The problem of finding a definite integral ∫ (upper limit π / 4, lower limit 0) x / 1 + cos2x DX

The problem of finding a definite integral ∫ (upper limit π / 4, lower limit 0) x / 1 + cos2x DX

∫ (upper limit π / 4, lower limit 0) x / (1 + cos2x) DX
= ∫ (upper limit π / 4, lower limit 0) x × / (1 / 2 × (secx) ^ 2) DX
= 1 / 2 ×∫ (upper limit π / 4, lower limit 0) x dtanx
The original function of = 1 / 2 × [π / 4 - ∫ (upper limit π / 4, lower limit 0) TaNx DX] TaNx is - ln| cosx|
＝π/8－ln2/4

Why is the maximum area of an ellipse inscribed rectangle 2Ab
A B semi major axis and semi minor axis & nbsp; thank you!

Using the parameter equation of ellipse, let P point coordinate (ACOS θ, bsin θ)
Rectangular area
S=4xy
=4*acosθ*bsinθ
=4ab*cosθsinθ
=2ab*sin(2θ)

To find the area of the largest rectangle in the inscribed rectangle of an ellipse by derivative method
Elliptic standard equation

It's very troublesome to use derivative
Let the ellipse be X & sup2; / A & sup2; + Y & sup2; / B & sup2; = 1
Take a point P (x, y), x > 0, Y > 0 in quadrant I of ellipse, then the area of inscribed rectangle s = 4xy
From the elliptic equation, y = B √ (1-x & sup2 / A & sup2;)
We can get s = 4bx √ (1-x & sup2 / A & sup2;)
Derivation of X on both sides: DS / DX = 4B (√ (A & sup2; - X & sup2;) - X & sup2; / √ (A & sup2; - X & sup2;)) / A
Let DS / DX = 0, x = √ 2A / 2
Then Smax = 2Ab
In fact, the parametric equation is the simplest
Let the equation of ellipse be x = asint, y = bCost
Take a point P (x, y) (x > 0, Y > 0) on the ellipse, then the area of the inscribed rectangle s = 4xy = 2absin2t
Since sin2t ∈ [- 1,1]
∴Smax=2ab