If ∫( 1-5 ²x ⁵x ²x)dx=f(x)+C where C is the constant of integration, then f((π)/(6))-f((π)/(4)) is equal to

Solution & Explanation

Core Logic

Split the integral into two terms:

I = ∫ (1)/( ⁵ x ² x) dx - ∫ (5 ² x)/( ⁵ x ² x) dx I = ∫ ( ² x)/( ⁵ x) dx - 5 ∫ (1)/( ⁵ x) dx
Step 1: Integration by Parts

Take the first integral I₁ = ∫ (1)/( ⁵ x) · ² x dx. Apply Integration by Parts taking u = (1)/( ⁵ x) and dv = ² x dx:

I₁ = (1)/( ⁵ x) · x - ∫ ( -(5 x)/( ⁶ x) ) x dx I₁ = ( x)/( ⁵ x) + 5 ∫ ( x)/( ⁶ x) · ( x)/( x) dx I₁ = ( x)/( ⁵ x) + 5 ∫ (1)/( ⁵ x) dx
Step 2: Re-substituting into Total Integral
I = ( ( x)/( ⁵ x) + 5 ∫ (1)/( ⁵ x) dx ) - 5 ∫ (1)/( ⁵ x) dx

The integral terms exactly cancel out:

I = ( x)/( ⁵ x) + C

Thus, f(x) = ( x)/( ⁵ x).

Step 3: Function Evaluation

Calculate f((π)/(6)):

f((π)/(6)) = ( (π/6))/( ⁵(π/6)) = 1/√(3)(1/2)⁵ = 32√(3)

Calculate f((π)/(4)):

f((π)/(4)) = ( (π/4))/( ⁵(π/4)) = 1(1/√(2))⁵ = (√(2))⁵ = 4√(2)

Difference:

f((π)/(6)) - f((π)/(4)) = 32√(3) - 4√(2) = 4√(3)(8 - √(6))
Pattern Recognition

Integrals of the form ∫ (complicated term 1 - complicated term 2) dx often cancel perfectly when Integration by Parts is applied to just one of the terms. Look for derivative relationships between the numerators.

Chapter Mix

Class 12 Mathematics: Integrals

Reference Study Guides

More Integrals Previous-Year Questions — Page 10

Q28 jee_main_2024_01_february_morning Properties of Definite Integrals
If ∫-π/2π/2 8√(2) x dx(1+ex)(1+ ⁴x)=απ+β ₑ(3+2√(2)), where α, β are integers, then α²+β² equals
Numerical Answer. Answer: 8 to 8

Solution

Related Formula

King's Property for symmetric integration limits:

∫₋ₐa f(x) dx = ∫₋ₐa f(-x) dx
Core Logic

Let the given definite integral be I:

I = ∫-π/2π/2 8√(2) x dx(1+ex)(1+ ⁴x) (1)

Apply King's property by replacing x with -x (since -π/2 + π/2 = 0):

I = ∫-π/2π/2 8√(2) (-x) dx(1+e(-x))(1+ ⁴(-x)) I = ∫-π/2π/2 8√(2) x dx(1+e- x)(1+ ⁴x) = ∫-π/2π/2 8√(2) x · ex dx(ex+1)(1+ ⁴x) (2)
Step 1: Simplify by Adding Expressions

Adding equations (1) and (2):

2I = ∫-π/2π/2 8√(2) x(1 + ex) dx(1+ex)(1+ ⁴x) 2I = ∫-π/2π/2 8√(2) x dx1+ ⁴x

Since the integrand is even, we can change the limits from 0 to π/2:

2I = 2 ∫₀π/2 8√(2) x dx1+ ⁴x I = ∫₀π/2 8√(2) x dx1+ ⁴x
Step 2: Substitution and Algebraic Deconstruction

Let x = t x dx = dt. Limits change from 0 to 1:

I = ∫₀¹ 8√(2) dt1+t⁴ = 4√(2) ∫₀¹ (2 dt)/(1+t⁴)

Dividing the numerator and denominator by t², we write it as two distinct expressions:

I = 4√(2) [ ∫₀¹ (1+(1)/(t²))/(t²+(1)/(t²)) dt - ∫₀¹ (1-(1)/(t²))/(t²+(1)/(t²)) dt ] I = 4√(2) [ ∫-∞⁰ (dz)/(z²+2) - ∫∞² (dk)/(k²-2) ]

where z = t - (1)/(t) and k = t + (1)/(t).

Step 3: Integrate and Evaluate Parameters

Evaluating standard anti-derivatives:

I = 4√(2) [ 1√(2) ⁻¹( z√(2)) ]-∞⁰ - 4√(2) [ 12√(2)ln| k-√(2)k+√(2)| ]∞² I = 4√(2)( π2√(2)) - 2 ln| 2-√(2)2+√(2)| = 2π - 2ln(√(2)-1)² I = 2π + 2ln(3+2√(2))

Matching parameters yields α = 2 and β = 2. Thus:

α² + β² = 2² + 2² = 8
Pattern Recognition

Sees: Exponential variables causing asymmetry in symmetric integration bounds. Shortcut: Using King's rule completely eliminates the confusing ex factor, leaving behind a straightforward rational trigonometric configuration.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Q10 jee_main_2024_29_january_evening Indefinite Integration
If ∫ (3)/(2) x + (3)/(2) x√( ³ x ³ x (x - θ)) d x = A √( θ x - θ) + B √( θ - θ x) + C, where C is the integration constant, then AB is equal to
  • A. 4sec(2θ)
  • B. 4 θ
  • C. 2 θ
  • D. 8secec(2θ)

Solution

Related Formula
(x - θ) = x θ - x θ
Core Logic

Let us partition the given integral I into two clean parts I₁ and I₂:

I = ∫ 3/2 x√( ³ x ³ x ( x θ - x θ)) dx + ∫ 3/2 x√( ³ x ³ x ( x θ - x θ)) dx

Factoring out appropriate powers of x and x from inside the roots:

I₁ = ∫ ² x√( x θ - θ) dx I₂ = ∫ ² x√( θ - x θ) dx
Step 1: Evaluating the Integrals

For I₁, substitute x θ - θ = t² ² x dx = (2t dt)/( θ):

I₁ = ∫ (2t dt)/(t θ) = (2t)/( θ) = 2 θ √( x θ - θ)

For I₂, substitute θ - x θ = z² ² x dx = (2z dz)/( θ):

I₂ = ∫ (2z dz)/(z θ) = (2z)/( θ) = 2 θ √( θ - x θ)
Step 2: Combining Coefficients

Comparing with the given expression, we find the coefficients:

A = 2 θ, B = 2 θ

Multiplying them together:

AB = 4 θ θ = (4)/( θ θ) = (8)/(2 θ θ) = 8 (2θ)
Pattern Recognition

When integrating roots of trigonometric functions involving (x - θ), try dividing/multiplying fields by ⁿ x or ⁿ x to force x / ² x templates.

Chapter Mix

Class 12 Mathematics: Integrals

Q24 jee_main_2024_29_january_evening Definite Integration
If ∫(π)/(6)(π)/(3)√(1 - 2x) dx = α +β √(2) +γ √(3) where α ,β and γ are rational numbers, then 3α + 4β - γ is equal to
Numerical Answer. Answer: 6 to 6

Solution

Related Formula
√(1 - 2x) = √(( x - x)²) = | x - x|
Core Logic

The absolute function | x - x| flips sign at x = (π)/(4) within the boundary range:

  • For x in [(π)/(6), (π)/(4)]: x ≥ x | x - x| = x - x
  • For x in [(π)/(4), (π)/(3)]: x ≥ x | x - x| = x - x
Step 1: Boundary Splitting Integration
I = ∫(π)/(6)(π)/(4) ( x - x) dx + ∫(π)/(4)(π)/(3) ( x - x) dx I = [ x + x ](π)/(6)(π)/(4) + [ - x - x ](π)/(4)(π)/(3) I = ( 1√(2) + 1√(2) - ((1)/(2) + √(3)2) ) + ( -(1)/(2) - √(3)2 - (- 1√(2) - 1√(2)) ) I = ( √(2) - 1 + √(3)2 ) + ( √(2) - 1 + √(3)2 ) = 2√(2) - 1 - √(3) = -1 + 2√(2) - √(3)
Step 2: Coefficient Matrix Matching

Comparing with the given baseline layout:

α = -1, β = 2, γ = -1

Evaluating the required target metrics:

3α + 4β - γ = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6
Pattern Recognition

Never forget that √(f(x)²) = |f(x)|. Skipping modulus checks inside definite root boundaries leads to incorrect answers.

Chapter Mix

Class 12 Mathematics: Integrals

Q jee_main_2024_27_jan_morning Properties of Definite Integrals
If (a, b) be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1), and I₁=∫ₐbx~sin(4x-x²)dx, I₂=∫ₐbsin(4x-x²)dx, then 36 I₁I₂ is equal to:
  • A. 72
  • B. 88
  • C. 80
  • D. 66

Solution

Related Formula
∫ₐ^b f(x) dx = ∫ₐ^b f(a+b-x) dx (King's Rule)
Core Logic

First, find the orthocentre (a,b) of Δ ABC with vertices A(1, 2), B(2, 3), and C(3, 1). Slope of AB = (3-2)/(2-1) = 1. The altitude from C onto AB must be perpendicular to AB, so its slope is -1. Equation of altitude from C(3,1):

y - 1 = -1(x - 3) ⇒ x + y = 4

The orthocentre (a, b) lies on all altitudes, including this one. Thus, it satisfies a + b = 4.

Step 1: Applying Definite Integral Properties

Given I₁ = ∫ₐ^b x (4x-x²) dx, let's rewrite the argument of sine:

4x - x² = x(4-x)

Apply King's Rule replacing x with (a+b-x). Since we proved a+b = 4, substitute x with (4-x):

I₁ = ∫ₐ^b (4-x) ((4-x)(4 - (4-x))) dx I₁ = ∫ₐ^b (4-x) ((4-x)x) dx I₁ = ∫ₐ^b (4-x) (4x-x²) dx
Step 2: Evaluating the Integral Ratio

Expand the newly formed integral:

I₁ = 4 ∫ₐ^b (4x-x²) dx - ∫ₐ^b x (4x-x²) dx

Notice that the second term is I₁ and the first integral is I₂:

I₁ = 4I₂ - I₁ ⇒ 2I₁ = 4I₂ ⇒ (I₁)/(I₂) = 2
Step 3: Final Output Evaluation

We need the value of 36 (I₁)/(I₂):

36 × 2 = 72
Pattern Recognition

Whenever you see ∫ₐ^b x · f(x(a+b-x)) dx, immediately apply King's Rule to factor out x. You rarely need the individual values of the integration bounds, only their sum.

Chapter Mix

Class 11 Maths: Straight Lines Class 12 Maths: Definite Integration

Q9 jee_main_2024_27_jan_morning Integration of Irrational Functions
If ∫₀¹ 1√(3+x)+√(1+x)dx=a+b√(2)+c√(3), where a, b, c are rational numbers, then 2a+3b-4c is equal to:
  • A. 4
  • B. 10
  • C. 7
  • D. 8

Solution

Related Formula
∫ xⁿ dx = xⁿ⁺¹n+1
Core Logic

To evaluate integrals with sum of square roots in the denominator, multiply and divide by the conjugate to rationalize it.

I = ∫₀¹ √(3+x)-√(1+x)(√(3+x)+√(1+x))(√(3+x)-√(1+x))dx I = ∫₀¹ √(3+x)-√(1+x)(3+x) - (1+x)dx I = (1)/(2) ∫₀¹ (√(3+x) - √(1+x)) dx
Step 1: Integration and Bounds Setup

Integrate the resulting expression:

I = (1)/(2) [ (3+x)3/23/2 - (1+x)3/23/2 ]₀¹ I = (1)/(2) · (2)/(3) [ (3+x)3/2 - (1+x)3/2 ]₀¹ I = (1)/(3) [ ((4)3/2 - (2)3/2) - ((3)3/2 - (1)3/2) ]
Step 2: Term Simplification

Evaluate the boundary powers: 43/2 = 8 23/2 = 2√(2) 33/2 = 3√(3) 13/2 = 1

Substitute back into the expression:

I = (1)/(3) [ 8 - 2√(2) - 3√(3) + 1 ] = (1)/(3) [ 9 - 2√(2) - 3√(3) ] I = 3 - (2)/(3)√(2) - √(3)
Step 3: Finding Co-efficients

Comparing with a+b√(2)+c√(3) yields: a = 3, b = -(2)/(3), c = -1

Compute 2a+3b-4c:

2(3) + 3(-(2)/(3)) - 4(-1)

6 - 2 + 4 = 8

Pattern Recognition

Whenever you see a sum of square roots in the denominator of an integrand, the immediate algorithmic next step is rationalization.

Chapter Mix

Class 12 Maths: Definite Integration

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