∫ (upper limit 1, lower limit-1) (2-x ^ 2) ^ (3 / 2) DX

∫ (upper limit 1, lower limit-1) (2-x ^ 2) ^ (3 / 2) DX

Integral: (2-x ^ 2) ^ (3 / 2) DX
=X (2-x ^ 2) ^ (3 / 2) - integral: XD (2-x ^ 2) ^ (3 / 2)
=X (2-x ^ 2) ^ (3 / 2) - integral: X (- 2x) * 3 / 2 * (2-x ^ 2) ^ (1 / 2) DX
=X (2-x ^ 2) ^ (3 / 2) + 3 integral: x ^ 2 (2-x ^ 2) ^ (1 / 2) DX
Now: integral: x ^ 2 (2-x ^ 2) ^ (1 / 2) DX
Let: x = radical (2) * cost, t [0, PI / 2]
T = arccos (radical (2) * x / 2)
So:
Integral: x ^ 2 (2-x ^ 2) ^ (1 / 2) DX
=Integral: 2 (cost) ^ 2 * radical (2) Sint * radical (2) * - Sint) DT
=-4 integral: (sintcost) ^ 2DT
=-Integral: (sin (2t)) ^ 2DT
=Integral: (COS (4T) - 1) DT
=1/4sin4t-t+C
x[0,1]
t[pi/2,pi/4]
The definite integral is:
pi/4
The upper and lower limits of the integral are: (0,1)
The definite integral is:
=X (2-x ^ 2) ^ (3 / 2) + 3 integral: x ^ 2 (2-x ^ 2) ^ (1 / 2) DX
=x(2-x^2)^(3/2)|(0,1)+3*pi/4
=1+3pi/4
Because the integrand is an even function, the final definite integral is:
I =2*(1+3pi/4)
=2+3pi/3

A definite integral problem!
∫ 1 / (1 + SiNx), (upper limit is π / 4, lower limit is 0)

∵1/(1+sinx)=1/[sin²(x/2)+cos²(x/2)+2sin(x/2)cos(x/2)]
=1/[sin(x/2)+cos(x/2)]²
=sec²(x/2)/[1+tan(x/2)]²
The original formula = ∫ (0, π / 4) sec & sup2; (x / 2) / [1 + Tan (x / 2)] & sup2; DX
=2∫(0,π/4)d(1+tan(x/2))/[1+tan(x/2)]²
={-2/[1+tan(x/2)]}|(0,π/4)
=2-√2