JEE Main Maths Integral Calculus Previous Year Questions With Solutions

Calculus is the study of change, it basically analyses things that change and is a significant part of Mathematics. Calculus is a branch of mathematics focused on limits, integrals, derivatives, functions and infinite series. Integral calculus and Differential calculus are the two main branches of this topic. This article covers indefinite integral, definite integral, integration using partial fractions, integration by parts and properties of a definite integral. The integral calculus questions from the previous years of JEE Main are present on this page, along with the detailed solution for each question. About 2-4 questions are asked from this topic in JEE Examination.

JEE Main Maths Integral Calculus Previous Year Questions With Solutions

Question 1: ∫{[sin8x − cos8x] / [1 − 2 sin2x cos2x]} dx = _________.

Solution:

∫{[sin8x − cos8x] / [1 − 2 sin2x cos2x]} dx

= ∫{[(sin4x + cos4x) * (sin4x − cos4x)] / [(sin2x + cos2x)2 − 2 sin2x cos2x]} dx

= ∫(sin4x – cos4x) dx

= ∫[sin2x + cos2x] * [sin2x – cos2x] dx

= ∫(sin2x + cos2x) dx

= ∫−cos2xdx

= [−sin2x / 2] + c

Question 2: ∫x2dx / (a + bx)2 = ___________.

Solution:

Put a + bx = t

⇒ x = [t − a] / [b] and dx = dt / [b]

I =∫([t − a] / b)2 * [1 / t2] * [dt / b]

= [1 / b2]∫(1 − (2a / t) + [a2 * t−2]) dt

= [1 / b2] * [(t − 2a log t) − (a2 / t)]

= [1 / b2] [(x + a / b) − [2a / b] * log (a + bx) − [a2 / b] * [1 / (a + bx)]

Question 3: ∫[x5 / √(1 + x3)] dx = ________.

Solution:

Put 1 + x3 = t2

⇒ 3x2 dx = 2tdt and x3 = t2 − 1

So, ∫[x5 / √(1 + x3)] dx = ∫{[x2 * x3] / √(1 + x3)} dx

= [2 / 3] ∫{[(t2 − 1) * t] dt / [t]}

= [2 / 3] ∫(t2 − 1) dt

= [2 / 3] [(t3 / 3) − t] + c

= [2 / 3] [{(1 + x3)3/2 / 3} − {(1 + x3)½}]+ c

Question 4: ∫tan32x sec2x dx = __________.

Solution:

∫tan32x sec2x dx = ∫[(sec2 2x − 1) sec2x * tan2x] dx

=∫sec32x tan2x dx −∫sec2x tan2x dx ……. (i)

Now, we take ∫sec32x tan2x dx

Put sec 2x = t

⇒ sec 2x tan 2x = dt/2, then it reduces to

[1 / 2] ∫t2 dt = t3 / 6

= [sec32x] / [6]

From (i), ∫sec32x tan2x dx −∫sec2x tan2x dx

= [sec3 2x / 6] − [sec2x / 2] + c

Trick: Let sec 2x = t, then sec 2x tan 2x dx = [1 / 2] dt

[1 / 2] ∫(t2 − 1) dt = [1 / 6]t3 − [1 / 2]t + c

= [sec3 2x / 6] − [sec 2x / 2] + c

Question 5: If ∫{[4ex + 6e−x] / [9ex − 4e−x]} dx = Ax + B log (9e2x − 4) + C, then find A, B and C.

Solution:

I =∫{[4ex + 6e−x] / [9ex − 4e−x]} dx

= [4 / 9]∫[9e2xdx] / [9e2x − 4] + 6∫[dx] / [9e2x − 4]

∫dx / [9e2x − 4] = [1 / 8] log (9e2x − 4) − [1 / 4] log3 − [1 / 4] x + constant

I = [35 / 36] log (9e2x − 4) − [3 / 2] x − [3 / 2] log3 + constant

Comparing with the given integral, we get

A = −3 / 2, B = 35 / 36, C = [−3 / 2] log3 + constant

Question 6: ∫x cos2x dx = ______.

Solution:

x cos2x dx = [1 / 2] ∫x (1 + cos2x) dx

= [x2 / 4] + [1 / 2] [(x sin2x) / (2) −∫(sin2x / 2) dx] + c

= [x2 / 4] + (x sin2x / 4) + (cos2x / 8) + c

Question 7: ∫x / [1 + x4] dx = ________.

Solution:

Put t = x2 ⇒ dt = 2x dx, therefore,

∫x / [1 + x4] dx = [1 / 2] ∫1 / [1 + t2] dt

= [1 / 2] tan−1 t + c

= [1 / 2] tan−1 x2 + c

Question 8: ∫[1 + x2] / √[1 − x2] dx = ________.

Solution:
Put x = sinθ ⇒ dx = cosθ dθ, then it reduces to

∫(1 + sin2θ) dθ = θ + [1 / 2]∫(1 − cos2θ) dθ

= [3θ / 2] − [1 / 2] sinθ * √[1 − sin2θ] + c

= [3 / 2] sin−1 x − [1 / 2]x √[1 − x2] + c

Question 9: ∫√(1 + sin [x / 2]) dx = _________.

Solution:

∫√(1 + sin [x / 2]) dx = ∫√(sin2 [x / 4] + cos2 [x / 4] + 2 sin [x / 4] cos [x / 4]) dx

= ∫(sin [x / 4] + cos [x / 4]) dx

= 4 (sin [x / 4] – cos [x / 4]) + c

Question 10: ∫[sinx] / [sin (x − α)] dx = ________.

Solution:

∫[sinx] / [sin (x − α)] dx =

∫[sin (x − α + α)] / [sin (x − α)] dx

= ∫{[(sin (x − α) cosα + cos (x − α) sinα] / [sin (x − α)]} dx

= ∫cosα dx +∫sinα * cot (x − α) dx

= x cosα + sinα * log sin (x − α) + c

Question 11: ∫(log x)2 dx = _______.

Solution:

∫(log x)2 dx

Put log x = t

⇒ et = x

⇒ dx = et dt, then it reduces to

∫t2 * et dt = t2 * et − 2t * et + 2et + c

= x (log x)2 − 2x log x + 2x + c

Question 12: ∫[cos2θ] * log ([cosθ + sinθ] / [cosθ − sinθ]) dθ = ___________.

Solution:

We know that

log ([cosθ + sinθ] / [cosθ − sinθ]) = log ([1 + tanθ] / [1 − tanθ]) = log tan (π / 4 + θ) ∫secθ dθ = log tan (π / 4 + θ / 2)

∫sec2θ dθ = [1 / 2] * log tan (π / 4 + θ)

2 sec2θ = [d / dθ] log tan (π / 4 + θ) …….(i)

Integrating the given expression by parts, we get

I = [1 / 2] sin2θ log tan (π / 4 + θ) − [1 / 2] ∫[sin2θ * 2 sec2θ] dθ by (i)

= [1 / 2] sin2θ log tan (π / 4 + θ) −∫tan2θ dθ

= [1 / 2] sin2θ log tan (π / 4 + θ) − [1 / 2] log sec2θ

Question 13: ∫dx / (sinx + sin2x) = _________.

Solution:

I = ∫dx / [sinx (1 + 2cosx)]

= ∫[sinx dx] / [sin2x * (1 + 2 cosx)]

= ∫sinx dx / {(1 − cosx) * (1 + cosx) * (1 + 2 cosx)}

Now differential coefficient of cosx is − sinx, which is given in numerator and hence, we make the substitution cosx = t ⇒ −sinx dx = dt

I = −∫dt / [(1 − t) (1 + t) (1 + 2t)]

We split the integrand into partial fractions

I = −∫[{1 / 6 (1 − t)} − {1 / 2 (1 + t)} + {4 / 3 (1 + 2t)}] dt

[1 / 6] log (1 − cosx) + [1 / 2] log (1 + cosx) − [2 / 3] log (1 + 2 cosx)

Question 14: For which of the following values of m, the area of the region bounded by the curve y = x − x2 and the line y = mx equals 9 / 2?

Solution:

The equation of curve is y = x − x2

⇒ x2 − x = −y

(x − [1 / 2])2 = − (y − [1 / 4])

This is a parabola whose vertex is (12, 14)

Hence, point of intersection of the curve and the line x − x2 = mx

⇒ x (1 − x − m) = 0 i.e., x =0 or x = 1 − m

[9 / 2] = ∫1−m0 (x − x2 − mx) dx

= ([x2 / 2] − [x3 / 3] − [mx2 / 2])1−m0

= [(1 − m)] * [(1 − m)2 / 2] − [(1 − m)3 / 3] = [(1 − m)3/ 6]

(1 − m)3 = [6 * 9] / [2] = 27

⇒ 1 − m = 271/3 = 3

⇒ m = −2

Also, (1 − m)3 − 33 = 0

(1 − m −3) [(1 − m)2 + 9 + (1 − m)3] = 0

(1 − m)2 + 3 (1 − m) + 9 = 0

m2 − 2m + 1 − 3m + 3 + 9 = 0

m2 − 5m + 13 = 0

m = (5 ± √[25 − 52]) / 2 i.e., m is imaginary

Hence, m = −2.

Question 15: Let f : R → R and g : R → R be continuous functions, then the value of the integral ∫π/2−π/2 [f (x) + f (−x)] [g (x) − g (−x)] dx = ___________.

Solution:

Let h(x) = {f (x) + f (−x)} {g (x) − g (−x)}

h (−x) = {f (−x) + f (x)} {g (−x) − g (x)}

= −{f (−x) + f (x)} {g (x) − g (−x)}

= − h (x)

Therefore, ∫π/2−π/2 h (x) dx = 0.