JEE Main Previous Year Questions (2019): Indefinite Integral | Chapter-wise Tests for JEE Main & Advanced PDF Download

[JEE Main MCQs]

Q1: Let a ∈ (0,π/2) be fixed. If the integral
= A(x) cos 2α + B(x) sin 2α  + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively :
(a) x – α and log_e|cos(x – α)|
(b) x + α and log_e|sin(x – α)|
(c) x + α and log_e|sin(x + α)|
(d) x – α and log_e|sin(x – α)|
Ans: (d)
To solve the given integral, first we simplify the expression in the integral as follows :

Simplifying further, this becomes :

By using the formula for sin(a + b) = sin a cos b + cos a sin b and sin(a - b) = sin a cos b - cos a sin b, we get: 

Now, let's use the substitution method. Let t = x - α, therefore x = y + α and dx = dt. Substituting these values into the integral, we get :


Now on comparing, we get
A(x) = x−α and B(x) = log_e⁡|sin⁡(x−α)

Q2: The integral  is equal to: (Here C is a constant of integration)
(a)
(b)
(c)
(d)
Ans: (c)


Q3: If ∫x⁵e^-x2 dx = g(x)e^-x2 + C where c is a constant of integration, then g(–1) is equal to :
(a) 1
(b) −1
(c) -(5/2)
(d) -(1/2)
Ans: (c)
Let x² = t



Q4: Ifwhere C is a constant of integration then:
where C is a constant of integration then :
(a) A = 1/54 and f(x) = 9(x–1)² 
(b) A = 1/54 and f(x) = 3(x–1)
(c) A = 1/81 and f(x) = 3(x–1)
(d) A = 1/27 and f(x) = 9(x–1)²
Ans: (b)

On Differentiating ...(i)
2(x – 1)dx = 18tanθ sec²θ dθ


then A = 1/54
f(x) = 3(x – 1)

Q5: If ∫^{esec x} (sec x tan x f(x) + sec x tan x + sex²x)dx = e^secx f(x) + c. Then f(x) is
(a) x sec x + tan x + 1/2
(b) sec x + xtan x - 1/2
(c) sec x - tan x - 1/2
(d) sec x + tan x + 1/2
Ans: (d)
Given

Differentiating both sides with respect to x,


From question we can not find the value of C. So we have to choose any random value of C where (sec ⁡ x + tan x) present.
Only in option D, (sec x + tan ⁡x) term present. So only possible option is D.

Q6: The integral ∫sec^2/3 x cosec^4/3 x dx is equal to (Hence C is a constant of integration)
(a) -3/4 tan^-4/3 x + C
(b) 3tan^–1/3x + C
(c) –3cot^–1/3x+ C
(d) - 3tan^–1/3x + C
Ans: (d)

Let cot x = t³ 
⇒ - cosec²x dx = 3t²dt


Q7: If where C is a constant of integration, then the function ƒ(x) is equal to
(a) 3/x²
(b)
(c)
(d)
Ans: (c)



Q8: is equal to
(where c is a constant of integration)
(a) 2x + sinx + 2sin2x + c
(b) x + 2sinx + sin2x + c
(c) x + 2sinx + 2sin2x + c
(d) 2x + sinx + sin2x + c
Ans: (b)

Use sin²x = 2sinxcosx and sin³x = 3sinx – 4sin³x

= x + sin²x + 2sinx + C

Q9: The integraldx is equal to : (where C is a constant of integration)
(a)
(b)
(c)
(d)
Ans: (a)


Q10: The integral ∫ cos(loge x) dx is equal to : (where C is a constant of integration)
(a) x/2 [sin(log_e x) − cos(log_e x)] + C
(b) x[cos(log_e x) + sin(log_e x)] + C
(c) x/2 [cos(log_e x) + sin(log_e x)] + C
(d) x[cos(log_e x) − sin(log_e x)] + C
Ans: (c)


Q11: If where C is a constant of integration, then f(x) is equal to :
(a) 2/3 (x − 4)
(b) 1/3 (x + 4)
(c) 1/3 (x + 1)
(d) 2/3 (x + 2)
Ans: (b)



Q12: If + C for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals :
(a)
(b)
(c)
(d)
Ans: (b)



Q13: Ifwhere C is a constant of inegration, then f(x) is equal to -
(a) − 2x³ − 1
(b) − 2x³ + 1
(c) 4x³ + 1
(d) − 4x³ − 1
Ans: (d)



Q14: Let n ≥ 2 be a natural number and 0 < θ < π/2. Then  is equal to (where C is a constant of integration)
(a)
(b)
(c)
(d)
Ans: (c)



Q15:
f(0) = 0 , then the value of f(1) is :
(a) -(1/2)
(b) -(1/4)
(c) 1/2
(d) 1/4
Ans: (d)



Q16: For x² ≠ nπ + 1, n ∈ N (the set of natural numbers), the integral

is equal to (where c is a constant of integration) 
(a)
(b)
(c)
(d)
Ans: (d)


⇒ x² - 1 = 2t
⇒ 2xdx = 2dt
⇒ xdx = dt
∴ ∫ tan t dt
= In |sec t| + C