JEE Main Study Notes for Integral Calculus: Integral Calculus is also known as integration. It helps to find out the antiderivatives of a function. This antiderivative is also called the integrals of a function. Integral Calculus helps to evaluate quantities like arc length, volume, area, work, etc. The topic of Integral Calculus carries a weight of around 3-4% in JEE Main.

• JEE Main Mathematics Question Paper has a total of 12 marks on the topic of Integral Calculus.
• JEE Main Study Notes for Integral Calculus include topics such as Indefinite Integral, Definite Integral, Integration using partial fractions, Integration by Parts, and Properties of Definite Integral.

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What is Integral Calculus?

Integral Calculus is that part of the calculus that deals with integration and its usage in the solution of differential equations and the determination of areas or volumes.;

Indefinite Integral

An integral of a particular form, without the upper and lower limits, is known as an indefinite integral. Indefinite integrals are usually represented as –

where c is an arbitrary constant.

Indefinite integrals are even known as Antiderivatives

Check:

• JEE Main Study Notes on Differential Calculus

Take a look at the following example:

The function F(x) = x3/3 is an indefinite integral of f(x) = x2. Since the derivative of constant is zero, x2 is sure to have an infinite number of indefinite integrals such as (x3/3) + 0, (x3/3) + 7, (x3/3) − 42, (x3/3) + 293, and so on. Therefore, all the indefinite integrals of x2 can be found by tweaking the value of C in F(x) = (x3/3) + C

• Suppose the integrand is a derivative of a known function, the corresponding indefinite integral can be assessed directly.
• Suppose the integrand is not a derivative of a known function, the integral can be assessed using the help of any of the three rules given below -

1) Integration by substitution or changing the independent variable

2) Integration by parts

3) Integration by partial fractions

• A few indefinite integrals can be assessed through direct substitutions:

1) Suppose an integral is of the form ∫ f(g(x)) g'(x) dx, then enter g(x) = t, if ∫ f(t) exists.

2) ∫ f'(x)/f(x) dx = ln |f (x)| + c, by entering f (x) = t => f' (x) dx = dt

=> ∫ dt/t = ln |t| + c = ln |f (x)| + c.

3) ∫ f'(x)√f(x) dx = 2 √f(x)+c, Enter f (x) = t

Then, ∫ dt/√t = 2√t + c = 2√f(x) + c.

• A few standard substitutions are given below:

1) For terms with the form x2 + a2 or √x2 + a2, enter x = a tanθ or a cotθ.

2) For terms with the form x2 - a2 or √x2 – a2 , enter x = a sec θ or a cosecθ.

3) For terms with the form a2 - x2 or √x2 + a2, enter x = a sin θ or a cosθ.

4) When both √a+x and √a–x are present, enter x = a cos θ.

5) For the form √(x–a)(b–x), enter x = a cos2θ + b sin2θ.

6) For the type (√x2+a2±x)n or (x±√x2–a2)n, enter the expression inside the bracket = t.

7) For 1/(x+a)n1 (x+b)n2, where n1,n2 ∈ N (and > 1), enter (x + a) = t (x + b).

• Suppose the integrand is of the form f(x)g(x), where g(x) is a function of the integral of f(x), enter the integral of f(x) = t.
• The integral product of two functions of x is assessed using integration by parts. Assume that u and v are two functions of x, then ∫uv dx = u∫v dx - ∫[du/dx ∫v dx]dx.

• While doing integration by parts, whether a function is u or v must be determined according to the ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent) method of integration.
• When both the functions are directly integrable, then the first function is selected in such a manner that the derivative of the function that is thus obtained under the integral sign is easily integrable.
• Suppose in the product of the two functions, one of the functions is not directly integrable like lnx, sin-1x, cos-1x, tan-1x, then it is taken as the first function and the remaining function gets taken as the second function.
• When there is no second function available, unity is taken as the second function. For example, in the integration of∫tan-1x dx, tan-1x is considered as the first function and 1 is the second function.
• In the integral ∫g(x)exdx, when g(x) can be expressed as g(x) = f(x) + f'(x), then ∫g(x)ex dx=∫ ex [f(x) + f'(x)] dx = exf(x) + c.
• A rational function P(x)/Q(x) is appropriate when the degree of polynomial Q(x) is higher when compared to the degree of the polynomial P(x).
• When the degree of P(x) is higher than or equal to the degree of Q(x), first P(x)/Q(x) = h(x) + P(x)/Q(x) is written, where h(x) is a polynomial and p(x) is a polynomial of a degree lower than the degree of the polynomial Q(x).
• Irrational functions of the form (ax+b)1/nand x can be evaluated easily by the substitution tn = ax+ b. Hence, ∫f(x, (ax + b)1/n)dx = ∫ f ((tn– b)/ a , t) ntn-1 / a dt.
• The integrals of the form ∫ R (sin x, cos x) dx can be answered by substituting tan(x/2) = t. Thus, the remaining the trigonometric functions can be substituted as sin x = 2t/(1+t2), cos x = (1-t2)/(1+t2) , x = 2 tan-1t, and hence, dx = dt/(1+t2).
• Most of the integrals of the form ∫ R (sin x, cos x) dx can be answered using the method given above, but in some situations, the substitution cot x/2 = t may be useful. A few other substitutions that can be used for certain cases include:

1) When R(-sin x, cos x) = -R(sin x, cos x), substitute cos x = t

2) When R(sin x, -cos x) = -R(sin x, cos x), substitute sin x = t

3) When R(-sin x, - cos x) = R(sin x, cos x), substitute tan x = t

• For integrals of the form ∫ (pcosx + q sinx + r) / (a cos x + b sin x + c) dx:

1) Represent the numerator as m (denominator) + l (differential coefficient of denominator) + n.

2) Then, compute m, l, and n through comparison of the coefficients of sin x, cos x, and constant term and split the integral into a sum of three integrals.

3) This can be represented as l∫ dx + m ∫d.c. of (Denominator) / denominator dx + n ∫ dx/ (a cos x + b sin x + c).

• Integrals of the form ∫ (pcos x + q sin x) / (a cos x + b sin x) dx can be answered through representing the numerator as l (denominator) + m(d.c. of denominator) and then find l and m as mentioned above.
• For integrals of the type ∫(sinmx cosnx)dx, where m, n ∈ natural numbers, the substitutions are given below will be helpful -

1) When one of them is odd, substitute for the term even power.

2) When both are odd, substitute any one of the terms.

3) When both are even, use only trigonometric identities.

• The different indefinite integral formulae that must be memorized because they are very helpful in solving problems are -

• 1. ∫ exdx = ex + c
• 2. ∫ 1/x dx = ln |x| + c
• 3. ∫ ax dx = ax/ ln a + c (a > 0)
• 4. ∫ cos x dx = sin x + c
• 5. ∫ sin x dx = - cos x + c
• 6. ∫ sec2 x dx = tan x + c
• 7. ∫ cosec x cot x dx = - cosec x + c
• 8. ∫sec x tan x dx = sec x + c
• 9. ∫ cosec2 x dx = - cot x + c

Must Read:

• JEE Main Study Notes for Trigonometry
• JEE Main Study Notes for Complex Number

Integration by parts

This method is used for integrating the product of two functions. When f(x) and g(x) are two integrable functions, then,

∫f(x) g(x) dx = f(x) ∫ g(x) dx – ∫ {d/dx (f(x)) . ∫g(x) dx} dx.

• To select the first function, the order given below is followed -

Inverse → Logarithmic → Algebraic → Trigonometric → Exponential

Integration using partial fractions

Suppose f(x) and g(x) are two polynomials and deg (f(x)) < deg (g(x)), then f(x)/g(x) is known as a proper rational fraction.

When deg (f(x)) ≥ deg (g(x)), then f(x)/g(x) is known as an improper rational fraction.

When f(x)/g(x) is an improper rational function, divide f(x) by g(x) and change it to a proper rational function such as f(x) / g(x) = l(x) + h(x)/g(x).

Any proper rational function f(x)/g(x) can be represented as the sum of rational functions that each have a factor of g(x). Every one of these factors is known as a partial fraction and the process of deriving them is described as the decomposition of f(x)./g(x) into partial fractions.

Check: JEE Main Study Notes for Probability

Definite Integral

If given a function f(x) which is continuous on the interval [a,b], the interval is divided into n subintervals with an equal width, Δx , and from each interval, a point x1* is chosen. Following this, the definite integral of f(x) from a to b would be -

Properties of Definite Integral;

If you add a minus sign on the interval, the limit of the definite integral will interchange.

The value of the integral will be zero if the upper and lower limit coincide.

A definite integral can be broken into parts across a sum or difference.

• ; where c is any number,

By this property, we can get the method of integrating a function over adjacent intervals, [a,c] and [c,b]. There is no need to be between a and b.

This property shows that as long as the function and limits are the same, the variable used for integration does not make any difference.

Example 1: Let f(x) = x/(1+xn)1/n for n ≥ 2 and g(x) = (1,2,3,4,… n times) (x). Then find the value of ∫xn-2 g(x) dx.

Solution: The value of f(x) is given to be f(x) = x/(1+xn)1/n

Then ff(x) = f(x)/(1 + f(x)n)1/n

= x/(1+2xn)1/n

So, fff(x) = x/(1+3xn)1/n

Hence, g(x) = (fofof…..n times)(x) = x/(1+nxn)1/n

Hence, I = ∫xn-2 g(x) dx

= ∫ xn-1dx/(1 + nxn)1/n

= 1/n2 ∫ n2xn-1dx/(1 + nxn)1/n

= 1/n2 ∫ [d/dx (1 + nxn)] / (1 + nxn)1/n . dx

Hence, I = [(1+ nxn)1-1/n ]/ n(n-1) + K.

Example 2: Evaluate the given expression

∫ dx/ [ x2(x4 + 1)]3/4

Solution: Let I = ∫ dx/ [x2(x4 + 1)]3/4

= ∫ dx/ x2x3(1+x-4)3/4

= ∫ dx/ x5(1+x-4)

Put 1+x-4 = z

Then -4x-5 dx = dz

Hence, I = ∫ -dz/4z3/4

= -1/4 ∫ dz/z3/4

= -z1/4 + c

Example 3: Evaluate ∫ (x +1)/ x(1+xex)2 dx.

Solution: Let I = ∫ (x +1)/ x(1+xex)2 dx

= ∫ ex (x +1)/ xex (1+xex)2 dx

Put (1+xex) = t, then (ex+xex) dx = dt

Hence, I = ∫ dt/ (t-1)t2 = ∫[1/(t-1) – 1/t – 1/t2] dt

= log |t-1| - log |t| +1/t + c

= log | [(t-1)/t] | +1/t + c

= log | xex/ (1+xex)| + 1/ (1+xex) + c

Solved Examples of Integral Calculus

Question 1: If f(x) = x/(1+xn)1/n for n ≥ 2 and g(x) = (fofofofof… n times) (x), find the value of ∫xn-2 g(x) dx.

Solution: The value of f(x) is given as f(x) = x/(1+xn)1/n

Then, ff(x) = f(x)/(1 + f(x)n)1/n

= x/(1+2xn)1/n

So, fff(x) = x/(1+3xn)1/n

Hence, g(x) = (fofof…..n times)(x) = x/(1+nxn)1/n

Hence, I = ∫xn-2 g(x) dx

= ∫ xn-1dx/(1 + nxn)1/n

= 1/n2 ∫ n2xn-1dx/(1 + nxn)1/n

= 1/n2 ∫ [d/dx (1 + nxn)] / (1 + nxn)1/n . dx

Hence, I = [(1+ nxn)1-1/n ]/ n(n-1) + K.

Question 2: Evaluate the given expression ∫ dx/ [ x2(x4 + 1)]3/4

Solution: Let I = ∫ dx/ [x2(x4 + 1)]3/4

= ∫ dx/ x2x3(1+x-4)3/4

= ∫ dx/ x5(1+x-4)

Put 1+x-4 = z

Then, -4x-5 dx = dz

Hence, I = ∫ -dz/4z3/4

= -1/4 ∫ dz/z3/4

= -z1/4 + c

Question 3: Evaluate ∫ (x +1)/ x(1+xex)2 dx.

Solution: Let I = ∫ (x +1)/ x(1+xex)2 dx

= ∫ ex (x +1)/ xex (1+xex)2 dx

Put (1+xex) = t, then (ex+xex) dx = dt

Hence, I = ∫ dt/ (t-1)t2 = ∫[1/(t-1) – 1/t – 1/t2] dt

= log |t-1| - log |t| +1/t + c

= log | [(t-1)/t] | +1/t + c

= log | xex/ (1+xex)| + 1/ (1+xex) + c

Question 4: ∫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 5: ∫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 6: ∫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

Check: JEE Main Study Notes on Properties of Determinants

JEE Main Integral Calculus Study Notes: Tips for Preparation

Below are the preparation tips for JEE Main Integral Calculus.

• The candidates should divide their schedules into different parts.
• Follow the books of R D Sharma, Arihant, Tata McGraw-Hill, etc.
• ;When it comes to math, memorizing concepts or rote learning should be avoided. This subject necessitates stronger practical and analytical abilities. It's critical that you grasp the concepts completely and have answers to both the "How" and "Why" questions.
• Students should begin their JEE Main preparations as soon as possible. Students usually begin in the eleventh grade. During this time, they will go over JEE Math syllabus for JEE Main exams. Learn about the crucial topics and plan your preparation accordingly.

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Tricks to Attempt Integral Calculus Questions

1. Try differentiating similar solutions for indefinite integration. You can differentiate all options within two minutes if you are quick at differentiation.;
2. Questions for definite integration are generally tough but try to use the King's rule - f(a+b-x) = f(x) integral from limits a to b. You could also use the Queen's rule to change the limits for trigonometric functions to half the initial value.

Points to Remember while Studying Mathematics for JEE Main

1. Focus on Algebra and Calculus
2. Prepare a list of all the formulae and study from it.
3. Revise applications of geometry and differential equations.
4. Solve mock tests regularly.
5. Use appropriate reference material like NCERT books for practice.

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