Integrals
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∫(cos 4x-1)/(cot x-tan x) dx is equal to
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∫(cos 4x-1)/(cot x-tan x) dx is equal to
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I = ∫ (cos 4x-1)/ (cot x-tan x) dx = ∫ (2 cos² 2x-1-1)/ (cos²x-sin²x)/ (sin x cos x) dx
= 2 ∫ (cos² 2x-1) sin x cos x) / (cos 2x) dx = ∫ (cos² 2x-1) sin 2x) / (cos 2x) dx
Put cos 2x = t
-2 sin 2x dx = dt
I = -1/2 ∫ (t²-1)/t dt = 1/2 ∫ (1/t - t) dt
= 1/2 (ln|t| - t²/2) + c
= 1/2 ln|cos 2x| - 1/4 cos² (2x)
Upvote 
∫[logₑ x] dx, x > 0 and [. ] is greatest integer function, is equal to
∫[logₑ x] dx, x > 0 and [. ] is greatest integer function, is equal to
I = ∫? ² [log? x] dx
By using graph
I = ∫? 0 dx + ∫? ² 1 dx
I = 0 + e² - e
Upvote sin-1(sin(2π/3)) + cos-1(cos(7π/6)) + tan-1(tan(3π/4)) is equal to:
sin-1(sin(2π/3)) + cos-1(cos(7π/6)) + tan-1(tan(3π/4)) is equal to:
sin? ¹ (√3/2) + cos? ¹ (-√3/2) + tan? ¹ (-1)
= π/3 + 5π/6 – π/4
= (4π+10π–3π)/12 = 11π/12
Upvote If the functions are defined as f(x) = √x and g(x) = √(1-x), then what is the common domain of the following functions: f + g, f - g, f/g, g/f, g - f where (f+g)(x) = f(x) + g(x), (f/g)(x) = f(x)/g(x)
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If the functions are defined as f(x) = √x and g(x) = √(1-x), then what is the common domain of the following functions: f + g, f - g, f/g, g/f, g - f where (f+g)(x) = f(x) + g(x), (f/g)(x) = f(x)/g(x)
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f (x) + g (x) = √x + √1-x. The domain requires x ≥ 0 and 1-x ≥ 0, so x ≤ 1. Domain is [0,1].
f (x) - g (x) = √x - √1-x. Domain is [0,1].
f (x)/g (x) = √x / √1-x. Requires x ≥ 0 and 1-x > 0, so x < 1. Domain is [0,1).
g (x)/f (x) = √1-x / √x. Requires 1-x ≥ 0 and x > 0. Domain is (0,1].
The common domain for all these functi
Upvote Which of the following statements is incorrect for the function g(α) for α ∈ R such that g(α) = ∫[from α to (π/2 - α)] (sin x / (cos x + sin x)) dx
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Which of the following statements is incorrect for the function g(α) for α ∈ R such that g(α) = ∫[from α to (π/2 - α)] (sin x / (cos x + sin x)) dx
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Kindly consider the following figure
Upvote Let {x} and [x] denote the fractional part of x and the greatest integer ≤ x respectively of a real number x. if ∫(from 0 to n){x}dx, ∫(from 0 to n)[x]dx and 10(n² – n), (n ∈ N, n > 1) are three consecutive terms of a G.P. then n is equal to
Let {x} and [x] denote the fractional part of x and the greatest integer ≤ x respectively of a real number x. if ∫(from 0 to n){x}dx, ∫(from 0 to n)[x]dx and 10(n² – n), (n ∈ N, n > 1) are three consecutive terms of a G.P. then n is equal to
Clearly, ∫ [0 to n] {x}dx = n/2
∫ [0 to n] [x]dx = 1 + 2 + 3 . . . n − 1
= n (n-1)/2
∴ (n (n-1)/2)² = n/2 {10n (n-1)}
Solving, n = 21
Upvote The integral ∫(from π/6 to π/3) tan³x · sin²3x(2sec²x · sin²3x + 3tanx · sin6x)dx is
The integral ∫(from π/6 to π/3) tan³x · sin²3x(2sec²x · sin²3x + 3tanx · sin6x)dx is
∫ [π/6 to π/3] (d/dx (tan? x) * sin³3x + tan? x * d/dx (sin³3x) dx
= 1/2 ∫ [π/6 to π/3] d/dx (tan? x * sin?3x) dx
= 1/2 [tan? x · sin?3x] from π/6 to π/3
= 1/2 [ (√3)? · 0 - 1/ (√3)? )]
= -1/18
Upvote If ∫ sin⁻¹(√(x/(1+x))) dx = A(x)tan⁻¹(√x) + B(x) + C, where C is a constant of integration, then the ordered pair (A(x), B(x)) can be:
If ∫ sin⁻¹(√(x/(1+x))) dx = A(x)tan⁻¹(√x) + B(x) + C, where C is a constant of integration, then the ordered pair (A(x), B(x)) can be:
I = ∫sin? ¹ (√x/√1+x)dx
∫tan? ¹ (√x)dx
= xtan? ¹√x - ∫ (1/ (1+x) * 1/ (2√x)xdx + C
= xtan? ¹√x - ∫ (t²/ (1+t²) * (t*2t dt)/ (2t) + C (x=t²)
= xtan? ¹√x - ∫ (t²/ (1+t²)dt + C = xtan? ¹√x - t + tan? ¹t + C = xtan? ¹√x - √x + tan? ¹√x + C
= (x+1)tan? ¹√x - √x + C => (Ax) = x+1 => B (x) = -√x
Upvote