1 −1 1 −1 1 1 −1 −1 1 1 1 1
(A) 6 , , 3 (B) 6 , 3 , 3 (C) , , 2 (D) 6 , 3 , 5
3 6 3

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𝟑𝒙+𝟏
48. If ∫ (𝒙−𝟏)(𝒙−𝟐)(𝒙−𝟑) 𝒅𝒙 = 𝑨 𝐥𝐨𝐠|𝒙 − 𝟏| + 𝑩 𝐥𝐨𝐠|𝒙 − 𝟐| + 𝑪 𝐥𝐨𝐠|𝒙 − 𝟑| + 𝑪.

Then A, B, C are respectively. [KCET 2020] (Average)
(A) 2, −7, 5 (B) 5, −7, −5 (C) 2, −7, −5 (D) 5, −7, 5
𝒅𝒙 𝟏
49. If ∫ (𝒙+𝟐)(𝒙𝟐 +𝟏) = 𝒂𝐥𝐨𝐠|𝟏 + 𝒙𝟐 | + 𝒃𝒕𝒂𝒏−𝟏 𝒙 + 𝟓 𝐥𝐨𝐠|𝒙 + 𝟐| + 𝑪, then [KCET 2022] (Difficult)
−1 2 1 −2
(A) 𝑎 = 10 , 𝑏 = 5 (B) 𝑎 = 10 , 𝑏 = 5
−1 −2 1 2
(C) 𝑎 = 10 , 𝑏 = 5 (D) 𝑎 = 10 , 𝑏 = 5
𝒙𝟐 +𝟏
50. The value of ∫ 𝒙𝟐 −𝟏 𝒅𝒙 = (Average)
𝑥+1 𝑥−1 𝑥−1
(A) log(𝑥 2 − 1) + 𝐶 (B) log (𝑥−1) + 𝐶 (C) log (𝑥+1) + 𝐶 (D) 𝑥 + log (𝑥+1) + 𝐶
𝟏
51. ∫ 𝒙 𝒅𝒙 = [KCET 2018] (Easy)
𝒆 +𝟏
𝑒 𝑥 +1 𝑒 𝑥 −1 𝑒𝑥 𝑒𝑥
(A) log ( 𝑒 𝑥 ) + 𝐶 (B) log ( 𝑒 𝑥 ) + 𝐶 (C) log (𝑒 𝑥 +1) + 𝐶 (D) log (𝑒 𝑥 −1) + 𝐶
𝒅𝒙 𝒙𝟔
52. If ∫ 𝒙+𝒙𝟕 = 𝒑(𝒙) then, ∫ 𝒙+𝒙𝟕 𝒅𝒙 = (Difficult)

(A) 𝑙𝑜𝑔|𝑥| + 𝑝(𝑥) + 𝐶 (B) 𝑙𝑜𝑔|𝑥| − 𝑝(𝑥) + 𝐶 (C) 𝑥 − 𝑝(𝑥) + 𝐶 (D) 𝑥 + 𝑝(𝑥) + 𝐶
Integration By Parts
53. If linear function 𝒇(𝒙) and 𝒈(𝒙) satisfy [KCET 2014] (Average)
∫[(𝟑𝒙 − 𝟏)𝒄𝒐𝒔𝒙 + (𝟏 − 𝟐𝒙)𝒔𝒊𝒏𝒙]𝒅𝒙 = 𝒇(𝒙)𝒄𝒐𝒔𝒙 + 𝒈(𝒙)𝒔𝒊𝒏𝒙 + 𝑪 , then
(A) 𝑔(𝑥) = 3(𝑥 − 1) (B) 𝑔(𝑥) = 3 + 𝑥 (C) 𝑓(𝑥) = 3𝑥 − 5 (D) 𝑓(𝑥) = 3(𝑥 − 1)
𝐥𝐨𝐠𝒙
54. ∫ 𝒙𝟐 𝒅𝒙 = (Easy)
1 1 1
(A) 2 (logx + 1) + 𝐶 (B)(log𝑥 + 1) + 𝐶 (C) 𝑥 (log𝑥 + 1) + 𝐶 (D) − 𝑥 (log𝑥 + 1) + 𝐶

55. If 𝑰𝒏 = ∫(𝒍𝒐𝒈𝒙)𝒏 𝒅𝒙 then 𝑰𝒏 + 𝒏𝑰𝒏−𝟏 = (Difficult)
(A) (𝑥𝑙𝑜𝑔𝑥)𝑛 + 𝐶 (B) 𝑥(𝑙𝑜𝑔𝑥)𝑛 + 𝐶 (C) 𝑛(𝑙𝑜𝑔𝑥)𝑛 + 𝐶 (D) (𝑙𝑜𝑔𝑥)𝑛−1 + 𝐶
56. ∫ 𝒙𝟑 𝐬𝐢𝐧 𝟑𝒙 𝒅𝒙 = [KCET 2019] (Average)
𝑥 3 cos 3𝑥 𝑥 2 sin 3𝑥 2𝑥 cos 3𝑥 2 sin 3𝑥 𝑥 3 cos 3𝑥 𝑥 2 sin 3𝑥 2𝑥 cos 3𝑥 2 sin 3𝑥
(A) − + + − +𝐶 (B) − − + − +𝐶
3 3 9 27 3 3 9 27
𝑥 3 cos 3𝑥 𝑥 2 sin 3𝑥 2𝑥 cos 3𝑥 2 sin 3𝑥 𝑥 3 cos 3𝑥 𝑥 2 sin 3𝑥 2𝑥 cos 3𝑥 2 sin 3𝑥
(C) − + − − +𝐶 (D) + − − +𝐶
3 3 9 27 3 3 9 27

57. ∫ 𝐭𝐚𝐧−𝟏 √𝒙 𝒅𝒙 is equal to (Difficult)
(A) (𝑥 + 1) tan−1 √𝑥 − √𝑥 + 𝐶 (B) 𝑥 tan−1 √𝑥 − √𝑥 + 𝐶
(C) √𝑥 − 𝑥 tan−1 √𝑥 + 𝐶 (D) √𝑥 − (𝑥 + 1) tan−1 √𝑥 + 𝐶
58. If ∫ 𝒇(𝒙)𝒅𝒙 = 𝝍(𝒙) then ∫ 𝒙𝟓 𝒇(𝒙𝟑 )𝒅𝒙 = (Difficult)
1 1
(A) 3 𝑥 3 𝜓(𝑥 3 ) − 3 ∫ 𝑥 3 𝜓(𝑥 3 )𝑑𝑥 + 𝐶 (B) 3 𝑥 3 𝜓(𝑥 3 ) − 3 ∫ 𝑥 2 𝜓(𝑥 3 )𝑑𝑥 + 𝐶
1 1
(C) 3 [𝑥 3 𝜓(𝑥 3 ) − 3 ∫ 𝑥 3 𝜓(𝑥 3 )𝑑𝑥] + 𝐶 (D) 3 [𝑥 3 𝜓(𝑥 3 ) − 3 ∫ 𝑥 2 𝜓(𝑥 3 )𝑑𝑥] + 𝐶

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𝟏+𝒔𝒊𝒏𝒙
59. ∫ 𝒆𝒔𝒊𝒏𝒙 ( 𝒔𝒆𝒄𝒙 ) 𝒅𝒙 = [KCET 2018] (Easy)

(A) 𝑒 𝑠𝑖𝑛𝑥 + 𝐶 (B) 𝑒 𝑠𝑖𝑛𝑥 𝑠𝑖𝑛𝑥 + 𝐶 (C) 𝑒 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 + 𝐶 (D) 𝑒 𝑠𝑖𝑛𝑥 (𝑠𝑖𝑛𝑥 + 1) + 𝐶
𝟏+𝒔𝒊𝒏𝒙
60. ∫ 𝒆𝒙 (𝟏+𝒄𝒐𝒔𝒙) 𝒅𝒙 = [KCET 2015, 2021] (Average)
𝑥
(A) 𝑒 𝑥 𝑡𝑎𝑛 (2) + 𝐶 (B) 𝑒 𝑥 𝑡𝑎𝑛𝑥 + 𝐶 (C) 𝑒 𝑥 (1 + 𝑐𝑜𝑠𝑥) + 𝐶 (D) 𝑒 𝑥 (1 + 𝑠𝑖𝑛𝑥) + 𝐶
𝒆𝒙 (𝒙𝟐 𝐭𝐚𝐧−𝟏 𝒙+𝐭𝐚𝐧−𝟏 𝒙+𝟏)𝒅𝒙
61. The value of ∫ = [KCET 2016] (Easy)
𝒙𝟐 +𝟏
−1 𝑥
(A) 𝑒 𝑥 tan−1 𝑥 + 𝐶 (B) tan−1 (𝑒 𝑥 ) + 𝐶 (C) tan−1 (𝑥 𝑒 ) + 𝐶 (D) 𝑒 tan +𝐶
(𝒙+𝟑)
62. ∫ (𝒙+𝟒)𝟐 𝒆𝒙 𝒅𝒙 = [KCET 2017] (Easy)
𝑒𝑥 𝑒𝑥 𝑒𝑥 1
(A) (𝑥+3) + 𝐶 (B) (𝑥+4) + 𝐶 (C) (𝑥+4)2 + 𝐶 (D) (𝑥+4)2 + 𝐶
𝒙𝒆𝒙 𝒅𝒙
63. The value of ∫ (𝟏+𝒙)𝟐 is equal to [KCET 2021] (Easy)
𝑒𝑥
(A) (1+𝑥) + 𝐶 (B) 𝑒 𝑥 (1 + 𝑥) + 𝐶 (C) 𝑒 𝑥 (1 + 𝑥)2 + 𝐶 (D) 𝑒 𝑥 (1 + 𝑥 2 ) + 𝐶

𝟏−𝒙 𝟐
64. ∫ 𝒆𝒙 (𝟏+𝒙𝟐 ) 𝒅𝒙 is equal to (Average)
𝑒 −𝑥 𝑒𝑥 𝑒𝑥 𝑒 −𝑥
(A) 1+𝑥 2 + 𝐶 (B) 1+𝑥 2 + 𝐶 (C) (1+𝑥 2 )2 + 𝐶 (D) (1+𝑥 2 )2 + 𝐶

𝐥𝐨𝐠𝒙−𝟏 𝟐
65. ∫ [𝟏+(𝐥𝐨𝐠𝒙)𝟐 ] 𝒅𝒙 = (Difficult)
𝑥 log𝑥 𝑥 𝑥𝑒 𝑥
(A) log𝑥+1 + 𝐶 (B) 1+(log𝑥)2 + 𝐶 (C) 1+(log𝑥)2 + 𝐶 (D) 1+𝑥 2 + 𝐶

66. The value of ∫ 𝒆𝒔𝒊𝒏𝒙 𝒔𝒊𝒏𝟐𝒙𝒅𝒙 is [KCET 2020] (Difficult)
(A) 2𝑒 𝑠𝑖𝑛𝑥 (𝑐𝑜𝑠𝑥 − 1) + 𝐶 (B) 2𝑒 𝑠𝑖𝑛𝑥 (𝑠𝑖𝑛𝑥 + 1) + 𝐶
(C) 2𝑒 𝑠𝑖𝑛𝑥 (𝑐𝑜𝑠𝑥 + 1) + 𝐶 (D) 2𝑒 𝑠𝑖𝑛𝑥 (𝑠𝑖𝑛𝑥 − 1) + 𝐶
𝒙+𝒔𝒊𝒏𝒙
67. ∫ 𝟏+𝒄𝒐𝒔𝒙 𝒅𝒙 is equal to (Average)
𝑥 𝑥
(A) log(1 + 𝑐𝑜𝑠𝑥) + 𝐶 (B) log(𝑥 + 𝑠𝑖𝑛𝑥) + 𝐶 (C) 𝑥 − 𝑡𝑎𝑛 2 + 𝐶 (D) 𝑥 𝑡𝑎𝑛 2 + 𝐶
−𝟏 𝒙 𝒙
68. ∫ 𝒆𝐭𝐚𝐧 (𝟏 + 𝟏+𝒙𝟐 ) 𝒅𝒙 = (Average)
1 −1 𝑥 1 −1 𝑥 −1 𝑥 −1 𝑥
(A) 2 𝑒 tan +𝐶 (B) 2 𝑥𝑒 tan +𝐶 (C) 𝑥𝑒 tan +𝐶 (D) 𝑒 tan +𝐶
𝒙𝟐 −𝒙+𝟏 −𝟏 𝒙 −𝟏 𝒙
69. If ∫ 𝒙𝟐 +𝟏 𝒆𝐜𝐨𝐭 𝒅𝒙 = 𝑨(𝒙)𝒆𝐜𝐨𝐭 + 𝑪, then 𝑨(𝒙) is equal to (Average)

(A) −𝑥 (B) 𝑥 (C) √1 − 𝑥 (D) √1 + 𝑥

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Integrals Of Some More Types

70. ∫ √𝒙𝟐 + 𝟐𝒙 + 𝟓 𝒅𝒙 is equal to [KCET 2017] (Easy)
1
(A) 2 (𝑥 + 1)√𝑥 2 + 2𝑥 + 5 + 2 log|𝑥 + 1 + √𝑥 2 + 2𝑥 + 5 | + 𝐶
1
(B) (𝑥 + 1)√𝑥 2 + 2𝑥 + 5 + 2 log|𝑥 + 1 + √𝑥 2 + 2𝑥 + 5 | + 𝐶

(C) (𝑥 + 1)√𝑥 2 + 2𝑥 + 5 + 2 log|𝑥 + 1 + √𝑥 2 + 2𝑥 + 5 | + 𝐶

(D) (𝑥 + 1)√𝑥 2 + 2𝑥 + 5 − 2 log|𝑥 + 1 + √𝑥 2 + 2𝑥 + 5 | + 𝐶

71. ∫ √𝟓 − 𝟐𝒙 + 𝒙𝟐 𝒅𝒙 = [KCET 2023] (Average)
(𝑥−1)
(A) √5 + 2𝑥 + 𝑥 2 + 2log|(𝑥 − 1) + √5 + 2𝑥 + 𝑥 2 | + 𝐶
2
(𝑥−1)
(B) √5 − 2𝑥 + 𝑥 2 + 2log|(𝑥 + 1) + √5 + 2𝑥 + 𝑥 2 | + 𝐶
2
(𝑥−1)
(C) √5 − 2𝑥 + 𝑥 2 + 2log|(𝑥 − 1) + √5 − 2𝑥 + 𝑥 2 | + 𝐶
2
(𝑥−1)
(D) √5 − 2𝑥 + 𝑥 2 + 2log|(𝑥 − 1) + √5 + 2𝑥 + 𝑥 2 | + 𝐶
2

72. ∫ √(𝒙 − 𝟑)(𝟓 − 𝒙) 𝒅𝒙 = (Difficult)
(𝑥+4) 1
(A) − √(𝑥 − 3)(5 − 𝑥) + sin−1(𝑥 − 4) + 𝐶
2 2
(𝑥+4) 1
(B) √(𝑥 − 3)(5 − 𝑥) + sin−1(𝑥 − 4) + 𝐶
2 2
(𝑥−4) 1
(C) − √(𝑥 − 3)(5 − 𝑥) + sin−1(𝑥 − 4) + 𝐶
2 2
(𝑥−4) 1
(D) √(𝑥 − 3)(5 − 𝑥) + sin−1(𝑥 − 4) + 𝐶
2 2

73. ∫ √𝟕𝒙 − 𝟏𝟎 − 𝒙𝟐 𝒅𝒙 = (Difficult)
(2𝑥−7) 9 2𝑥−7
(A) √7𝑥 − 10 − 𝑥 2 + 8 sin−1 ( 3 ) + 𝐶
4
(2𝑥−7) 9 2𝑥−7
(B) − √7𝑥 − 10 − 𝑥 2 + 8 sin−1 ( 3 ) + 𝐶
4
(2𝑥−7) 9 2𝑥−7
(C) − √7𝑥 − 10 − 𝑥 2 − 8 sin−1 ( 3 ) + 𝐶
4
(2𝑥−7) 9 2𝑥−7
(D) √7𝑥 − 10 − 𝑥 2 − 8 sin−1 ( 3 ) + 𝐶
4

74. ∫ √𝟒𝒙𝟐 + 𝟗 𝒅𝒙 = (Easy)
𝑥 9
(A) 2 √4𝑥 2 + 9 − 4 log|2𝑥 + √4𝑥 2 + 9| +𝐶
𝑥 9
(B) 2 √4𝑥 2 + 9 + 4 log|2𝑥 + √4𝑥 2 + 9| +𝐶
𝑥 9
(C) − 2 √4𝑥 2 + 9 − 4 log|2𝑥 + √4𝑥 2 + 9| +𝐶
𝑥 9
(D) − 2 √4𝑥 2 + 9 + 4 log|2𝑥 + √4𝑥 2 + 9| +𝐶

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DEFINITE INTEGRAL
Fundamental Theorem Of Calculus
𝒃−𝟔
75. ∫𝒂−𝟔 𝒇(𝒙 + 𝟔)𝒅𝒙 is equal to [KCET 2026] (Easy)
𝑏 𝑏 𝑏 𝑏
(A) ∫𝑎 𝑓(𝑥 − 6)𝑑𝑥 (B) ∫𝑎 𝑓(𝑥 + 6)𝑑𝑥 (C) ∫𝑎 𝑓(𝑥)𝑑𝑥 (D) ∫𝑎 𝑓(−𝑥)𝑑𝑥
𝟎
76. ∫−𝟐[𝒙𝟑 + 𝟑𝒙𝟐 + 𝟑𝒙 + 𝟑 + (𝒙 + 𝟏)𝐜𝐨𝐬(𝒙 + 𝟏)]𝒅𝒙 = [KCET 2023] (Average)
(A) 0 (B) 1 (C) 3 (D) 4
𝝅
77. If ∫𝟎𝟐 √𝟏 − 𝒔𝒊𝒏𝟐𝒙 𝒅𝒙 is equal to (Difficult)

(A) 2√2 (B) 2(√2 + 1) (C) 2(√2 − 1) (D) 2
𝟐𝝅 𝒙
78. The value of ∫𝟎 √𝟏 + 𝒔𝒊𝒏 (𝟐) 𝒅𝒙 is [KCET 2025] (Average)

(A) 0 (B) 2 (C) 4 (D) 8
𝟏 𝒙 𝒆𝒙
79. ∫𝟎 (𝟐+𝒙)𝟑 𝒅𝒙 is equal to [KCET 2022] (Average)
𝑒 1 𝑒 1 𝑒 1 𝑒 1
(A) 9 − 4 (B) 9 + 4 (C) 27 + 8 (D) 27 − 8
𝟏

80. The value of ∫ 𝐜𝐨𝐬 −𝟏 𝒙 𝒅𝒙 is equal to
𝟐
𝟏 [KCET 2020] (Difficult)
−
𝟐

𝜋2 𝜋
(A) 2 (B) 2 (C) 1 (D) 𝜋
𝟑
81. ∫−𝟑 𝐜𝐨𝐭 −𝟏 𝒙 𝒅𝒙 is equal to [KCET 2019] (Average)
(A) 6 𝜋 (B) 3 𝜋 (C) 3 (D) 0
Evaluation Of Definite Integrals By Substitution
𝟏 𝒅𝒙
82. ∫𝟎 𝒆𝒙+𝒆−𝒙 is equal to [KCET 2018] (Easy)
𝜋 𝜋 𝜋
(A) 4 − tan−1(𝑒) (B) tan−1(𝑒) − 4 (C) tan−1(𝑒) + 4 (D) tan−1(𝑒)
𝝅
𝒄𝒐𝒔𝒙 𝒔𝒊𝒏𝒙
83. ∫𝟎𝟐 𝟏+𝒔𝒊𝒏𝒙 𝒅𝒙 is equal to [KCET 2022] (Average)

(A) log2 − 1 (B) −log2 (C) log2 (D) 1 − log2
𝝅
84. ∫𝟎 √𝒔𝒊𝒏𝜽𝒄𝒐𝒔𝟑 𝜽𝒅𝜽 is equal to
𝟐 [KCET 2022] (Average)
8 7 8 7
(A) 23 (B) 23 (C) 21 (D) 21
𝝅
𝒅𝒙
85. ∫𝟎 𝒂𝟐 𝒔𝒊𝒏𝟐 𝒙+𝒃𝟐 𝒄𝒐𝒔𝟐 𝒙 =
𝟐 [KCET 2017] (Difficult)
𝜋 𝜋𝑏 𝜋𝑎 𝜋𝑎
(A) 2𝑎𝑏 (B) 4𝑎 (C) 2𝑏 (D) 4𝑏
𝟏⁄𝟐 𝒅𝒙
86. ∫𝟎 is equal to [KCET 2018] (Difficult)
(𝟏+𝒙𝟐 )√𝟏−𝒙𝟐

1 2 2 3 √2 3 √2 √3
(A) tan−1 √3 (B) tan−1 ( ) (C) 2 tan−1 (2) (D) 2 tan−1 ( 2 )
√2 √2 √2
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𝟏 𝟏+𝒙
87. ∫𝟎 √𝟏−𝒙 𝒅𝒙 is equal to [KCET 2019] (Average)
𝜋 𝜋 1 𝜋
(A) 2 (B) 2 − 1 (C) 2 (D) 2 + 1
𝟏 𝐥𝐨𝐠(𝟏+𝒙)
88. The value of ∫𝟎 𝒅𝒙 is [KCET 2020] (Average)
𝟏+𝒙𝟐
𝜋 𝜋 1 𝜋
(A) 8 log2 (B) 2 log2 (C) 2 (D) 4 log2
𝝅
89. If 𝑰𝒏 = ∫𝟎𝟒 𝒕𝒂𝒏𝒏 𝒙𝒅𝒙 , where 𝒏 is any positive integer, then 𝑰𝟏𝟎 + 𝑰𝟖 is
equal to [KCET 2021] (Difficult)
1 1 1
(A) 7 (B) 8 (C) 9 (D) 9

Some Properties Of Definite Integrals
𝟓
90. ∫−𝟓|𝒙 + 𝟐|𝒅𝒙 is equal to [KCET 2017] (Easy)
(A) 27 (B) 28 (C) 29 (D) 30
91. If [𝒙] is the greatest integer function not greater than x , then
𝟖
∫𝟎 [𝒙]𝒅𝒙 is equal to [KCET 2022] (Easy)
(A) 20 (B) 28 (C) 29 (D) 30
𝟑.𝟓
92. ∫𝟎.𝟐 [𝒙]𝒅𝒙 is equal to [KCET 2017] (Average)
(A) 3 (B) 3.5 (C) 4 (D) 4.5
𝟐
93. ∫𝟎 [𝒙𝟐 ]𝒅𝒙 is equal to [KCET 2019] (Difficult)

(A) 5 − √2 + √3 (B) 5 − √2 − √3 (C) −5 − √2 − √3 (D) 5 + √2 − √3
𝟓
94. ∫𝟏 (|𝒙 − 𝟑| + |𝟏 − 𝒙|)𝒅𝒙 is equal to [KCET 2024] (Average)
5
(A) 12 (B) 6 (C) 21 (D) 10
𝟖 √𝟏𝟎−𝒙
95. ∫𝟐 𝒅𝒙 = [KCET 2016] (Easy)
√𝒙+√𝟏𝟎−𝒙

(A) 10 (B) 8 (C) 0 (D) 3
𝟖 𝟓√𝟏𝟎−𝒙
96. ∫𝟐 𝒅𝒙 = [KCET 2023] (Easy)
𝟓√𝒙 +𝟓√𝟏𝟎−𝒙

(A) 6 (B) 5 (C) 4 (D) 3
𝟑𝝅
𝒅𝒙
97. ∫ 𝝅
𝟖
is equal to (Easy)
𝟏+𝒄𝒐𝒕𝒙
𝟖
𝜋 𝜋 𝜋
(A) 8 (B) 4 (C) 2 (D) 2𝜋
𝝅 𝟐
98. If 𝒇(𝒙) = 𝒇(𝝅 + 𝒆 − 𝒙) 𝒂𝒏𝒅 ∫𝒆 𝒇(𝒙)𝒅𝒙 = 𝒆+𝝅 , then
𝝅
∫𝒆 𝒙𝒇(𝒙)𝒅𝒙 is equal to [KCET 2014] (Average)
𝜋+𝑒 𝜋−𝑒
(A) (B) (C) 𝜋 − 𝑒 (D) 1
2 2

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𝟑
99. ∫𝟏 (𝒙 − 𝟏)(𝒙 − 𝟐)(𝒙 − 𝟑)𝒅𝒙 is equal to (Easy)
(A) 0 (B) 1 (C) 2 (D) 3
𝟑𝝅
𝒙
100. The value of ∫ 𝝅
𝟒
𝒅𝒙 is equal to (Average)
𝟏+𝒔𝒊𝒏𝒙
𝟒

(A) 𝜋 (B) 2𝜋 (C) (√2 − 1)𝜋 (D) (√2 + 1)𝜋
𝒃 𝒙𝒏
101. If ∫𝒂 𝒙𝒏 +(𝟏𝟔−𝒙)𝒏 𝒅𝒙 = 𝟔, then (Average)

(A) 𝑎 = 4, 𝑏 = 12, 𝑛 ∈ 𝑅 (B) 𝑎 = 2, 𝑏 = 14, 𝑛 ∈ 𝑅
(C) 𝑎 = −4, 𝑏 = 20, 𝑛 ∈ 𝑅 (D) 𝑎 = 2, 𝑏 = 8, 𝑛 ∈ 𝑅
𝝅
𝒔𝒊𝒏𝟏𝟎𝟎𝟎 𝒙
102. ∫𝟎 𝒔𝒊𝒏𝟏𝟎𝟎𝟎 𝒙+𝒄𝒐𝒔𝟏𝟎𝟎𝟎 𝒙 𝒅𝒙 =
𝟐 [KCET 2016] (Easy)
𝜋 𝜋
(A) 1000 (B) 1 (C) 4 (D) 4
𝝅
𝒕𝒂𝒏𝟕 𝒙
103. ∫𝟎𝟐 𝒄𝒐𝒕𝟕 𝒙+𝒕𝒂𝒏𝟕 𝒙 𝒅𝒙 = [KCET 2017] (Easy)
𝜋 𝜋 𝜋 𝜋
(A) 6 (B) 4 (C) 3 (D) 2
𝟒𝟎𝟒𝟐 √𝒙𝒅𝒙
104. The value of ∫𝟎 is equal to [KCET 2021] (Easy)
√𝒙+√𝟒𝟎𝟒𝟐−𝒙

(A) 2021 (B) 4042 (C) 8084 (D) 1010
𝝅 𝒙𝒕𝒂𝒏𝒙
105. ∫𝟎 𝒔𝒆𝒄𝒙 𝒄𝒐𝒔𝒆𝒄𝒙 𝒅𝒙 = [KCET 2023] (Average)
𝜋 𝜋 𝜋2 𝜋2
(A) 2 (B) 4 (C) 2 (D) 4
𝝅
𝒔𝒊𝒏𝒙+𝒄𝒐𝒔𝒙
106. ∫𝟎𝟒 𝐥𝐨𝐠 ( ) 𝒅𝒙 = [KCET 2015] (Easy)
𝒄𝒐𝒔𝒙
𝜋 𝜋 𝜋
(A) 4 log2 (B) 2 log2 (C) 8 log2 (D) log2
𝝅
𝟑𝒔𝒆𝒄𝒙+𝟓𝒄𝒐𝒔𝒆𝒄𝒙
107. ∫𝟎𝟐 𝒔𝒆𝒄𝒙+𝒄𝒐𝒔𝒆𝒄𝒙 𝒅𝒙 = (Easy)
𝜋
(A) 2 (B) 𝜋 (C) 2𝜋 (D) 3𝜋
𝟏 𝟏
108. The value of ∫𝟎 𝒍𝒐𝒈 (𝒙 − 𝟏) 𝒅𝒙 [KCET 2025] (Average)
1
(A) 0 (B) log 𝑒 (2) (C) log 𝑒 2 (D) 1
𝟏𝟎𝟎𝟎
109. If [∙] is the greatest integer function, then ∫𝟎 𝒆𝒙−[𝒙] 𝒅𝒙 is equal to (Easy)
𝑒 1000 −1 𝑒 1000 −1 𝑒−1
(A) (B) (C) 1000(𝑒 − 1) (D) 1000
1000 𝑒−1
𝝅
110.The value of the integral ∫ 𝒍𝒐𝒈(𝒔𝒆𝒄𝜽 − 𝒕𝒂𝒏𝜽) 𝒅𝜽 is 𝟒
𝝅 [KCET 2014] (Average)
−
𝟒
𝜋 𝜋 𝜋
(A) 2 (B) 0 (C) 4 (D) − 4

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𝝅
111. The value of ∫𝟒𝝅 𝒔𝒊𝒏𝟏𝟎𝟑 𝒙 ∙ 𝒄𝒐𝒔𝟏𝟎𝟏 𝒙 𝒅𝒙 is [KCET 2016] (Easy)
−
𝟒

𝜋 103 𝜋 101
(A) ( ) (B) ( ) (C) 2 (D) 0
4 4
𝟏
112. The value of ∫−𝟏 𝒔𝒊𝒏𝟓 𝒙 𝒄𝒐𝒔𝟒 𝒙 𝒅𝒙 is [KCET 2025] (Easy)
𝜋 𝜋
(A) 𝜋 (B) 2 (C) 0 (D) − 2
𝝅
113. ∫−𝝅(𝟏 − 𝒙𝟐 ) 𝐬𝐢𝐧 𝒙 𝒄𝒐𝒔𝟐 𝒙 𝒅𝒙 is equal to [KCET 2024] (Easy)
𝜋2 𝜋3
(A) 𝜋 − 3 (B) 2𝜋 − 𝜋 3 (C) 𝜋 − 2 (D) 0
𝝅
𝒙 𝟐
𝟒 𝒆 𝒔𝒆𝒄 𝒙𝒅𝒙
114. ∫ 𝝅 is equal to (Average)
− 𝒆𝟐𝒙 −𝟏
𝟒

(A) 1 (B) 0 (C) 2 (D) 𝑒
115. One of the possible functions 𝒇(𝒙) which satisfies
𝟐
∫−𝟐 𝒇(𝒙)𝒅𝒙 = 𝟎 is [KCET 2025] (Average)
2+𝑥
(A) log 𝑒 (2−𝑥) (B) sin(2 + 𝑥) (C) 2𝑥 3 + 2𝑥 + 1 (D) 2𝑥𝑡𝑎𝑛𝑥
∞ 𝟐
116. ∫−∞ 𝒙𝒆−𝒙 𝒅𝒙 is equal to (Easy)
2 1 2 2 2 2
(A) 𝑒 −𝑥 (B) − 2 𝑒 −𝑥 + 𝑥𝑒 −𝑥 (C) 3 𝑒 −𝑥 (D) 0
𝝅
𝒅𝒙
117.∫ 𝟒
𝝅 is equal to [KCET 2015] (Average)
− 𝟏+𝒄𝒐𝒔𝟐𝒙
𝟒

(A) 2 (B) 0 (C) 1 (D) 4
𝝅
𝒅𝒙
118. ∫𝟐𝝅 𝒆𝐬𝐢𝐧 𝒙 +𝟏 is equal to [KCET 2017] (Average)
−
𝟐
𝜋 𝜋
(A) 1 (B) 0 (C) 2 (D) − 2
𝝅
𝟐 𝒄𝒐𝒔𝒙
119. The value of ∫ 𝝅 𝒅𝒙 is [KCET 2020] (Average)
− 𝟏+𝒆𝒙
𝟐

(A) −2 (B) 2 (C) 0 (D) 1
𝟐
120. ∫−𝟐|𝒙 𝐜𝐨𝐬 𝝅𝒙|𝒅𝒙 is equal to [KCET 2018] (Difficult)
8 4 2 1
(A) 𝜋 (B) 𝜋 (C) 𝜋 (D) 𝜋

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INTEGRALS
1 2 3 4 5 6 7 8 9 10
D A B C D C B B A B
11 12 13 14 15 16 17 18 19 20
D A D D C C D A D A
21 22 23 24 25 26 27 28 29 30
B C C B A A C C A B
31 32 33 34 35 36 37 38 39 40
B D C A B C A D B A
41 42 43 44 45 46 47 48 49 50
C D A B A C C A A D
51 52 53 54 55 56 57 58 59 60
C B A D B A A D B A
61 62 63 64 65 66 67 68 69 70
A B A B C D D C B A
71 72 73 74 75 76 77 78 79 80
C D A B C D C D A B
81 82 83 84 85 86 87 88 89 90
B B D C A A D A C C
91 92 93 94 95 96 97 98 99 100
B D B A D D A D A C
101 102 103 104 105 106 107 108 109 110
B C B A D C C A D B
111 112 113 114 115 116 117 118 119 120
D C D B A D C C D A

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