Integrals

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) + c

I = ∫? ² [log? x] dx
By using graph
I = ∫? 0 dx + ∫? ² 1 dx
I = 0 + e² - e

sin? ¹ (√3/2) + cos? ¹ (-√3/2) + tan? ¹ (-1)
= π/3 + 5π/6 – π/4
= (4π+10π–3π)/12 = 11π/12

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 functional forms to be considered is (0,1).

Kindly consider the following figure

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

∫ [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