y ′ = 1 − x 2 y'=\sqrt{1-x^2} y ′ = 1 − x 2
y = ∫ 1 − x 2 d x y=\int\sqrt{1-x^2}dx y = ∫ 1 − x 2 d x
x = s i n u , u = a r c s i n x , d x = c o s u d u x=sinu, u=arcsinx,dx=cosudu x = s in u , u = a rcs in x , d x = cos u d u
y = ∫ c o s 2 u d u = ∫ c o s 2 u + 1 2 d u = c o s u s i n u + u 2 y=\int cos^2udu=\int \frac {cos2u+1}{2}du=\frac {cosusinu+u}{2} y = ∫ co s 2 u d u = ∫ 2 cos 2 u + 1 d u = 2 cos u s in u + u
y = a r c s i n x + x 1 − x 2 2 + C y=\frac {arcsinx+x\sqrt{1-x^2}}{2}+C y = 2 a rcs in x + x 1 − x 2 + C
y = ∑ n = 0 ∞ a n x n y=\sum^{\infin}_{n=0} a_nx^n y = ∑ n = 0 ∞ a n x n
y ′ = ∑ n = 1 ∞ n a n x n − 1 = 1 − x 2 y'=\sum^{\infin}_{n=1} na_nx^{n-1}=\sqrt{1-x^2} y ′ = ∑ n = 1 ∞ n a n x n − 1 = 1 − x 2
c + 2 ∑ n = 0 ∞ a n x n = a r c s i n x + x ∑ n = 1 ∞ n a n x n − 1 c+2\sum^{\infin}_{n=0} a_nx^n=arcsinx+x\sum^{\infin}_{n=1} na_nx^{n-1} c + 2 ∑ n = 0 ∞ a n x n = a rcs in x + x ∑ n = 1 ∞ n a n x n − 1
c + 2 ∑ n = 0 ∞ a n x n = a r c s i n x + ∑ n = 1 ∞ n a n x n c+2\sum^{\infin}_{n=0} a_nx^n=arcsinx+\sum^{\infin}_{n=1} na_nx^n c + 2 ∑ n = 0 ∞ a n x n = a rcs in x + ∑ n = 1 ∞ n a n x n
a r c s i n x = c + 2 ∑ n = 0 ∞ a n x n − ∑ n = 1 ∞ n a n x n arcsinx=c+2\sum^{\infin}_{n=0} a_nx^n-\sum^{\infin}_{n=1} na_nx^n a rcs in x = c + 2 ∑ n = 0 ∞ a n x n − ∑ n = 1 ∞ n a n x n
a r c s i n x = c + 2 a 0 + ∑ n = 1 ∞ ( 2 − n ) a n x n arcsinx=c+2a_0+\sum^{\infin}_{n=1}(2-n)a_nx^n a rcs in x = c + 2 a 0 + ∑ n = 1 ∞ ( 2 − n ) a n x n
a r c s i n 0 = 0 ⟹ c + 2 a 0 = 0 arcsin0=0\implies c+2a_0=0 a rcs in 0 = 0 ⟹ c + 2 a 0 = 0
Answer:
a r c s i n x = ∑ n = 1 ∞ ( 2 − n ) a n x n arcsinx=\sum^{\infin}_{n=1}(2-n)a_nx^n a rcs in x = ∑ n = 1 ∞ ( 2 − n ) a n x n
( a r c s i n x ) ′ = ∑ n = 1 ∞ n ( 2 − n ) a n x n − 1 = 1 1 − x 2 (arcsinx)'=\sum ^{\infin}_{n=1}n(2-n)a_nx^{n-1}=\frac {1}{\sqrt{1-x^2}} ( a rcs in x ) ′ = ∑ n = 1 ∞ n ( 2 − n ) a n x n − 1 = 1 − x 2 1
( a r c s i n 0 ) ′ = 1 = a 1 (arcsin0)'=1=a_1 ( a rcs in 0 ) ′ = 1 = a 1
( a r c s i n 0 ) ′ ′ ′ = 1 ⟹ a 3 = − 1 / 6 (arcsin0)'''=1\implies a_3=-1/6 ( a rcs in 0 ) ′′′ = 1 ⟹ a 3 = − 1/6
By same way we can get a 4 , a 5 , . . . a_4,a_5,... a 4 , a 5 , ...
So:
π 6 = a r c s i n ( 1 / 2 ) = 1 2 + 1 6 ⋅ 1 2 3 + . . . \frac {\pi}{6}=arcsin(1/2)=\frac {1}{2}+\frac {1}{6}\cdot\frac {1}{2^3}+... 6 π = a rcs in ( 1/2 ) = 2 1 + 6 1 ⋅ 2 3 1 + ...
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