JEE Advanced Maths Indefinite Integration Previous Year Questions with Solutions

Indefinite Integration JEE Advanced previous year questions with solutions are given on this page. These are given in a detailed manner so that you can easily understand the solutions. These are designed by our specialized experts. The term integration refers to the summation of discrete data. Aspirants who are preparing for the JEE Advanced are recommended to go through these solutions, so that they can understand the pattern of questions and difficulty level.

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JEE Advanced Maths Indefinite Integration Previous Year Questions with Solutions

Question 1: The integral ∫sec²x/(sec x + tan x)^9/2 dx equals (for some arbitrary constant k)

Solution:

Let I = ∫ sec²x/(sec x + tan x)^9/2 dx

= ∫sec x .sec x. /(sec x + tan x)^9/2 dx

Put sec x + tan x = t

Differentiating, we get

(sec x tan x + sec²x )dx = dt

sec x (tan x + sec x) dx = dt

sec x dx = dt/t

1/t = 1/(sec x + tan x)

Multiply numerator and denominator by (sec x – tan x), we get

sec – tan x = 1/t

2 sec x = t + 1/t

sec x = ½ (t + 1/t)

So I = ½ ∫[(t + 1/t)dt/t]/t^9/2

= ½ ∫ [t^-9/2 + t^-13/2] dt

= ½ [ (-2/7)t^-7/2 – (2/11)t^-11/2]+ C

= -[ (t^-7/2/7) + t^-11/2/11] + C

= -[ (sec x + tan x)^-7/2/7 + (sec x + tan x)^-11/2/11]+ C

Question 2: The integral ∫ dx/(x+4)^8/7(x-3)^6/7 is equal to: Where C is the constant of integration:

(a) [(x-3)/(x+4)]^1/7 + C

(b) -[(x-3)/(x+4)]^-1/7 + C

(c) ½ [(x-3)/(x+4)]^3/7 + C

(d) (-1/13) [(x-3)/(x+4)]^-13/7 + C

Solution:

Let I = ∫ dx/(x+4)^8/7(x-3)^6/7

= ∫ [(x-3)/(x+4)]^-6/7 1/(x+4)² dx

Let [(x-3)/(x+4)] = t⁷

Differentiating w.r.t. x

7/(x+4)² dx = 7t⁶ dt

I = ∫t^-6 t⁶ dt

= t + c

= [(x-3)/(x+4)]^1/7 + C

Hence option a is the answer.

Question 3: The integral ∫(2x³-1)/(x⁴ + x) dx is equal to (here C is the constant of integration)

(a) ½ log |(x³ + 1)/x²| + C

(b) ½ log |(x³ + 1)/x³| + C

(c) log |(x³ + 1)/x| + C

(d) none of these

Solution:

Let I = ∫(2x³-1)/(x⁴ + x) dx

Divide by x²

= ∫(2x-x^-2)/(x² + x^-1) dx

Put x² + x^-1 = u

=> (2x – x^-2)dx = du

=> I = ∫du/u

= log |u|+ C

= log |x² + x^-1 | + C

= log |(x³ + 1)/x| + C

Hence option c is the answer.

Question 4: The integral ∫sec^2/3x cosec^4/3 x dx is equal to

(a) -3 (tan x)^-1/3 + C

(b) -(¾) (tan x)^-4/3 + C

(c) -3 (cot x)^-1/3 + C

(d) 3 (tan x)^-1/3 + C

Solution:

Let I = ∫sec^2/3x cosec^4/3 x dx

= ∫sec²x dx/tan^4/3 x

Put tan x = z

sec²x dx = dz

So I = ∫dz/z^4/3

= ∫z^-4/3 dz

= -3z^-1/3 + C

= -3 (tan x)^-1/3 + C

Hence option a is the answer.

Question 5: The integral ∫cos (log_e x) dx is equal to (where C is the constant of integration)

(a) (x/2)[ sin (log_ex) – cos (log_e x)] + C

(b) x [ cos (log_ex) + sin (log_e x)] + C

(c) (x/2)[ cos (log_ex) + sin (log_e x)] + C

(d) x [ cos (log_ex) – sin (log_e x)] + C

Solution:

Let I = ∫cos (log_e x) dx

= cos (log_e x) x – ∫-sin (log_e x).x/x dx

= x cos (log_e x) + sin (log_e x). x- ∫cos (log_e x) dx

= x cos (log_e x) + sin (log_e x). x – I

2I = x (cos (log_e x) + sin (log_e x))

I = (x/2) (cos (log_e x) + sin (log_e x)) + C

Hence option c is the answer.

Question 6: Evaluate ∫[e^x/√(4-e^2x)] dx

Solution:

Let I = ∫ [e^x/√(4-e^2x)] dx

= ∫ [e^x/√(2²– (e^x)²)] dx

put e^x = t

e^x dx = dt

So I = ∫ dt/√(2² – t²)

= sin^-1(t/2) + C

= sin^-1 (e^x/2) + C

Question 7: Evaluate ∫(√tan x + √cot x) dx

Solution:

∫(√tan x + √cot x) dx = ∫(√(sin x/cos x) + √(cos x/sin x)dx

= √2 ∫(sin x + cos x)/√sin 2x dx

Put sin x – cos x = t

=> (cos x + sin x) dx = dt

Also (sin x – cos x)² = t²

1 – sin 2x = t²

=> sin 2x = 1-t²

I = √2 ∫ dt/ √(1-t²)

= √2 sin^-1 t + C

= √2 sin^-1(sin x – cos x) + C

Question 8: Evaluate the following ∫dx/x²(x⁴ + 1)^3/4

(a) -[(x⁴ + 1)/x⁴]^1/4 + C

(b) [(x⁴ + 1)]^1/4 + C

(c) -[(x⁴ + 1)]^1/4 + C

(d) [(x⁴ + 1)/x⁴]^1/4 + C

Solution:

Let I = ∫ dx/x²(x⁴ + 1)^3/4

= ∫ dx/x⁵(1 + 1/x⁴)^3/4

Let t = 1 + 1/x⁴

dt = -4x^-5 dx

= -(4/x⁵) dx

So I = (-¼) ∫ dt/t^3/4

= -t^1/4 + C

= -[(x⁴ + 1)/x⁴]^1/4 + C

Hence option a is the answer.

Question 9: Evaluate the following ∫(e^{log x}+ sin x) cos x dx

(a) x sin x + cos x – ¼ cos 2x + C

(b) x sin x – cos x – ½ cos 2x + C

(c) ½ x sin x – cos x – ¼ cos 2x + C

(d) none of these

Solution:

∫(e^{log x}+ sin x) cos x dx = ∫ x cos x + ½ ∫ sin 2x dx

= x sin x – ∫sin x dx + ½ (-cos 2x)/2

= x sin x + cos x – ¼ cos 2x + C

Hence option a is the answer.

Question 10: Evaluate ∫ (x-1)e^x/(x+1)³ dx

(a) 0

(b) (x+1)²/e^x+ C

(c) e^x/(x+1)² + C

(d) none of these

Solution:

Let I = ∫ [(x-1)e^x/(x+1)³] dx

= ∫[(x+1-2)e^x/(x+1)³] dx

= ∫[1/(x+1)² – 2/(x+1)³]e^x dx

= e^x/(x+1)² + C (since ∫e^x(f(x) + f’(x)) dx = e^x f(x))

Hence option c is the answer.

Indefinite Integration – JEE Advanced Questions

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