Integrals: CBSE Class 12 Previous Year Questions
40 different questions from Integrals (NCERT Chapter 7) asked in CBSE Class 12 Maths board exams 2026.
Pick a mark group to practise, each with answers and step-by-step solutions.
Most asked Integrals questions
If ∫ b 2 + c 2 x 2 3 a x d x = A log b 2 + c 2 x 2 + K , then the value of A is
(A) 3a(B) 2 b 2 3 a (C) b 2 c 2 3 a (D) 2 c 2 3 a
Show answer & solution
Answer: (D) 2 c 2 3 a
Put t = b 2 + c 2 x 2 , so d t = 2 c 2 x d x . ∫ b 2 + c 2 x 2 3 a x d x = 2 c 2 3 a ∫ t d t = 2 c 2 3 a log ∣ b 2 + c 2 x 2 ∣ + K .So A = 2 c 2 3 a .
The value of ∫ − 1 1 x 2 + 2∣ x ∣ + 1 x 3 d x is
(A) 0(B) log 2 (C) 2 log 2 (D) 2 1 log 2
Show answer & solution
Answer: (A) 0
Let f ( x ) = x 2 + 2∣ x ∣ + 1 x 3 . Then f ( − x ) = x 2 + 2∣ x ∣ + 1 − x 3 = − f ( x ) , so f is odd. For an odd function, ∫ − a a f ( x ) d x = 0 . So the value is 0.
∫ − 1 1 ( 1 − ∣ x ∣ ) d x is equal to :
(A) 2 ∫ 0 1 ( 1 + x ) d x (B) 2 ∫ − 1 0 ( 1 + x ) d x (C) 0 (D) 2 ∫ − 1 0 ( 1 − x ) d x
Show answer & solution
Answer: (B) 2 ∫ − 1 0 ( 1 + x ) d x
1 − ∣ x ∣ is even, so the integral is 2 ∫ − 1 0 ( 1 − ∣ x ∣ ) d x .For − 1 ≤ x ≤ 0 , ∣ x ∣ = − x , so 1 − ∣ x ∣ = 1 + x . Integral = 2 ∫ − 1 0 ( 1 + x ) d x (value 1).
∫ 25 − 16 x 2 d x is equal to :
(A) 5 1 sin − 1 4 x + C (B) 25 1 sin − 1 16 x + C (C) 4 1 sin − 1 5 4 x + C (D) 16 1 sin − 1 5 4 x + C
Show answer & solution
Answer: (C) 4 1 sin − 1 5 4 x + C
Put 4 x = t , d x = 4 d t . ∫ 25 − 16 x 2 d x = 4 1 ∫ 5 2 − t 2 d t = 4 1 sin − 1 5 t + C .= 4 1 sin − 1 5 4 x + C .
If ∫ 0 1 e x + e − x d x = tan − 1 e + k , then the value of k is :
(A) e(B) 4 π (C) 0(D) − 4 π
Show answer & solution
Answer: (D) − 4 π
e x + e − x 1 = e 2 x + 1 e x . Put e x = t , e x d x = d t ; limits 1 to e .∫ 1 e 1 + t 2 d t = tan − 1 e − tan − 1 1 = tan − 1 e − 4 π .So k = − 4 π .
Find ∫ x − 2 x + 2 d x
Show answer & solution
Answer: x 2 − 4 + 2 log x + x 2 − 4 + C
x − 2 x + 2 = x 2 − 4 x + 2 .∫ x 2 − 4 x d x = x 2 − 4 (put t = x 2 − 4 ).∫ x 2 − 4 2 d x = 2 log x + x 2 − 4 .So the integral = x 2 − 4 + 2 log x + x 2 − 4 + C .
Find : ∫ ( x 2 + 9 ) ( x 2 + 16 ) x 2 d x
Show answer & solution
Answer: − 7 3 tan − 1 3 x + 7 4 tan − 1 4 x + C
Put x 2 = t for the partial fractions: ( t + 9 ) ( t + 16 ) t = t + 9 A + t + 16 B . t = − 9 : A = 7 − 9 ; t = − 16 : B = − 7 − 16 = 7 16 .So ( x 2 + 9 ) ( x 2 + 16 ) x 2 = − 7 9 ⋅ x 2 + 9 1 + 7 16 ⋅ x 2 + 16 1 . Integral = − 7 9 ⋅ 3 1 tan − 1 3 x + 7 16 ⋅ 4 1 tan − 1 4 x + C = − 7 3 tan − 1 3 x + 7 4 tan − 1 4 x + C .
If I 1 = ∫ − π /4 π /4 1 + c o s 2 x d x and I 2 = ∫ − 1/2 1/2 ∣ x ∣ d x , then show that I 1 − 4 I 2 = 0 .
Show answer & solution
Answer: Proved.
I 1 = ∫ − π /4 π /4 2 c o s 2 x d x = 2 1 ∫ − π /4 π /4 sec 2 x d x = 2 1 [ tan x ] − π /4 π /4 = 2 1 ( 1 + 1 ) = 1 .∣ x ∣ is even, so I 2 = 2 ∫ 0 1/2 x d x = 2 ⋅ 2 1 ⋅ 4 1 = 4 1 .I 1 − 4 I 2 = 1 − 4 ⋅ 4 1 = 0 . Hence proved.
Evaluate : ∫ 12 π 12 5 π 1 + c o t x d x
Show answer & solution
Answer: 6 π
I = ∫ π /12 5 π /12 s i n x + c o s x s i n x d x .Using ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x with a + b = 2 π : I = ∫ π /12 5 π /12 c o s x + s i n x c o s x d x . Adding: 2 I = ∫ π /12 5 π /12 1 d x = 12 5 π − 12 π = 3 π . I = 6 π .
Evaluate : ∫ 6 − π 2 π ( sin ∣ x ∣ + cos ∣ x ∣ ) d x
Show answer & solution
For x < 0 : sin ∣ x ∣ + cos ∣ x ∣ = − sin x + cos x ; for x ≥ 0 : sin x + cos x . ∫ − π /6 0 ( cos x − sin x ) d x = [ sin x + cos x ] − π /6 0 = 1 − ( − 2 1 + 2 3 ) = 2 3 − 3 .∫ 0 π /2 ( sin x + cos x ) d x = [ − cos x + sin x ] 0 π /2 = 1 − ( − 1 ) = 2 .Total = 2 3 − 3 + 2 = 2 7 − 3 .
Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on
MonoMath CBSE →