Calculus solution: (4-x ^ 2) DX under ∫ radical Calculus solution: (4-x ^ 2) DX under ∫ radical The integral number is 2 above and 0 below At least give me an idea. I don't just want answers. I don't see the root sign on the second floor.

Calculus solution: (4-x ^ 2) DX under ∫ radical Calculus solution: (4-x ^ 2) DX under ∫ radical The integral number is 2 above and 0 below At least give me an idea. I don't just want answers. I don't see the root sign on the second floor.

Let x = 2siny, then y is between 0 and PI / 2
∫ under the root sign (4-x ^ 2) DX [0, PI / 2]
=[2y|0,pi/2]+[sin(2y)|0,pi/2]
=pi

Calculus: the definite integral under the root sign (8-2y Square) in (negative root sign 2, root sign 2)

　　∫[-√2→√2] √(8-2y ²) dy
=√2∫[-√2→√2] √(4-y ²) dy
Let y = 2sinu, then √ (4-y) ²)= 2cosu,dy=2cosudu,u：-π/4→π/4
=√2∫[-π/4→π/4] 4cos ² u du
Using parity symmetry
=8√2∫[0→π/4] cos ² u du
=4√2∫[0→π/4] (1+cos2u) du
=4√2(u+(1/2)sin2u) |[0→π/4]
=4√2(π/4+1/2)
=√2π+2√2

It is known that cos (x-45 degrees) = root of ten, and X belongs to (90 degrees, 135 degrees). Find the value of SiN x and sin (2x + 60 degrees)

Cos (x-45 degrees) = root of ten sign two
cosxcos45+sinxsin45=√2/10
cosx*√2/2+sinx*√2/2=√2/10
(cosx+sinx)=1/5
cos^2x+2cosxsinx+sin^2x=1/25
sin2x=-24/25
X belongs to (90 degrees, 135 degrees)
2X belongs to (180 degrees, 270 degrees)
cos2x=-√1-(-24/25)^2=-7/25
Sin (2x + 60 degrees) = sin2xcos60 + cos2xsin60
=-24/25*1/2+(-7/25)* √3/2
=-(24+7√3)/50
(cosx+sinx)=1/5
cosx=1/5-sinx
cos^2x=(1/5-sinx)^2
1-sin^2x=1/25-1/10sinx+sin^2x
sin^2x-1/20sinx-24/50=0
Let t = SiNx, then it becomes
t^2-1/20t-24/50=0
t^2-1/20t+(1/10)^2-(1/10)^2-24/50=0
(t-1/10)^2=49/100
T-1 / 10 = 7 / 10 or T-1 / 10 = - 7 / 10
T = 4 / 5 or T = - 3 / 5
SiNx = 4 / 5 or SiNx = - 3 / 5
X belongs to (90 degrees, 135 degrees)
sinx>0
So SiNx = 4 / 5

Find ∫ cos ² X DX and ∫ sin ² x dx

1)=
x/2 + sin(2*x)/4
+C
2)x/2 - sin(2*x)/4+C

∫ （ cos ² x-sin ² x/sin ² xcos ² x） dx＝? Integral

∫ (cos ² x - sin ² x)/(sin ² xcos ² x) dx
= ∫ cos2x/[(1/2) ² sin ² 2x] dx
= 2∫ 1/sin ² 2x d(sin2x)
= - 2/sin2x + C
= - 2csc2x + C

∫﹙2x ²－ 3x﹚／﹙x＋1﹚dx

∫(2x ²- 3x)/(x+1)dx
=∫[2x(x+1)-5(x+1)+5]/(x+1)dx
=∫ [2x-5+5/(x+1)] dx
=x ²- 5x+5ln(x+1)+C