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Integral Calculus carries one of the highest weightages in JEE Main Mathematics. Around 3 questions are generally asked from this chapter carrying a total of 12 marks. Integral Calculus is considered to be a very hard topic but also a scoring one. It is highly recommended that you refer to NCERT books in the beginning and then progress to Arihant. Regular practice will help you master solving Integral Calculus questions. **Check JEE Main Mathematics Syllabus**

The chapter on Integral Calculus carries a weight of around 3-4% in JEE Main.

Questions are asked from topics such as indefinite integral, definite integral, integration using partial fractions, integration by parts, and properties of definite integral.

Read the article to discover the required preparation notes on Integral Calculus that will aid you in acing this section.

**Must Read:**

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

Integral calculus can be used in the figure given above to find out the area between the two curves, F1 and F2.

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.

**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

In order to write P(x)/ Q(x) in partial fractions, make Q(x) in the form Q(x) = (x – a)k ... (x2 + αx + β)r where the binomials are varied, and then set -

where A1, A2, ...,Ak, M1, M2, ......, Mr, N1, N2, ...... ,Nr are real constants that are to be determined.

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).

In integrals of the form:

1) Substitute px + q = l (that is the differential coefficient of ax2 + bx + c) + m.

2) Find l and m through comparison of the coefficient of x and constant term on both sides of the identity.

3) The question will now reduce to the sum of two integrals that can be easily integrated.

For integrals of the type:

1) Substitute ax2 + bx + c = M (px2 + qx + r) + N (2px + q) + R.

2) Find M, N, and R.

3) The integration lowers to integration of three independent functions.

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.

For integrals of the form:

To compute the indefinite integrals of the form:

1) Write ax2 +bx+c = A1 (dx +e) ( 2fx +g) +B1( dx +e) +C1, where A1, B1 and C1 are constants that can be found through comparison of the coefficient of similar terms on both the sides.

2) The integral given will now reduce to the form -

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 given below will be helpful -

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

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

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 decomposition of f(x)./g(x) into partial fractions.

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

Suppose , then the equation f(x) = 0 contains a minimum if one root lying in (a, b) only if f is a continuous function in (a, b).

Suppose the function f is the same, then,

dx, where c is any point that lies outside or inside [a, b].

This will prove to be true only if f is piecewise continuous in (a, b).

if f(x) = -f(-x). Here, f is an odd function.

if f(x) = f(-x). Here, f is an even function.

. Here, f is a periodic function with period a.

. Here, f(a+x) = f(x). a is the period of the function f.

If f(x) ≤ φ[x] for a ≤ x ≤ b then

Gamma Function:

Suppose n is a positive rational number, then the improper integral

is described as the gamma function.

Γ(n+1) = n!

Γ1 = 1

Γ0 = ∞

Γ(1/2) = √π

Suppose f(x) ≥ 0 on the interval [a, b], then

Walli’s Formula:

Here, K = , when both m and n are even (m, n ϵ N)= 1 otherwise

Leibnitz’s Rule:

Suppose h(x) and g(x) are differentiable functions of x, then

For a monotonically increasing function in (a, b)

. Here, f(x) is a continuous function on [a, b] and F’(x) = f(x).

Suppose f(x) = f(a – x), then

Suppose f(x) = – f (a – x), then

Suppose f(x) is a periodic function with period T, then

In the result given above, if n = 1, then,

The definite integral f(x)dx is indeed a limiting case of the addition of an infinite series, given that f(x) is continuous on [a, b] i.e.,

where h = b – a/n.

The inverse is also true, meaning, if there is an infinite series in the form given above, it can be represented as a definite integral.

The method to represent the infinite series as definite integral is -

1. Represent the series given in the form Σ 1/n f (r/n).

2. Then, the limit will be its sum when n → ∞, i.e. limn→∞ h Σ 1/n f(r/n).

3. Replace r/n with x, 1/n with dx, and limn→∞ Σ with the sign of ∫.

4. For the first and the last terms of r, the lower and the upper limit integration are the limiting values of r/n respectively,

Some particular case of the above are:

Vital results:

Given below is a list of important rules that form the basis of solving definite integral numerical problems -

1) .

The limits can be interchanged on any definite integral. This can be done by simple adding a minus sign on the integral.

2) .

The value of the integral is zero when the upper and lower limits coincide.

(3) , where c is any number.

A constant can be removed from the integral sign for both, definite and indefinite integrals.

(4) .

Across a sum or difference, a definite integral can be divided into parts.

(5) , where c is any number.

This property explains how a function can be integrated over adjacent intervals, [a,c] and [c,b].

Note: It is not mandatory for c to be between a and b.

(6)

This property proves that as long as the limits and function are the same, the variable that is utilized for integration does not create any difference.

**Ques 1. Solve **

**Solution:**

=

=

=

**Ques 2. 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.

**Ques 3. 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

**Ques 4. 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

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.

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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