Math Analysis Homework - Week 7
Class 1
T1
f ( x ) = 1 + x + x 2 1 − x + x 2 \begin{gathered}
f(x)=\dfrac{1+x+x^2}{1-x+x^2}
\end{gathered} f ( x ) = 1 − x + x 2 1 + x + x 2 的四阶配亚诺余项麦克劳林级数是?
f ( x ) = 1 + 2 x x 2 − x + 1 \begin{gathered}
f(x)=1+\dfrac{2x}{x^2-x+1} \\
\end{gathered} f ( x ) = 1 + x 2 − x + 1 2 x 考虑
1 1 − ( x − x 2 ) = 1 + ( x − x 2 ) + ( x − x 2 ) 2 + ( x − x 2 ) 3 + o ( ( x − x 2 ) 3 ) = 1 + x − x 2 + x 2 − 2 x 3 + x 3 + o ( x 3 ) = 1 + x − x 3 + o ( x 3 ) f ( x ) = 1 + 2 x + 2 x 2 − 2 x 4 + o ( x 4 ) \begin{gathered}
\dfrac{1}{1-(x-x^2)} \\
=1+(x-x^2)+(x-x^2)^2+(x-x^2)^3+o((x-x^2)^3) \\
=1+x-x^2+x^2-2x^3+x^3+o(x^3) \\
=1+x-x^3+o(x^3) \\
f(x)=1+2x+2 x^2-2x^4+o(x^4)
\end{gathered} 1 − ( x − x 2 ) 1 = 1 + ( x − x 2 ) + ( x − x 2 ) 2 + ( x − x 2 ) 3 + o (( x − x 2 ) 3 ) = 1 + x − x 2 + x 2 − 2 x 3 + x 3 + o ( x 3 ) = 1 + x − x 3 + o ( x 3 ) f ( x ) = 1 + 2 x + 2 x 2 − 2 x 4 + o ( x 4 )
T2
Solve a , b a,b a , b
lim x → 0 ( a + b cos x ) sin x − x x 5 = C ≠ 0 \begin{gathered}
\lim_{x \to 0} \dfrac{(a+b\cos x)\sin x-x}{x^5}=C\ne 0
\end{gathered} x → 0 lim x 5 ( a + b cos x ) sin x − x = C = 0
( a + b cos x ) sin x − x = ( a + b − b x 2 2 + b x 4 24 + o ( x 5 ) ) ( x − x 3 6 + x 5 120 ) − x \begin{gathered}
(a+b\cos x)\sin x-x \\
=(a+b-\dfrac{bx^2}{2}+\dfrac{bx^4}{24}+o(x^5))(x-\dfrac{x^3}{6} +\dfrac{x^5}{120} )-x
\end{gathered} ( a + b cos x ) sin x − x = ( a + b − 2 b x 2 + 24 b x 4 + o ( x 5 )) ( x − 6 x 3 + 120 x 5 ) − x 要求其0 0 0 4 4 4 5 5 5
{ a + b = 1 − a + b 6 − b 2 = 0 b 24 + a + b 120 + b 12 ≠ 0 \begin{gathered}
\begin{cases}
a+b=1 \\
-\dfrac{a+b}{6} -\dfrac{b}{2} =0 \\
\dfrac{b}{24} +\dfrac{a+b}{120}+\dfrac{b}{12} \ne 0
\end{cases}
\end{gathered} ⎩ ⎨ ⎧ a + b = 1 − 6 a + b − 2 b = 0 24 b + 120 a + b + 12 b = 0 得
{ a = 4 3 b = − 1 3 \begin{gathered}
\begin{cases}
a=\dfrac{4}{3} \\
b=-\dfrac{1}{3}
\end{cases}
\end{gathered} ⎩ ⎨ ⎧ a = 3 4 b = − 3 1
啥叫x的五阶无穷小啊... 如果理解成o ( x 5 ) o(x^5) o ( x 5 )
T3
lim x → + ∞ ( x 6 + x 5 6 − x 6 − x 5 6 ) \begin{gathered}
\lim_{x \to +\infty} (\sqrt[ 6 ]{ x^6+x^5 } -\sqrt[ 6 ]{ x^6-x^5 } )
\end{gathered} x → + ∞ lim ( 6 x 6 + x 5 − 6 x 6 − x 5 )
let t = 1 x A n s = lim t → 0 1 + t 6 − 1 − t 6 t = lim t → 0 ( 1 + t ) − 5 6 6 + ( 1 − t ) − 5 6 6 = 1 3 \begin{gathered}
\text{let } t=\dfrac{1}{x} \\
Ans=\lim_{t \to 0} \dfrac{\sqrt[6]{1+t}-\sqrt[ 6 ]{ 1-t } }{t} \\
=\lim_{t \to 0} \dfrac{(1+t)^{-\frac56}}{6}+\dfrac{(1-t)^{-\frac56}}{6} \\
= \dfrac{1}{3}
\end{gathered} let t = x 1 A n s = t → 0 lim t 6 1 + t − 6 1 − t = t → 0 lim 6 ( 1 + t ) − 6 5 + 6 ( 1 − t ) − 6 5 = 3 1
T4
α > − 1 lim n → ∞ ∏ i = 1 n ( 1 + i n α + 2 ) n α \begin{gathered}
\alpha>-1 \\
\lim_{n \to \infty} \prod _{i = 1} ^{n} (1+\dfrac{i}{n^{\alpha+2}} )^{n^\alpha}
\end{gathered} α > − 1 n → ∞ lim i = 1 ∏ n ( 1 + n α + 2 i ) n α
= exp lim n → ∞ n α ∑ i = 1 n ln ( 1 + i n α + 2 ) = exp lim n → ∞ n α ∑ i = 1 n i n α + 2 = exp lim n → ∞ n α 1 2 n α = e \begin{gathered}
=\exp \lim_{n \to \infty} n^\alpha \sum _{i = 1} ^{n} \ln(1+\dfrac{i}{n^{\alpha+2}} ) \\
=\exp \lim_{n \to \infty} n^\alpha \sum _{i = 1} ^{n} \dfrac{i}{n^{\alpha+2}} \\
=\exp \lim_{n \to \infty} n^\alpha \dfrac{1}{2n^\alpha} \\
=\sqrt e
\end{gathered} = exp n → ∞ lim n α i = 1 ∑ n ln ( 1 + n α + 2 i ) = exp n → ∞ lim n α i = 1 ∑ n n α + 2 i = exp n → ∞ lim n α 2 n α 1 = e
Class 2
T1
x 1 = sin x 0 > 0 , x n + 1 = sin x n ⟹ lim n → ∞ n 3 x n = 1 \begin{gathered}
x_1=\sin x_0>0,x_{n+1}=\sin x_n \\
\implies \lim_{n \to \infty} \sqrt{ \dfrac{n}{3} } x_n=1
\end{gathered} x 1 = sin x 0 > 0 , x n + 1 = sin x n ⟹ n → ∞ lim 3 n x n = 1
apparently lim n → ∞ x n = 0 lim n → ∞ 1 x n 2 n = lim n → ∞ 1 sin 2 ( x n ) − 1 x n 2 = lim n → ∞ x n 2 − sin 2 ( x n ) x n 2 sin 2 x n = lim n → ∞ x n 2 − ( x n − x n 3 6 + O ( x 5 ) ) 2 x n 2 ( x n − x n 3 6 + O ( x 5 ) ) 2 = lim n → ∞ x n 4 3 + o ( x n 4 ) x n 4 + o ( x n 4 ) = 1 3 ⟹ lim n → ∞ n 3 x n = 1 \begin{gathered}
\text{apparently } \lim_{n \to \infty} x_n=0 \\
\lim_{n \to \infty} \dfrac{1}{x_n^2n} \\
=\lim_{n \to \infty} \dfrac{1}{\sin^2(x_n)} - \dfrac{1}{x_n^2} \\
=\lim_{n \to \infty} \dfrac{x_n^2-\sin^2(x_n)}{x_n^2\sin^2x_n} \\
=\lim_{n \to \infty} \dfrac{x_n^2-(x_n-\dfrac{x_n^3}{6} +O(x^5))^2}{x_n^2(x_n-\dfrac{x_n^3}{6} +O(x^5))^2} \\
=\lim_{n \to \infty} \dfrac{\dfrac{x_n^4}{3} +o(x_n^4)}{x_n^4+o(x_n^4)} =\dfrac{1}{3} \\
\implies \lim_{n \to \infty} \sqrt{ \dfrac{n}{3} } x_n=1
\end{gathered} apparently n → ∞ lim x n = 0 n → ∞ lim x n 2 n 1 = n → ∞ lim sin 2 ( x n ) 1 − x n 2 1 = n → ∞ lim x n 2 sin 2 x n x n 2 − sin 2 ( x n ) = n → ∞ lim x n 2 ( x n − 6 x n 3 + O ( x 5 ) ) 2 x n 2 − ( x n − 6 x n 3 + O ( x 5 ) ) 2 = n → ∞ lim x n 4 + o ( x n 4 ) 3 x n 4 + o ( x n 4 ) = 3 1 ⟹ n → ∞ lim 3 n x n = 1
T2
f ∈ D 2 [ a , b ] , f + ′ ( a ) = f − ′ ( b ) = 0 ⟹ ∃ ξ ∈ ( a , b ) s . t . ∣ f ′ ′ ( ξ ) ∣ ≥ 4 ( b − a ) 2 ∣ f ( b ) − f ( a ) ∣ \begin{gathered}
f\in D^2[a,b],f'_+(a)=f'_-(b)=0 \\
\implies \exists \xi \in (a,b) \ s.t.\
\vert f''(\xi) \vert \ge \dfrac{4}{(b-a)^2} \vert f(b)-f(a) \vert
\end{gathered} f ∈ D 2 [ a , b ] , f + ′ ( a ) = f − ′ ( b ) = 0 ⟹ ∃ ξ ∈ ( a , b ) s . t . ∣ f ′′ ( ξ ) ∣ ≥ ( b − a ) 2 4 ∣ f ( b ) − f ( a ) ∣
g ( x ) = { f ( x ) , x ∈ [ a , b ] f ( a ) + f + ′ ′ ( a ) 2 ( x − a ) 2 , x ∈ ( − ∞ , a ) f ( b ) + f − ′ ′ ( b ) 2 ( x − b ) 2 , ∈ ( b , + ∞ ) g ( x ) = g ( x 0 ) + ( x − x 0 ) g ′ ( x 0 ) + ( x − x 0 ) 2 g ′ ′ ( ξ ) 2 x = a + b 2 , x 0 = a , b ⟹ g ( a + b 2 ) = g ( a ) + ( a − b ) 2 g ′ ′ ( ξ 1 ) 8 g ( a + b 2 ) = g ( b ) + ( a − b ) 2 g ′ ′ ( ξ 2 ) 8 ⟹ ∣ g ′ ′ ( ξ 1 ) − g ′ ′ ( ξ 2 ) 2 ∣ = 4 ( b − a ) 2 ∣ g ( b ) − g ( a ) ∣ max ∣ g ′ ′ ( ξ 1 ) ∣ , ∣ g ′ ′ ( ξ 2 ) ∣ ≥ ∣ g ′ ′ ( ξ 1 ) − g ′ ′ ( ξ 2 ) 2 ∣ ⟹ max ∣ g ′ ′ ( ξ 1 ) ∣ , ∣ g ′ ′ ( ξ 2 ) ∣ ≥ 4 ( b − a ) 2 ∣ g ( b ) − g ( a ) ∣ ⟹ max ∣ f ′ ′ ( ξ 1 ) ∣ , ∣ f ′ ′ ( ξ 2 ) ∣ ≥ 4 ( b − a ) 2 ∣ f ( b ) − f ( a ) ∣ \begin{gathered}
g(x)=\begin{cases}
f(x),x\in [a,b] \\
f(a)+\dfrac{f''_+(a)}{2}(x-a)^2,x\in (-\infty,a) \\
f(b)+\dfrac{f''_-(b)}{2}(x-b)^2,\in (b,+\infty)
\end{cases} \\
g(x)=g(x_0)+(x-x_0)g'(x_0)+\dfrac{(x-x_0)^2g''(\xi)}{2} \\
x=\dfrac{a+b}{2},x_0=a,b \implies \\
g(\dfrac{a+b}{2})=g(a)+\dfrac{(a-b)^2g''(\xi_1)}{8} \\
g(\dfrac{a+b}{2})=g(b)+\dfrac{(a-b)^2g''(\xi_2)}{8} \\
\implies \vert \dfrac{g''(\xi_1)-g''(\xi_2)}{2} \vert =\dfrac{4}{(b-a)^2} \vert g(b)-g(a) \vert \\
\max \vert g''(\xi_1) \vert ,\vert g''(\xi_2) \vert \ge \vert \dfrac{g''(\xi_1)-g''(\xi_2)}{2} \vert \\
\implies \max \vert g''(\xi_1) \vert ,\vert g''(\xi_2) \vert\ge \dfrac{4}{(b-a)^2} \vert g(b)-g(a) \vert \\
\implies \max \vert f''(\xi_1) \vert ,\vert f''(\xi_2) \vert\ge \dfrac{4}{(b-a)^2} \vert f(b)-f(a) \vert
\end{gathered} g ( x ) = ⎩ ⎨ ⎧ f ( x ) , x ∈ [ a , b ] f ( a ) + 2 f + ′′ ( a ) ( x − a ) 2 , x ∈ ( − ∞ , a ) f ( b ) + 2 f − ′′ ( b ) ( x − b ) 2 , ∈ ( b , + ∞ ) g ( x ) = g ( x 0 ) + ( x − x 0 ) g ′ ( x 0 ) + 2 ( x − x 0 ) 2 g ′′ ( ξ ) x = 2 a + b , x 0 = a , b ⟹ g ( 2 a + b ) = g ( a ) + 8 ( a − b ) 2 g ′′ ( ξ 1 ) g ( 2 a + b ) = g ( b ) + 8 ( a − b ) 2 g ′′ ( ξ 2 ) ⟹ ∣ 2 g ′′ ( ξ 1 ) − g ′′ ( ξ 2 ) ∣ = ( b − a ) 2 4 ∣ g ( b ) − g ( a ) ∣ max ∣ g ′′ ( ξ 1 ) ∣ , ∣ g ′′ ( ξ 2 ) ∣ ≥ ∣ 2 g ′′ ( ξ 1 ) − g ′′ ( ξ 2 ) ∣ ⟹ max ∣ g ′′ ( ξ 1 ) ∣ , ∣ g ′′ ( ξ 2 ) ∣ ≥ ( b − a ) 2 4 ∣ g ( b ) − g ( a ) ∣ ⟹ max ∣ f ′′ ( ξ 1 ) ∣ , ∣ f ′′ ( ξ 2 ) ∣ ≥ ( b − a ) 2 4 ∣ f ( b ) − f ( a ) ∣
T3
f ∈ C 2 [ a , b ] , f ( a ) = f ( b ) = 0 ⟹ { max x ∈ [ a , b ] ∣ f ( x ) ∣ ≤ 1 8 ( b − a ) 2 max x ∈ [ a , b ] ∣ f ′ ′ ( x ) ∣ max x ∈ [ a , b ] ∣ f ′ ( x ) ∣ ≤ 1 2 ( b − a ) max x ∈ [ a , b ] ∣ f ′ ′ ( x ) ∣ \begin{gathered}
f\in C^2[a,b],f(a)=f(b)=0 \\
\implies \begin{cases}
\max_{x\in [a,b]}\vert f(x) \vert \le \dfrac{1}{8} (b-a)^2\max_{x\in [a,b]}\vert f''(x) \vert \\
\max_{x\in [a,b]} \vert f'(x) \vert \le \dfrac{1}{2} (b-a)\max_{x\in [a,b]}\vert f''(x) \vert
\end{cases}
\end{gathered} f ∈ C 2 [ a , b ] , f ( a ) = f ( b ) = 0 ⟹ ⎩ ⎨ ⎧ max x ∈ [ a , b ] ∣ f ( x ) ∣ ≤ 8 1 ( b − a ) 2 max x ∈ [ a , b ] ∣ f ′′ ( x ) ∣ max x ∈ [ a , b ] ∣ f ′ ( x ) ∣ ≤ 2 1 ( b − a ) max x ∈ [ a , b ] ∣ f ′′ ( x ) ∣
(1):
let x 0 s . t . ∣ f ( x 0 ) ∣ = max ∣ f ( x ) ∣ if x 0 ∈ { a , b } : Obviously f ( x 0 ) ∈ ( a , b ) , f ′ ( x 0 ) = 0 f ( x ) = f ( x 0 ) + f ′ ′ ( ξ ) 2 ( x − x 0 ) 2 ⟹ { 0 = f ( a ) = f ( x 0 ) + f ′ ′ ( ξ 1 ) 2 ( a − x 0 ) 2 0 = f ( b ) = f ( x 0 ) + f ′ ′ ( ξ 2 ) 2 ( b − x 0 ) 2 ( b − x 0 ) + ( x 0 − a ) = b − a ⟹ min ( a − x 0 ) 2 , ( b − x 0 ) 2 ≤ ( a − b 2 ) 2 ⟹ ∣ f ( x ) ∣ ≤ 1 8 ( b − a ) 2 ( max ∣ f ′ ′ ( ξ 1 ) ∣ , ∣ f ′ ′ ( ξ 2 ) ∣ ) ≤ 1 8 ( b − a ) 2 max x ∈ [ a , b ] ∣ f ′ ′ ( x ) ∣ \begin{gathered}
\text{let } x_0 \ s.t.\
\vert f(x_0)\vert =\max\vert f(x)\vert \\
\text{if } x_0\in \{ a,b \}: \text{Obviously} \\
f(x_0)\in (a,b),f'(x_0)=0 \\
f(x)=f(x_0)+\dfrac{f''(\xi)}{2} (x-x_0)^2 \\
\implies \begin{cases}
0=f(a)=f(x_0)+\dfrac{f''(\xi_1)}{2} (a-x_0)^2 \\
0=f(b)=f(x_0)+\dfrac{f''(\xi_2)}{2} (b-x_0)^2
\end{cases} \\
(b-x_0)+(x_0-a)=b-a \\
\implies \min (a-x_0)^2,(b-x_0)^2 \le (\dfrac{a-b}{2} )^2 \\
\implies \vert f(x)\vert \le \dfrac{1}{8} (b-a)^2(\max \vert f''(\xi_1)\vert ,\vert f''(\xi_2)\vert) \\
\le \dfrac{1}{8} (b-a)^2\max_{x\in [a,b]}\vert f''(x) \vert
\end{gathered} let x 0 s . t . ∣ f ( x 0 ) ∣ = max ∣ f ( x ) ∣ if x 0 ∈ { a , b } : Obviously f ( x 0 ) ∈ ( a , b ) , f ′ ( x 0 ) = 0 f ( x ) = f ( x 0 ) + 2 f ′′ ( ξ ) ( x − x 0 ) 2 ⟹ ⎩ ⎨ ⎧ 0 = f ( a ) = f ( x 0 ) + 2 f ′′ ( ξ 1 ) ( a − x 0 ) 2 0 = f ( b ) = f ( x 0 ) + 2 f ′′ ( ξ 2 ) ( b − x 0 ) 2 ( b − x 0 ) + ( x 0 − a ) = b − a ⟹ min ( a − x 0 ) 2 , ( b − x 0 ) 2 ≤ ( 2 a − b ) 2 ⟹ ∣ f ( x ) ∣ ≤ 8 1 ( b − a ) 2 ( max ∣ f ′′ ( ξ 1 ) ∣ , ∣ f ′′ ( ξ 2 ) ∣ ) ≤ 8 1 ( b − a ) 2 x ∈ [ a , b ] max ∣ f ′′ ( x ) ∣ (2):
let M = max ∣ f ′ ′ ( x ) ∣ { 0 = f ( a ) = f ( x ) + f ′ ( x ) ( a − x ) + f ′ ′ ( ξ 1 ) 2 ( a − x ) 2 0 = f ( b ) = f ( x ) + f ′ ( x ) ( b − x ) + f ′ ′ ( ξ 2 ) 2 ( b − x ) 2 ⟹ f ′ ( x ) ( b − a ) = f ′ ′ ( ξ 1 ) 2 ( a − x ) 2 − f ′ ′ ( ξ 2 ) 2 ( b − x ) 2 ∣ f ′ ( x ) ∣ ≤ 1 2 ( b − a ) ∣ f ′ ′ ( ξ 1 ) ( a − x ) 2 − f ′ ′ ( ξ 2 ) ( b − x ) 2 ∣ ≤ 1 2 ( b − a ) M ∣ ( a − x ) 2 + ( b − x ) 2 ∣ ≤ 1 2 ( b − a ) M ( a − b ) 2 = 1 2 ( b − a ) max x ∈ [ a , b ] ∣ f ′ ′ ( x ) ∣ \begin{gathered}
\text{let } M=\max \vert f''(x) \vert \\
\begin{cases}
0=f(a)=f(x)+f'(x)(a-x)+\dfrac{f''(\xi_1)}{2} (a-x)^2 \\
0=f(b)=f(x)+f'(x)(b-x)+\dfrac{f''(\xi_2)}{2} (b-x)^2 \\
\end{cases} \\
\implies
f'(x)(b-a)=\dfrac{f''(\xi_1)}{2} (a-x)^2-\dfrac{f''(\xi_2)}{2} (b-x)^2 \\
\vert f'(x) \vert \le \dfrac{1}{2(b-a)} \vert f''(\xi_1)(a-x)^2-f''(\xi_2)(b-x)^2 \vert \\
\le \dfrac{1}{2(b-a)}M\vert (a-x)^2+(b-x)^2 \vert \\
\le \dfrac{1}{2(b-a)}M(a-b)^2 \\ \\
=\dfrac{1}{2} (b-a)\max_{x\in [a,b]}\vert f''(x) \vert
\end{gathered} let M = max ∣ f ′′ ( x ) ∣ ⎩ ⎨ ⎧ 0 = f ( a ) = f ( x ) + f ′ ( x ) ( a − x ) + 2 f ′′ ( ξ 1 ) ( a − x ) 2 0 = f ( b ) = f ( x ) + f ′ ( x ) ( b − x ) + 2 f ′′ ( ξ 2 ) ( b − x ) 2 ⟹ f ′ ( x ) ( b − a ) = 2 f ′′ ( ξ 1 ) ( a − x ) 2 − 2 f ′′ ( ξ 2 ) ( b − x ) 2 ∣ f ′ ( x ) ∣ ≤ 2 ( b − a ) 1 ∣ f ′′ ( ξ 1 ) ( a − x ) 2 − f ′′ ( ξ 2 ) ( b − x ) 2 ∣ ≤ 2 ( b − a ) 1 M ∣ ( a − x ) 2 + ( b − x ) 2 ∣ ≤ 2 ( b − a ) 1 M ( a − b ) 2 = 2 1 ( b − a ) x ∈ [ a , b ] max ∣ f ′′ ( x ) ∣
T4
{ f ( x ) , g ( x ) ∈ C + ∞ ( − 1 , 1 ) ∀ n ∈ N , ∣ f ( n ) ( x ) − g ( n ) ( x ) ∣ ≤ n ! ∣ x ∣ ⟹ f ( x ) = g ( x ) , x ∈ ( − 1 , 1 ) \begin{gathered}
\begin{cases}
f(x),g(x)\in C^{+\infty}(-1,1) \\
\forall n\in{\mathbb N},\vert f^{(n)}(x)-g^{(n)}(x) \vert \le n! \vert x \vert
\end{cases} \\
\implies f(x)=g(x),x\in (-1,1)
\end{gathered} { f ( x ) , g ( x ) ∈ C + ∞ ( − 1 , 1 ) ∀ n ∈ N , ∣ f ( n ) ( x ) − g ( n ) ( x ) ∣ ≤ n ! ∣ x ∣ ⟹ f ( x ) = g ( x ) , x ∈ ( − 1 , 1 )
F ( x ) = f ( x ) − g ( x ) ∣ F ( n ) ( x ) ∣ ≤ n ! ∣ x ∣ n = 0 ⟹ ∣ F ( x ) ∣ ≤ ∣ x ∣ , F ( 0 ) = 0 ∣ F ( x ) ∣ = ∣ ∑ i = 0 n F ( i ) ( 0 ) i ! x i + F ( n + 1 ) ( ξ ) ( n + 1 ) ! x n + 1 ∣ ≤ ∣ F n + 1 ( ξ ) ( n + 1 ) ! x n + 1 ∣ ≤ ∣ x n + 2 ∣ F ( x ) = lim n → ∞ ∣ F ( x ) ∣ ≤ lim n → ∞ ∣ x n + 2 ∣ = 0 \begin{gathered}
F(x)=f(x)-g(x) \\
\vert F^{(n)}(x) \vert \le n!\vert x \vert \\ \\
n=0 \implies \vert F(x) \vert \le \vert x \vert,F(0)=0 \\
\vert F(x) \vert={\left \vert \sum _{i = 0} ^{n} \dfrac{F^{(i)}(0)}{i!} x^i+\dfrac{F^{(n+1)}(\xi)}{(n+1)!} x^{n+1} \right \vert} \\
\le \vert \dfrac{F^{n+1}(\xi)}{(n+1)!}x^{n+1} \vert \\
\le \vert x^{n+2} \vert \\
F(x)=\lim_{n \to \infty} \vert F(x) \vert \le \lim_{n \to \infty} \vert x^{n+2} \vert =0 \\
\end{gathered} F ( x ) = f ( x ) − g ( x ) ∣ F ( n ) ( x ) ∣ ≤ n ! ∣ x ∣ n = 0 ⟹ ∣ F ( x ) ∣ ≤ ∣ x ∣ , F ( 0 ) = 0 ∣ F ( x ) ∣ = i = 0 ∑ n i ! F ( i ) ( 0 ) x i + ( n + 1 )! F ( n + 1 ) ( ξ ) x n + 1 ≤ ∣ ( n + 1 )! F n + 1 ( ξ ) x n + 1 ∣ ≤ ∣ x n + 2 ∣ F ( x ) = n → ∞ lim ∣ F ( x ) ∣ ≤ n → ∞ lim ∣ x n + 2 ∣ = 0
T5
{ f ( x ) ∈ D 2 [ 0 , 1 ] f ( 0 ) = f ( 1 ) = 0 min x ∈ [ 0 , 1 ] f ( x ) = − 1 ⟹ ∃ ξ , f ′ ′ ( ξ ) ≥ 8 \begin{gathered}
\begin{cases}
f(x)\in D^2[0,1] \\
f(0)=f(1)=0 \\
\min_{x\in [0,1]}f(x)=-1
\end{cases} \\
\implies \exists \xi,f''(\xi)\ge 8
\end{gathered} ⎩ ⎨ ⎧ f ( x ) ∈ D 2 [ 0 , 1 ] f ( 0 ) = f ( 1 ) = 0 min x ∈ [ 0 , 1 ] f ( x ) = − 1 ⟹ ∃ ξ , f ′′ ( ξ ) ≥ 8
let x 0 s . t . ∣ f ( x 0 ) ∣ = max ∣ f ( x ) ∣ if x 0 ∈ { 0 , 1 } : Obviously f ( x 0 ) ∈ ( 0 , 1 ) , f ′ ( x 0 ) = 0 f ( x ) = f ( x 0 ) + f ′ ′ ( ξ ) 2 ( x − x 0 ) 2 ⟹ { 0 = f ( 0 ) = f ( x 0 ) + f ′ ′ ( ξ 1 ) 2 x 0 2 0 = f ( 1 ) = f ( x 0 ) + f ′ ′ ( ξ 2 ) 2 ( 1 − x 0 ) 2 ( − x 0 ) + ( x 0 − 1 ) = 0 + 1 ⟹ min x 0 2 , ( 1 − x 0 ) 2 ≤ ( 0 + 1 2 ) 2 ⟹ ∣ f ( x ) ∣ ≤ 1 8 ( max ∣ f ′ ′ ( ξ 1 ) ∣ , ∣ f ′ ′ ( ξ 2 ) ∣ ) ⟹ let f ( x ) = − 1 , ∃ ξ , f ′ ′ ( ξ ) ≥ 8 \begin{gathered}
\text{let } x_0 \ s.t.\
\vert f(x_0)\vert =\max\vert f(x)\vert \\
\text{if } x_0\in \{ 0,1 \}: \text{Obviously} \\
f(x_0)\in (0,1),f'(x_0)=0 \\
f(x)=f(x_0)+\dfrac{f''(\xi)}{2} (x-x_0)^2 \\
\implies \begin{cases}
0=f(0)=f(x_0)+\dfrac{f''(\xi_1)}{2} x_0^2 \\
0=f(1)=f(x_0)+\dfrac{f''(\xi_2)}{2} (1-x_0)^2
\end{cases} \\
(-x_0)+(x_0-1)=0+1 \\
\implies \min x_0^2,(1-x_0)^2 \le (\dfrac{0+1}{2} )^2 \\
\implies \vert f(x)\vert \le \dfrac{1}{8} (\max \vert f''(\xi_1)\vert ,\vert f''(\xi_2)\vert) \\
\implies \text{let } f(x)=-1,\exists \xi,f''(\xi)\ge 8
\end{gathered} let x 0 s . t . ∣ f ( x 0 ) ∣ = max ∣ f ( x ) ∣ if x 0 ∈ { 0 , 1 } : Obviously f ( x 0 ) ∈ ( 0 , 1 ) , f ′ ( x 0 ) = 0 f ( x ) = f ( x 0 ) + 2 f ′′ ( ξ ) ( x − x 0 ) 2 ⟹ ⎩ ⎨ ⎧ 0 = f ( 0 ) = f ( x 0 ) + 2 f ′′ ( ξ 1 ) x 0 2 0 = f ( 1 ) = f ( x 0 ) + 2 f ′′ ( ξ 2 ) ( 1 − x 0 ) 2 ( − x 0 ) + ( x 0 − 1 ) = 0 + 1 ⟹ min x 0 2 , ( 1 − x 0 ) 2 ≤ ( 2 0 + 1 ) 2 ⟹ ∣ f ( x ) ∣ ≤ 8 1 ( max ∣ f ′′ ( ξ 1 ) ∣ , ∣ f ′′ ( ξ 2 ) ∣ ) ⟹ let f ( x ) = − 1 , ∃ ξ , f ′′ ( ξ ) ≥ 8
T6
{ f ( x ) ∈ D n ( x 0 − δ , x 0 + δ ) ∀ i ∈ [ 2 , n − 1 ] , f ( i ) ( x 0 ) = 0 f ( n ) ( x 0 ) ≠ 0 , f ( n ) ( x ) is continuous at x 0 0 < ∣ h ∣ < δ ⟹ f ( x 0 + h ) − f ( x 0 ) = h f ′ ( x 0 + θ h ) , θ ∈ ( 0 , 1 ) ⟹ lim h → 0 θ = ( 1 n ) 1 n − 1 \begin{gathered}
\begin{cases}
f(x)\in D^n(x_0-\delta,x_0+\delta) \\
\forall i \in [2,n-1],f^{(i)}(x_0)=0 \\
f^{(n)}(x_0)\ne 0,f^{(n)}(x) \text{ is continuous at } x_0 \\
0<\vert h \vert <\delta \implies f(x_0+h)-f(x_0)=hf'(x_0+\theta h),\theta\in (0,1) \\
\end{cases} \\
\implies \lim_{h \to 0} \theta = (\dfrac{1}{n})^{\frac{1}{n-1} }
\end{gathered} ⎩ ⎨ ⎧ f ( x ) ∈ D n ( x 0 − δ , x 0 + δ ) ∀ i ∈ [ 2 , n − 1 ] , f ( i ) ( x 0 ) = 0 f ( n ) ( x 0 ) = 0 , f ( n ) ( x ) is continuous at x 0 0 < ∣ h ∣ < δ ⟹ f ( x 0 + h ) − f ( x 0 ) = h f ′ ( x 0 + θ h ) , θ ∈ ( 0 , 1 ) ⟹ h → 0 lim θ = ( n 1 ) n − 1 1
f ( x 0 + h ) − f ( x 0 ) = f ( n ) ( ξ 1 ) h n n ! f ′ ( x 0 + θ h ) = f ( n ) ( ξ 2 ) ( θ h ) n − 1 ( n − 1 ) ! f ( x 0 + h ) − f ( x 0 ) = h f ′ ( x 0 + θ h ) ⟹ f ( n ) ( ξ 2 ) θ n − 1 ( n − 1 ) ! = f ( n ) ( ξ 1 ) n ! ⟹ θ = ( f ( n ) ( ξ 1 ) f ( n ) ( ξ 2 ) ) 1 n − 1 ( 1 n ) 1 n − 1 lim h → 0 θ = ( 1 n ) 1 n − 1 \begin{gathered}
f(x_0+h)-f(x_0)=\dfrac{f^{(n)}(\xi_1)h^n}{n!} \\
f'(x_0+\theta h)=f^{(n)}(\xi_2)\dfrac{(\theta h)^{n-1}}{(n-1)!} \\
f(x_0+h)-f(x_0)=hf'(x_0+\theta h) \\
\implies f^{(n)}(\xi_2)\dfrac{\theta^{n-1}}{(n-1)!}=\dfrac{f^{(n)}(\xi_1)}{n!} \\
\implies \theta = (\dfrac{f^{(n)}(\xi_1)}{f^{(n)}(\xi_2)})^{\frac1{n-1}}(\dfrac{1}{n} )^{\frac1{n-1}} \\
\lim_{h \to 0} \theta = (\dfrac{1}{n}) ^{\frac1{n-1}}
\end{gathered} f ( x 0 + h ) − f ( x 0 ) = n ! f ( n ) ( ξ 1 ) h n f ′ ( x 0 + θ h ) = f ( n ) ( ξ 2 ) ( n − 1 )! ( θ h ) n − 1 f ( x 0 + h ) − f ( x 0 ) = h f ′ ( x 0 + θ h ) ⟹ f ( n ) ( ξ 2 ) ( n − 1 )! θ n − 1 = n ! f ( n ) ( ξ 1 ) ⟹ θ = ( f ( n ) ( ξ 2 ) f ( n ) ( ξ 1 ) ) n − 1 1 ( n 1 ) n − 1 1 h → 0 lim θ = ( n 1 ) n − 1 1
Class 3
T1
∫ ( 2 − x 3 x 2 ) 2 d x \begin{gathered}
\int (\dfrac{2-x^3}{x^2} )^2dx
\end{gathered} ∫ ( x 2 2 − x 3 ) 2 d x
= ∫ 4 x 4 d x + ∫ x 6 x 4 d x − ∫ 4 x 3 x 4 d x = − 4 3 x 3 + x 3 3 − 4 ln x + C \begin{gathered}
=\int \dfrac{4}{x^4} dx+\int \dfrac{x^6}{x^4} dx-\int \dfrac{4x^3}{x^4}dx \\
=-\dfrac{4}{3x^3} +\dfrac{x^3}{3} -4\ln x+C
\end{gathered} = ∫ x 4 4 d x + ∫ x 4 x 6 d x − ∫ x 4 4 x 3 d x = − 3 x 3 4 + 3 x 3 − 4 ln x + C
T2
∫ cos 2 x sin x − cos x d x \begin{gathered}
\int \dfrac{\cos 2x}{\sin x-\cos x} dx
\end{gathered} ∫ sin x − cos x cos 2 x d x
= ∫ cos 2 x − sin 2 x sin x − cos x d x = ∫ ( − cos x − sin x ) d x = cos x − sin x + C \begin{gathered}
=\int \dfrac{\cos^2 x-\sin ^2 x}{\sin x-\cos x} dx \\
=\int (-\cos x-\sin x)dx \\
=\cos x-\sin x+C
\end{gathered} = ∫ sin x − cos x cos 2 x − sin 2 x d x = ∫ ( − cos x − sin x ) d x = cos x − sin x + C
T3
∫ tan 2 x d x \begin{gathered}
\int \tan^2 xdx
\end{gathered} ∫ tan 2 x d x
= ∫ ( sec 2 x − 1 ) d x = tan x − x + C \begin{gathered}
=\int (\sec^2x-1)dx \\
=\tan x-x+C
\end{gathered} = ∫ ( sec 2 x − 1 ) d x = tan x − x + C
T4
∫ ( 2 x + 3 x ) 2 d x \begin{gathered}
\int (2^x+3^x)^2dx \\
\end{gathered} ∫ ( 2 x + 3 x ) 2 d x
= ∫ ( 4 x + 2 × 6 x + 9 x ) d x = 2 2 x − 1 ln 2 + 2 × 6 x ln 6 + 9 x 2 ln 3 + C \begin{gathered}
=\int (4^x+2\times 6^x+9^x)dx \\
=\dfrac{2^{2x-1}}{\ln 2}+\dfrac{2\times 6^x}{\ln 6} +\dfrac{9^x}{2\ln 3} +C
\end{gathered} = ∫ ( 4 x + 2 × 6 x + 9 x ) d x = ln 2 2 2 x − 1 + ln 6 2 × 6 x + 2 ln 3 9 x + C
T5
∫ 1 x 4 ( 1 + x 2 ) d x \begin{gathered}
\int \dfrac{1}{x^4(1+x^2)} dx
\end{gathered} ∫ x 4 ( 1 + x 2 ) 1 d x
= ∫ − x 2 + 1 x 4 d x + ∫ 1 1 + x 2 d x = ∫ − 1 x 2 d x + ∫ 1 x 4 d x + ∫ 1 1 + x 2 d x = 1 x − 1 3 x 3 + arctan ( x ) + C \begin{gathered}
=\int \dfrac{-x^2+1}{x^4} dx+\int \dfrac{1}{1+x^2} dx \\
=\int -\dfrac{1}{x^2} dx+\int \dfrac{1}{x^4} dx+\int \dfrac{1}{1+x^2} dx \\
=\dfrac{1}{x} -\dfrac{1}{3x^3} +\arctan(x)+C
\end{gathered} = ∫ x 4 − x 2 + 1 d x + ∫ 1 + x 2 1 d x = ∫ − x 2 1 d x + ∫ x 4 1 d x + ∫ 1 + x 2 1 d x = x 1 − 3 x 3 1 + arctan ( x ) + C
T6
∫ x cos x d x \begin{gathered}
\int x\cos xdx \\
\end{gathered} ∫ x cos x d x
= ∫ x d sin x = x sin x − ∫ sin x d x = x sin x + cos x + C \begin{gathered}
=\int xd\sin x \\
=x\sin x-\int \sin xdx \\
=x\sin x+\cos x+C
\end{gathered} = ∫ x d sin x = x sin x − ∫ sin x d x = x sin x + cos x + C
T7
∫ e x cos x d x \begin{gathered}
\int e^x\cos xdx \\
\end{gathered} ∫ e x cos x d x
∫ e x cos x d x = ∫ e x d sin x = e x sin x − ∫ e x sin x = e x sin x + e x cos x − ∫ e x cos x ⟹ ∫ e x cos x d x = e x ( sin x + cos x ) 2 + C \begin{gathered}
\int e^x\cos xdx
=\int e^xd\sin x \\
=e^x\sin x-\int e^x\sin x \\
=e^x\sin x+e^x\cos x-\int e^x\cos x \\
\implies \int e^x\cos xdx=\dfrac{e^x(\sin x+\cos x)}{2} +C
\end{gathered} ∫ e x cos x d x = ∫ e x d sin x = e x sin x − ∫ e x sin x = e x sin x + e x cos x − ∫ e x cos x ⟹ ∫ e x cos x d x = 2 e x ( sin x + cos x ) + C
T8
∫ x n ln x d x \begin{gathered}
\int x^n\ln xdx
\end{gathered} ∫ x n ln x d x
∫ x n ln x d x = x n + 1 n + 1 ln x − ∫ x n n + 1 d x = x n + 1 n + 1 ln x − x n + 1 ( n + 1 ) 2 \begin{gathered}
\int x^n\ln xdx \\
=\dfrac{x^{n+1}}{n+1} \ln x-\int \dfrac{x^n}{n+1} dx \\
=\dfrac{x^{n+1}}{n+1} \ln x-\dfrac{x^{n+1}}{(n+1)^2}
\end{gathered} ∫ x n ln x d x = n + 1 x n + 1 ln x − ∫ n + 1 x n d x = n + 1 x n + 1 ln x − ( n + 1 ) 2 x n + 1
T9
∫ x arctan x d x \begin{gathered}
\int x\arctan xdx
\end{gathered} ∫ x arctan x d x
= 1 2 ∫ arctan x d x 2 = 1 2 x 2 arctan x − 1 2 ∫ ( 1 − 1 1 + x 2 ) d x = 1 2 x 2 arctan x − 1 2 x + 1 2 arctan x + C \begin{gathered}
=\dfrac{1}{2} \int \arctan x dx^2 \\
=\dfrac{1}{2} x^2\arctan x-\dfrac{1}{2} \int (1-\dfrac{1}{1+x^2}) dx \\
=\dfrac{1}{2} x^2\arctan x-\dfrac{1}{2} x+\dfrac{1}{2} \arctan x+C
\end{gathered} = 2 1 ∫ arctan x d x 2 = 2 1 x 2 arctan x − 2 1 ∫ ( 1 − 1 + x 2 1 ) d x = 2 1 x 2 arctan x − 2 1 x + 2 1 arctan x + C
T10
∫ e arctan x ( 1 + x 2 ) 3 2 d x \begin{gathered}
\int \dfrac{e^{\arctan x}}{(1+x^2)^\frac32} dx \\
\end{gathered} ∫ ( 1 + x 2 ) 2 3 e a r c t a n x d x
let x = tan t a n s = ∫ e t cos t d t = e t 2 ( sin t + cos t ) + C = e arctan x 2 ( 1 + x 1 + x 2 ) + C \begin{gathered}
\text{let }x=\tan t \\
ans=\int e^t \cos tdt \\
=\dfrac{e^t}{2} (\sin t+\cos t)+C \\
=\dfrac{e^{\arctan x}}{2} (\dfrac{1+x}{\sqrt {1+x^2}} )+C
\end{gathered} let x = tan t an s = ∫ e t cos t d t = 2 e t ( sin t + cos t ) + C = 2 e a r c t a n x ( 1 + x 2 1 + x ) + C
T11
∫ arctan x x 2 d x \begin{gathered}
\int \dfrac{\arctan x}{x^2} dx
\end{gathered} ∫ x 2 arctan x d x
= − arctan x x + ∫ 1 x ( 1 + x 2 ) d x = − arctan x x + ∫ d x 2 2 x 2 ( 1 + x 2 ) = − arctan x x + 1 2 ln x 2 1 + x 2 + C \begin{gathered}
=-\dfrac{\arctan x}{x} +\int \dfrac{1}{x(1+x^2)} dx \\
=-\dfrac{\arctan x}{x} +\int \dfrac{dx^2}{2x^2(1+x^2)} \\
=-\dfrac{\arctan x}{x}+\dfrac{1}{2} \ln\dfrac{x^2}{1+x^2} +C
\end{gathered} = − x arctan x + ∫ x ( 1 + x 2 ) 1 d x = − x arctan x + ∫ 2 x 2 ( 1 + x 2 ) d x 2 = − x arctan x + 2 1 ln 1 + x 2 x 2 + C
T12
∫ cos ln x d x \begin{gathered}
\int \cos \ln xdx
\end{gathered} ∫ cos ln x d x
= t = ln x ∫ e t cos t d t = e t 2 ( sin t + cos t ) + C = x 2 ( sin ln x + cos ln x ) + C \begin{gathered}
\xlongequal{t=\ln x}\int e^t\cos tdt \\
=\dfrac{e^t}{2} (\sin t+\cos t)+C \\
=\dfrac{x}{2} (\sin \ln x+\cos \ln x)+C
\end{gathered} t = l n x ∫ e t cos t d t = 2 e t ( sin t + cos t ) + C = 2 x ( sin ln x + cos ln x ) + C
T13
∫ e 5 x d x = e 5 x 5 + C \begin{gathered}
\int e^{5x}dx \\
=\dfrac{e^{5x}}{5} +C
\end{gathered} ∫ e 5 x d x = 5 e 5 x + C
T14
∫ d x cos 2 7 x \begin{gathered}
\int \dfrac{dx}{\cos^2 7x}
\end{gathered} ∫ cos 2 7 x d x
= ∫ sec 2 ( 7 x ) d x = tan ( 7 x ) 7 + C \begin{gathered}
=\int \sec^2(7x)dx \\
=\dfrac{\tan(7x)}{7} +C
\end{gathered} = ∫ sec 2 ( 7 x ) d x = 7 tan ( 7 x ) + C
T15
∫ cos 3 x sin x d x \begin{gathered}
\int \cos^3 x\sin xdx
\end{gathered} ∫ cos 3 x sin x d x
= ∫ 1 + cos 2 x 2 sin 2 x 2 d x = − ∫ 1 + cos 2 x 8 d ( cos 2 x ) = − cos 2 x 8 − cos 2 2 x 16 \begin{gathered}
=\int \dfrac{1+\cos 2x}{2} \dfrac{\sin 2x}{2} dx \\
=-\int \dfrac{1+\cos 2x}{8} d(\cos 2x) \\
=-\dfrac{\cos 2x}{8} -\dfrac{\cos^2 2x}{16}
\end{gathered} = ∫ 2 1 + cos 2 x 2 sin 2 x d x = − ∫ 8 1 + cos 2 x d ( cos 2 x ) = − 8 cos 2 x − 16 cos 2 2 x
T16
∫ tan x + 1 cos 2 x d x \begin{gathered}
\int \dfrac{\sqrt{ \tan x+1 } }{\cos^2 x} dx
\end{gathered} ∫ cos 2 x tan x + 1 d x
= ∫ tan x + 1 d tan x = 2 3 ( 1 + tan x ) 3 2 + C \begin{gathered}
=\int \sqrt{ \tan x+1 } d\tan x \\
=\dfrac{2}{3} (1+\tan x)^{\frac32}+C
\end{gathered} = ∫ tan x + 1 d tan x = 3 2 ( 1 + tan x ) 2 3 + C
T17
∫ tan 4 x d x = ∫ tan 2 x ( sec 2 x − 1 ) d x = ∫ tan 2 d tan x − ∫ ( sec 2 x − 1 ) d x = tan 3 x 3 − tan x + x + C \begin{gathered}
\int \tan^4 xdx \\
=\int \tan^2 x(\sec^2 x-1)dx \\
=\int \tan^2 d\tan x-\int (\sec^2 x-1)dx \\
=\dfrac{\tan^3 x}{3} -\tan x+x+C
\end{gathered} ∫ tan 4 x d x = ∫ tan 2 x ( sec 2 x − 1 ) d x = ∫ tan 2 d tan x − ∫ ( sec 2 x − 1 ) d x = 3 tan 3 x − tan x + x + C
T18
∫ sin x x d x \begin{gathered}
\int \dfrac{\sin \sqrt x}{\sqrt x} dx
\end{gathered} ∫ x sin x d x
= ∫ 2 sin x d x = − 2 cos x + C \begin{gathered}
=\int 2\sin \sqrt xd\sqrt x \\
=-2\cos \sqrt x+C
\end{gathered} = ∫ 2 sin x d x = − 2 cos x + C
T19
∫ 1 + 1 − x 2 1 − 1 − x 2 d x \begin{gathered}
\int \dfrac{1+\sqrt{ 1-x^2 } }{1-\sqrt{ 1-x^2 } } dx
\end{gathered} ∫ 1 − 1 − x 2 1 + 1 − x 2 d x
= ∫ ( 1 + 1 − x 2 ) 2 x 2 d x = − 2 x − x + ∫ 2 1 − x 2 x 2 d x let x = cos t = − 2 x − x − ∫ 2 sin 2 t cos 2 t d t = − 2 x − x − ∫ 2 sec 2 t d t + ∫ 2 d t = − 2 x − x − 2 tan t + 2 t + C = − 2 x − x − 2 1 − x 2 x + 2 arccos x \begin{gathered}
=\int \dfrac{(1+\sqrt{1-x^2})^2}{x^2} dx \\
=-\dfrac{2}{x} -x+\int \dfrac{2\sqrt{1-x^2}}{x^2} dx \\
\text{let } x=\cos t \\
=-\dfrac{2}{x} -x-\int \dfrac{2\sin^2 t}{\cos^2 t}dt \\
=-\dfrac{2}{x} -x-\int 2\sec^2 tdt+\int 2dt \\
=-\dfrac{2}{x} -x-2\tan t+2t+C \\
=-\dfrac{2}{x} -x-\dfrac{2\sqrt{1-x^2}}{x} +2\arccos x
\end{gathered} = ∫ x 2 ( 1 + 1 − x 2 ) 2 d x = − x 2 − x + ∫ x 2 2 1 − x 2 d x let x = cos t = − x 2 − x − ∫ cos 2 t 2 sin 2 t d t = − x 2 − x − ∫ 2 sec 2 t d t + ∫ 2 d t = − x 2 − x − 2 tan t + 2 t + C = − x 2 − x − x 2 1 − x 2 + 2 arccos x
T20
∫ sin x + cos x sin x − cos x 3 d x \begin{gathered}
\int \dfrac{\sin x+\cos x}{\sqrt[ 3 ]{ \sin x-\cos x } } dx
\end{gathered} ∫ 3 sin x − cos x sin x + cos x d x
= ∫ ( sin x − cos x ) − 1 3 d ( sin x − cos x ) = 3 2 ( sin x − cos x ) 2 3 \begin{gathered}
=\int (\sin x-\cos x)^{-\frac13}d(\sin x-\cos x) \\
=\dfrac{3}{2} (\sin x-\cos x)^{\frac23}
\end{gathered} = ∫ ( sin x − cos x ) − 3 1 d ( sin x − cos x ) = 2 3 ( sin x − cos x ) 3 2
T21
∫ ln ( x + 1 + x 2 ) 1 + x 2 d x \begin{gathered}
\int \sqrt{ \dfrac{\ln (x+\sqrt{ 1+x^2 } )}{1+x^2} } dx
\end{gathered} ∫ 1 + x 2 ln ( x + 1 + x 2 ) d x
= ∫ ln ( x + 1 + x 2 ) d ( ln ( x + 1 + x 2 ) ) = 2 3 ln ( x + 1 + x 2 ) 3 2 \begin{gathered}
=\int \sqrt{ \ln(x+\sqrt{1+x^2}) } d(\ln(x+\sqrt{1+x^2})) \\
=\dfrac{2}{3}\ln(x+\sqrt{1+x^2})^{\frac32}
\end{gathered} = ∫ ln ( x + 1 + x 2 ) d ( ln ( x + 1 + x 2 )) = 3 2 ln ( x + 1 + x 2 ) 2 3
T22
Calculate the recurrence relation of
I n = ∫ x n 1 − x 2 d x \begin{gathered}
I_n=\int \dfrac{x^n}{\sqrt{ 1-x^2 } } dx
\end{gathered} I n = ∫ 1 − x 2 x n d x
x = sin t I n = ∫ sin n t d t I n = − sin n − 1 t cos t + ∫ ( n − 1 ) sin n − 2 t cos 2 t = − sin n − 1 t cos t + ∫ ( n − 1 ) sin n − 2 t ( 1 − sin 2 t ) d t = − sin n − 1 t cos t + ( n − 1 ) I n − 2 − ( n − 1 ) I n ⟹ I n = − 1 n sin n − 1 t cos t + n − 1 n I n − 2 \begin{gathered}
x=\sin t \\
I_n=\int \sin^n tdt \\
I_n=-\sin^{n-1}t\cos t+\int (n-1)\sin^{n-2}t\cos^2 t \\
=-\sin^{n-1}t\cos t+\int(n-1)\sin^{n-2}t(1-\sin^2 t)dt \\
=-\sin^{n-1}t\cos t+(n-1)I_{n-2}-(n-1)I_n \\
\implies I_n=-\dfrac{1}{n} \sin^{n-1}t\cos t+\dfrac{n-1}{n} I_{n-2}
\end{gathered} x = sin t I n = ∫ sin n t d t I n = − sin n − 1 t cos t + ∫ ( n − 1 ) sin n − 2 t cos 2 t = − sin n − 1 t cos t + ∫ ( n − 1 ) sin n − 2 t ( 1 − sin 2 t ) d t = − sin n − 1 t cos t + ( n − 1 ) I n − 2 − ( n − 1 ) I n ⟹ I n = − n 1 sin n − 1 t cos t + n n − 1 I n − 2