Calculus is the Study of progress, it essentially examinations things that change and is a critical piece of Mathematics. Calculus is a part of Mathematics focused on limits points, integrals, functions, and infinite series. Integral calculus and Differential calculus are the two principal parts of this subject. This article covers indefinite integral, integration using partial fractions, definite integral, integration by parts, and properties of a definite integral. The integral calculus inquiries from the last years of JEE Main are available on this page, alongside the definite answer for each question. Around 2-4 inquiries are asked from this subject in JEE Main. Dowload JEE Main Sample Papers

Integral Calculus Previous Years’ Questions with Solutions

1. Question: ∫{[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

2. Question: ∫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)]

3. Question: ∫[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

4. Question: ∫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

5. Question: 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]

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

6. Question: ∫x cos2x dx = ______.

Solution:

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

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

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

Solution:

Set t = x2 ⇒ dt = 2x dx, so,

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

= [1 / 2] tan−1 t + c

= [1 / 2] tan−1 x2 + c

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

Solution:

Set x = sinθ ⇒ dx = cosθ dθ, then it decreases 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

9. Question: ∫√(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

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

Solution:

∫[sinx] / [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

11. Question: ∫(log x)2 dx = _______.

Solution:

∫(log x)2 dx

Put log x = t

⇒ et = x

⇒ dx = et dt, then it decreases to

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

12. Question: ∫[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)

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

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

Solution:

I = ∫dx / [sinx (1 + 2cosx)]

= ∫[sinx dx] / [sin2x * (1 + 2 cosx)]

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

Currently differential constant of cosx is − sinx, which is given in numerator and henceforth, we make the replacement 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)

14. Question: 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 calculation 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.

Integral Calculus – Quick Formulas for Revision

1. ∫xn dx = \frac{x^{n+1}}{n+1}+cn+1xn+1+c , n ≠ -1
2. \int \frac{1}{x}\: dx = \log_{e}\left | x \right |+c∫x1dx=loge∣x∣+c
3. ∫ex dx = ex+c
4. ∫ax dx = \frac{a^{x}}{\log_{e}a}+clogeaax+c
5. ∫ cos x dx = sin x + c
6. ∫ sec2 x dx = tan x + c
7. ∫ cosec2 x dx = -cot x + c
8. ∫ cos x dx = sin x + c
9. ∫ sin x dx = -cos x + c
10. ∫ sec x dx = log \left | \\sec \: x + tan \: x\right |+c∣secx+tanx∣+c
11. ∫ cosec x cot x dx = – cosec x + c
12. ∫ cot x dx = \log \left | \sin x \right |+clog∣sinx∣+c
13. ∫ sec x tan x dx = sec x + c
14. ∫ cosec x dx = log \left | \\cosec \: x – \cot \: x\right |+c∣cosecx–cotx∣+c
15. \int \frac{1}{\sqrt{a^{2}-x^{2}}} \; dx = \sin ^{-1}(\frac{x}{a})+c∫a2−x21dx=sin−1(ax)+c
16. \int -\frac{1}{\sqrt{a^{2}-x^{2}}} \; dx = \cos ^{-1}(\frac{x}{a})+c∫−a2−x21dx=cos−1(ax)+c
17. \int \frac{1}{{a^{2}+x^{2}}} \; dx = \frac{1}{a}\tan ^{-1}(\frac{x}{a})+c∫a2+x21dx=a1tan−1(ax)+c
18. \int -\frac{1}{{a^{2}+x^{2}}} \; dx = \frac{1}{a}\cot ^{-1}(\frac{x}{a})+c∫−a2+x21dx=a1cot−1(ax)+c
19. \int \frac{1}{x\sqrt{x^{2}-a^{2}}}\; dx = \frac{1}{a}\sec ^{-1}(\frac{x}{a})+c∫xx2−a21dx=a1sec−1(ax)+c
20. \int -\frac{1}{x\sqrt{x^{2}-a^{2}}}\; dx = \frac{1}{a}\: cosec ^{-1}\left ( \frac{x}{a} \right )+c∫−xx2−a21dx=a1cosec−1(ax)+c

Theorems on Integration

1. \int (f(x)\pm g(x))\: dx = \int f(x)\: dx\pm g(x)\: dx∫(f(x)±g(x))dx=∫f(x)dx±g(x)dx
2. ∫ f(x) dx = g(x)+c ⇒ ∫ f(ax+b) dx = \frac{g(ax+b)}{a}+cag(ax+b)+c

Integration by substitutions:

f(x) = t, then f’(x) dx = dt

Integration by Part

∫( f(x) g(x)) dx = f(x) ∫g(x) dx – \int\left ( \frac{d}{dx}(f(x))\int (g(x))dx\right ) dx∫(dxd(f(x))∫(g(x))dx)dx

1. When you search integral ∫g(x) dx so, it will not contain an arbitrary constant.
2. The special of f(x) and g(x) is decided by ILATE rule.

Integration of Type

\int \frac{dx}{ax^{2}+bx+c}∫ax2+bx+cdx , \int \frac{dx}{\sqrt{ax^{2}+bx+c}}∫ax2+bx+cdx , \int \sqrt{ax^{2}+bx+c}\; dx∫ax2+bx+cdx