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:** âˆ«{[sin^{8}x âˆ’ cos^{8}x] / [1 âˆ’ 2 sin^{2}x cos^{2}x]} dx = _________.

**Solution: **

âˆ«{[sin^{8}x âˆ’ cos^{8}x] / [1 âˆ’ 2 sin^{2}x cos^{2}x]} dx

= âˆ«{[(sin^{4}x + cos^{4}x) * (sin^{4}x âˆ’ cos^{4}x)] / [(sin^{2}x + cos^{2}x)^{2} âˆ’ 2 sin^{2}x cos^{2}x]} dx

= âˆ«(sin^{4}x – cos^{4}x) dx

= âˆ«[sin^{2}x + cos^{2}x] * [sin^{2}x – cos^{2}x] dx

= âˆ«(sin^{2}x – cos^{2}x ) dx

= âˆ«âˆ’cos2xdx

= [âˆ’sin2x / 2] + c

**Question 2:** âˆ«x^{2}dx / (a + bx)^{2} = ___________.

**Solution:**

Put a + bx = t

â‡’ x = [t âˆ’ a] / [b] and dx = dt / [b]

I =âˆ«([t âˆ’ a] / b)^{2 }* [1 / t^{2}] * [dt / b]

= [1 / b^{3}]âˆ«(1 âˆ’ (2a / t) + [a^{2} * t^{âˆ’2}]) dt

= [1 / b^{3}] * [(t âˆ’ 2a log t) âˆ’ (a^{2} / t)]

= [1 / b^{3}] [(a + bx) âˆ’ 2a * log (a + bx) âˆ’ (a^{2} / (a + bx))] + c

**Question 3: **âˆ«[x^{5} / âˆš(1 + x^{3})] dx = ________.

**Solution:**

Put 1 + x^{3 }= t^{2}

â‡’ 3x^{2 }dx = 2tdt and x^{3 }= t^{2} âˆ’ 1

So, âˆ«[x^{5} / âˆš(1 + x^{3})] dx = âˆ«{[x^{2} * x^{3}] / âˆš(1 + x^{3})} dx

= [2 / 3] âˆ«{[(t^{2 }âˆ’ 1) * t] dt / [t]}

= [2 / 3] âˆ«(t^{2 }âˆ’ 1) dt

= [2 / 3] [(t^{3} / 3) âˆ’ t] + c

= [2 / 3] [{(1 + x^{3})^{3/2} / 3} âˆ’ {(1 + x^{3})^{Â½}}]+ c

**Question 4:** âˆ«tan^{3 }2x sec 2x dx = __________.

**Solution: **

âˆ«tan^{3}2x sec 2x dx = âˆ«tan^{2}2x tan 2x sec 2x dx

= âˆ«[(sec^{2 }2x âˆ’ 1) tan 2x sec 2x ] dx ..(i)

Let sec 2x = t, then 2 sec 2x tan 2x dx = dt

=> sec 2x tan 2x dx = (1/2) dt

So (i) becomes

[1 / 2] âˆ«(t^{2 }âˆ’ 1) dt = (1/2) (t

^{3}/3 – t) + c

= (1/2) [(1/3)sec^{3} 2x – sec 2x] + c

**Question 5:** Let f : R â†’ R and g : R â†’ R be continuous functions, then the value of the integral âˆ«^{Ï€/2}_{âˆ’Ï€/2 }[f (x) + f (âˆ’x)] [g (x) âˆ’ g (âˆ’x)] dx = ___________.

**Solution:**

Let h(x) = {f (x) + f (âˆ’x)} {g (x) âˆ’ g (âˆ’x)}

h (âˆ’x) = {f (âˆ’x) + f (x)} {g (âˆ’x) âˆ’ g (x)}

= âˆ’{f (âˆ’x) + f (x)} {g (x) âˆ’ g (âˆ’x)}

= âˆ’ h (x)

Therefore, âˆ«^{Ï€/2}_{âˆ’Ï€/2 }h (x) dx = 0. (odd function)

**Question 6:** âˆ«x cos^{2}x dx = ______.

**Solution: **

âˆ«x cos^{2}x dx = [1 / 2] âˆ«x (1 + cos2x) dxÂ (replace cos^{2}x by (1/2)(1+cos 2x))

= [x^{2} / 4] + [1 / 2] [(x sin2x) / (2) âˆ’âˆ«(sin2x / 2) dx] + c

= [x^{2} / 4] + (x sin2x / 4) + (cos2x / 8) + c

**Question 7:** âˆ«x / [1 + x^{4}] dx = ________.

**Solution:**

Put t = x^{2} â‡’ dt = 2x dx, therefore,

âˆ«x / [1 + x^{4}] dx = [1 / 2] âˆ«1 / [1 + t^{2}] dt

= [1 / 2] tan^{âˆ’1 }t + c

= [1 / 2] tan^{âˆ’1 }x^{2 }+ c

**Question 8:** âˆ«[1 + x^{2}] / âˆš[1 âˆ’ x^{2}] dx = ________.

**Solution:**

Put x = sinÎ¸ â‡’ dx = cosÎ¸ dÎ¸, then it reduces to

âˆ«(1 + sin^{2}Î¸) dÎ¸ = Î¸ + [1 / 2]âˆ«(1 âˆ’ cos2Î¸) dÎ¸

= [3Î¸ / 2] âˆ’ [1 / 2] sinÎ¸ * âˆš[1 âˆ’ sin^{2}Î¸] + c

= [3 / 2] sin^{âˆ’1 }x âˆ’ [1 / 2]x âˆš[1 âˆ’ x^{2}] + c

**Question 9: **âˆ«âˆš(1 + sin [x / 2]) dx = _________.

**Solution:**

âˆ«âˆš(1 + sin [x / 2]) dx = âˆ«âˆš(sin^{2 }[x / 4] + cos^{2} [x / 4] + 2 sin [x / 4] cos [x / 4]) dx

= âˆ«(sin [x / 4] + cos [x / 4]) dx

= 4 (sin [x / 4] – cos [x / 4]) + c

**Question 10:** âˆ«[sinx] / [sin (x âˆ’ Î±)] dx = ________.

**Solution: **

âˆ«[sinx] / [sin (x âˆ’ Î±)] dx =

âˆ«[sin (x âˆ’ Î± + Î±)] / [sin (x âˆ’ Î±)] dx

= âˆ«{[(sin (x âˆ’ Î±) cosÎ± + cos (x âˆ’ Î±) sinÎ±] / [sin (x âˆ’ Î±)]} dx

= âˆ«cosÎ± dx +âˆ«sinÎ± * cot (x âˆ’ Î±) dx

= x cosÎ± + sinÎ± * log |sin (x âˆ’ Î±)| + c

**Question 11:** âˆ«(log x)^{2} dx = _______.

**Solution: **

âˆ«(log x)^{2} dx

Put log x = t

â‡’ e^{t} = x

â‡’ dx = e^{t} dt, then it reduces to

âˆ«t^{2} * e^{t} dt = t^{2 }* e^{t }âˆ’ 2t * e^{t }+ 2e^{t} + c

= x (log x)^{2 }âˆ’ 2x log x + 2x + c

**Question 12:** âˆ«[cos2Î¸] * log ([cosÎ¸ + sinÎ¸] / [cosÎ¸ âˆ’ sinÎ¸]) dÎ¸ = ___________.

**Solution:**

We know that

log ([cosÎ¸ + sinÎ¸] / [cosÎ¸ âˆ’ sinÎ¸]) = log ([1 + tanÎ¸] / [1 âˆ’ tanÎ¸]) = log tan (Ï€ / 4 + Î¸)

âˆ«secÎ¸ dÎ¸ = log tan (Ï€ / 4 + Î¸ / 2)

âˆ«sec2Î¸ dÎ¸ = [1 / 2] * log tan (Ï€ / 4 + Î¸)

2 sec2Î¸ = [d / dÎ¸] log tan (Ï€ / 4 + Î¸) â€¦….(i)

Integrating the given expression by parts, we get

I = [1 / 2] sin2Î¸ log tan (Ï€ / 4 + Î¸) âˆ’ [1 / 2] âˆ«[sin2Î¸ * 2 sec2Î¸] dÎ¸ by (i)

= [1 / 2] sin2Î¸ log tan (Ï€ / 4 + Î¸) âˆ’âˆ«tan2Î¸ dÎ¸

= [1 / 2] sin2Î¸ log tan (Ï€ / 4 + Î¸) âˆ’ [1 / 2] log sec2Î¸

**Question 13: **âˆ«dx / (sinx + sin2x) = _________.

**Solution:**

I = âˆ«dx / [sinx (1 + 2cosx)]

= âˆ«[sinx dx] / [sin2x * (1 + 2 cosx)]

= âˆ«sinx dx / {(1 âˆ’ cosx) * (1 + cosx) * (1 + 2 cosx)}

Now differential coefficient of cosx is âˆ’ sinx, which is given in numerator and hence, we make the substitution cosx = t â‡’ âˆ’sinx dx = dt

I = âˆ’âˆ«dt / [(1 âˆ’ t) (1 + t) (1 + 2t)]

We split the integrand into partial fractions

I = âˆ’âˆ«[{1 / 6 (1 âˆ’ t)} âˆ’ {1 / 2 (1 + t)} + {4 / 3 (1 + 2t)}] dt

[1 / 6] log (1 âˆ’ cosx) + [1 / 2] log (1 + cosx) âˆ’ [2 / 3] log (1 + 2 cosx) + c**Question 14: **For which of the following values of m, the area of the region bounded by the curve y = x âˆ’ x^{2} and the line y = mx equals 9 / 2?

**Solution:**

The equation of curve is y = x âˆ’ x^{2}

Given line y = mx

The point of intersection of the curve and the line x âˆ’ x^{2 }= mx

â‡’ x (1 âˆ’ x âˆ’ m) = 0 i.e., x =0 or x = 1 âˆ’ m

[9 / 2] = âˆ«^{1âˆ’m}

_{0 }(x âˆ’ x

^{2}âˆ’ mx) dx

= ([x^{2} / 2] âˆ’ [x^{3} / 3] âˆ’ [mx^{2} / 2])^{1âˆ’m}_{0}

= [(1 âˆ’ m)] * [(1 âˆ’ m)^{2 }/ 2] âˆ’ [(1 âˆ’ m)^{3 }/ 3] = [(1 âˆ’ m)^{3}/ 6]

(1 âˆ’ m)^{3} = [6 * 9] / [2] = 27

â‡’ 1 âˆ’ m = 27^{1/3 }= 3

â‡’ m = âˆ’2

When m> 1, 1-m < 0.

[(1-m)(x^{2}/2) – x

^{3}/3]

_{1-m}

^{0}= 9/2

-(1-m)^{3} (1/2Â –Â 1/3) = 9/2

=> -(1-m)^{3} = 27

=>(1-m) = -3

so m = 4

Hence, m = âˆ’2, 4.

## Integral Calculus – Top JEE Questions

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JEE Advanced Maths Indefinite Integration Previous Year Questions with Solutions

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