Kỹ thuật đổi biến trong Tích Phân (VDC)

Vấn đề 2. Kỹ thuật đổi biến.
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Lời Giải:
Đặt $t=\ln (x^2+1),$ suy ra $\mathrm{\,d}t=\dfrac{2x\mathrm{\,d}x}{x^2+1} \to \dfrac{x\mathrm{\,d}x}{x^2+1}=\dfrac{\mathrm{\,d}t}{2}$.\\
Đổi cận: $\begin{cases}&x=0 \to t=0\\&x=\sqrt{\mathrm{e}^{2017}-1} \to t=2017\end{cases}.$\\
Khi đó $I=\dfrac{1}{2}\displaystyle\int\limits_0^{2017} f(t)\mathrm{\,d}t=\dfrac{1}{2}\displaystyle\int\limits_0^{2017} f(x)\mathrm{\,d}x=\dfrac{1}{2} \cdot 2=1$.
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Lời Giải:
Xét $\displaystyle\int\limits_1^9 \dfrac{f\left(\sqrt{x}\right)}{\sqrt{x}}\mathrm{\,d}x=4$. Đặt $t=\sqrt{x} \Rightarrow t^2=x,$ suy ra $2t\mathrm{\,d}t=\mathrm{d}x$.\\
Đổi cận $\begin{cases}&x=1 \to t=1\\&x=9 \to t=3\end{cases}$.
Suy ra $4=\displaystyle\int\limits_1^9 \dfrac{f\left(\sqrt{x}\right)}{\sqrt{x}}\mathrm{\,d}x=2\displaystyle\int\limits_1^3 f(t)2\mathrm{\,d}t \to \displaystyle\int\limits_1^3 f(t)\mathrm{\,d}t=2$.\\
Xét $\displaystyle\int\limits_0^{\tfrac{\pi}{2}} f\left(\sin x\right)\cos x\mathrm{\,d}x=2$. Đặt $u=\sin x,$ suy ra $\mathrm{\,d}u=\cos x\mathrm{\,d}x$.\\
Đổi cận $\begin{cases}&x=0 \to u=0\\&x=\dfrac{\pi}{2} \to u=1\end{cases}$.
Suy ra $2=\displaystyle\int\limits_0^{\tfrac{\pi}{2}} f\left(\sin x\right)\cos x\mathrm{\,d}x=\displaystyle\int\limits_0^1 f(t)\mathrm{\,d}t$.\\
Vậy $I=\displaystyle\int\limits_0^3 f(x)\mathrm{\,d}x=\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x+\displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x=4$.
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Lời Giải:
Xét $\displaystyle\int\limits_0^{\tfrac{\pi}{4}} f\left(\tan x\right)\mathrm{d}x=4$.\\
Đặt $t=\tan x,$ suy ra $\mathrm{\,d}t=\dfrac{1}{\cos^2x}\mathrm{\,d}x=\left(\tan^2x+1\right)\mathrm{d}x \to \mathrm{\,d}x=\dfrac{\mathrm{\,d}t}{1+t^2}$.\\
Đổi cận: $\begin{cases}&x=0 \to t=0\\&x=\dfrac{\pi}{4} \to t=1\end{cases}$.
Khi đó $4=\displaystyle\int\limits_0^{\tfrac{\pi}{4}} f\left(\tan x\right)\mathrm{d}x=\displaystyle\int\limits_0^1 \dfrac{f(t)}{t^2+1}\mathrm{\,d}t=\displaystyle\int\limits_0^1 \dfrac{f(x)}{x^2+1}\mathrm{\,d}x$.\\
Từ đó suy ra $I=\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=\displaystyle\int\limits_0^1 \dfrac{f(x)}{x^2+1}\mathrm{\,d}x+\displaystyle\int\limits_0^1 \dfrac{x^2f(x)}{x^2+1}\mathrm{\,d}x=4+2=6$.
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Lời Giải:
Xét $A=\displaystyle\int\limits_0^{\tfrac{\pi}{4}} \tan x \cdot f\left(\cos^2x\right)\mathrm{d}x=1$. Đặt $t=\cos^2x$.\\
Suy ra $\mathrm{\,d}t=-2\sin x\cos x\mathrm{\,d}x=-2\cos^2x\tan x\mathrm{\,d}x=-2t \cdot \tan x\mathrm{\,d}x \to \tan x\mathrm{\,d}x=-\dfrac{\mathrm{\,d}t}{2t}$.\\
Đổi cận: $\begin{cases}&x=0 \to t=1\\&x=\dfrac{\pi}{4} \to t=\dfrac{1}{2}\end{cases}.$\\
Khi đó $1=A=-\dfrac{1}{2}\displaystyle\int\limits_1^{\tfrac{1}{2}} \dfrac{f(t)}{t}\mathrm{\,d}t=\dfrac{1}{2}\displaystyle\int\limits_{\tfrac{1}{2}}^1 \dfrac{f(t)}{t}\mathrm{\,d}t=\dfrac{1}{2}\displaystyle\int\limits_{\tfrac{1}{2}}^1 \dfrac{f(x)}{x}\mathrm{\,d}x \to \displaystyle\int\limits_{\tfrac{1}{2}}^1 \dfrac{f(x)}{x}\mathrm{\,d}x=2$.\\
Xét $B=\displaystyle\int\limits_\mathrm{e}^{\mathrm{e}^2} \dfrac{f\left(\ln^2x\right)}{x\ln x}\mathrm{\,d}x=1$. Đặt $u=\ln^2x$.\\
Suy ra $\mathrm{\,d}u=\dfrac{2\ln x}{x}\mathrm{\,d}x=\dfrac{2\ln^2x}{x\ln x}\mathrm{\,d}x=\dfrac{2u}{x\ln x}\mathrm{\,d}x \to \dfrac{\mathrm{\,d}x}{x\ln x}=\dfrac{\mathrm{\,d}u}{2u}$.\\
Đổi cận: $\begin{cases}&x=e \to u=1\\&x=\mathrm{e}^2 \to u=4\end{cases}.$\\
Khi đó $1=B=\dfrac{1}{2}\displaystyle\int\limits_1^4 \dfrac{f(u)}{u}\mathrm{\,d}u=\dfrac{1}{2}\displaystyle\int\limits_1^4 \dfrac{f(x)}{x}\mathrm{\,d}x \to \displaystyle\int\limits_1^4 \dfrac{f(x)}{x}\mathrm{\,d}x=2$.\\
Xét tích phân cần tính $I=\displaystyle\int\limits_{\tfrac{1}{2}}^2 \dfrac{f(2x)}{x}\mathrm{\,d}x$.\\
Đặt $v=2x$,
suy ra $\begin{cases}&\mathrm{d}x=\dfrac{1}{2}\mathrm{\,d}v\\&x=\dfrac{v}{2}\end{cases}$.
Đổi cận: $\begin{cases}&x=\dfrac{1}{4} \to v=\dfrac{1}{2}\\&x=2 \to v=4\end{cases}.$\\
Khi đó $I=\displaystyle\int\limits_{\tfrac{1}{2}}^4 \dfrac{f(v)}{v}\mathrm{\,d}v=\displaystyle\int\limits_{\tfrac{1}{2}}^4 \dfrac{f(x)}{x}\mathrm{\,d}x=\displaystyle\int\limits_{\tfrac{1}{2}}^1 \dfrac{f(x)}{x}\mathrm{\,d}x+\displaystyle\int\limits_1^4 \dfrac{f(x)}{x}\mathrm{\,d}x=2+2=4$.
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Lời Giải:
Đặt $x=\dfrac{1}{t},$ suy ra $\mathrm{\,d}x=-\dfrac{1}{t^2}\mathrm{\,d}t$.
Đổi cận: $\begin{cases}&x=\dfrac{1}{2} \to t=2\\&x=2 \to t=\dfrac{1}{2}\end{cases}.$\\
Khi đó $I=\displaystyle\int\limits_2^{\tfrac{1}{2}} \dfrac{f\left(\dfrac{1}{t}\right)}{\dfrac{1}{t^2}+1} \cdot \left(-\dfrac{1}{t^2}\right)\mathrm{d}t=\displaystyle\int\limits_{\tfrac{1}{2}}^2 \dfrac{f\left(\dfrac{1}{t}\right)}{t^2+1}\mathrm{\,d}t=\displaystyle\int\limits_{\tfrac{1}{2}}^2 \dfrac{f\left(\dfrac{1}{x}\right)}{x^2+1}\mathrm{\,d}x$.\\
Suy ra $2I=\displaystyle\int\limits_{\tfrac{1}{2}}^2 \dfrac{f(x)}{x^2+1}\mathrm{\,d}x+\displaystyle\int\limits_{\tfrac{1}{2}}^2 \dfrac{f\left(\dfrac{1}{x}\right)}{x^2+1}\mathrm{\,d}x=\displaystyle\int\limits_{\tfrac{1}{2}}^2 \dfrac{f(x)+f\left(\dfrac{1}{x}\right)}{x^2+1}\mathrm{\,d}x=\displaystyle\int\limits_{\tfrac{1}{2}}^2 \dfrac{x^2+\dfrac{1}{x^2}+2}{x^2+1}\mathrm{\,d}x$\\
$=\displaystyle\int\limits_{\tfrac{1}{2}}^2 \dfrac{x^2+1}{x^2}\mathrm{\,d}x=\displaystyle\int\limits_{\tfrac{1}{2}}^2 \left(1+\dfrac{1}{x^2}\right)\mathrm{d}x=\left(x-\dfrac{1}{x}\right)\bigg|_{\tfrac{1}{2}}^2 =3 \to I=\dfrac{3}{2}$.
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Lời Giải:
Đặt $t=-x \to \mathrm{\,d}x=-\mathrm{\,d}t$.
Đổi cận: $\begin{cases}&x=-\dfrac{3\pi}{2} \to t=\dfrac{3\pi}{2}\\&x=\dfrac{3\pi}{2} \to t=-\dfrac{3\pi}{2}\end{cases}.$\\
Khi đó $I=-\displaystyle\int\limits_{\tfrac{3\pi}{2}}^{-\tfrac{3\pi}{2}} f(-t)\mathrm{\,d}t=\displaystyle\int\limits_{-\tfrac{3\pi}{2}}^{\tfrac{3\pi}{2}} f(-t)\mathrm{\,d}t=\displaystyle\int\limits_{-\tfrac{3\pi}{2}}^{\tfrac{3\pi}{2}} f(-x)\mathrm{\,d}x$.\\
Suy ra $2I=\displaystyle\int\limits_{-\tfrac{3\pi}{2}}^{\tfrac{3\pi}{2}} \left[f(t)+f(-t)\right]\mathrm{d}t=\displaystyle\int\limits_{-\tfrac{3\pi}{2}}^{\tfrac{3\pi}{2}} \sqrt{2+2\cos 2t}\mathrm{\,d}t=\displaystyle\int\limits_{-\tfrac{3\pi}{2}}^{\tfrac{3\pi}{2}} 2\left|\cos t\right|\mathrm{\,d}t\overset{CASIO}=12 \to I=6$.
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Lời Giải:
Đặt $x=t^5+4t+3,$ suy ra $\mathrm{\,d}x=(5t^4+4)\mathrm{\,d}t$.
Đổi cận $\begin{cases}&x=-2 \to t=-1\\&x=8 \to t=1\end{cases}.$\\
Khi đó $\displaystyle\int\limits_{-2}^8 f(x)\mathrm{\,d}x=\displaystyle\int\limits_{-1}^1 f(t^5+4t+3)(5t^4+4)\mathrm{\,d}t=\displaystyle\int\limits_{-1}^1 (2t+1)(5t^4+4)\mathrm{\,d}t=10$.
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Lời Giải:
Từ giả thiết $m \cdot f(x)+n \cdot f(1-x)=g(x)$, lấy tích phân hai vế ta được\\
$\displaystyle\int\limits_0^1 \left[m \cdot f(x)+n \cdot f(1-x)\right]\mathrm{d}x=\displaystyle\int\limits_0^1 g(x)\mathrm{\,d}x$\\
Suy ra $m+n\displaystyle\int\limits_0^1 f(1-x)\mathrm{\,d}x=1$ (do $\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=\displaystyle\int\limits_0^1 g(x)\mathrm{\,d}x=1$ ).$(1)$\\
Xét tích phân $\displaystyle\int\limits_0^1 f(1-x)\mathrm{\,d}x$. Đặt $t=1-x$, suy ra $\mathrm{\,d}t=-\mathrm{\,d}x$.
Đổi cận: $\begin{cases}&x=0 \to t=1\\&x=1 \to t=0\end{cases}.$\\
Khi đó $\displaystyle\int\limits_0^1 f(1-x)\mathrm{\,d}x=-\displaystyle\int\limits_1^0 f(t)\mathrm{\,d}t=\displaystyle\int\limits_0^1 f(t)\mathrm{\,d}t=\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=1$.$(2)$\\
Từ $(1)$ và $(2)$, suy ra $m+n=1$.
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Lời Giải:
Ta có $f'(x)=f'(1-x) \to f(x)=-f(1-x)+C$.
Suy ra $f(0)=-f(1)+C \to \xrightarrow{f(0)=1, f(1)=41}C=42$.
Suy ra $f(x)=-f(1-x)+42 \to f(x)+f(1-x)=42$
$ \to \displaystyle\int\limits_0^1 \left[f(x)+f(1-x)\right]\mathrm{d}x=\displaystyle\int\limits_0^1 42\mathrm{\,d}x=42$.$(1)$
Vì $f'(x)=f'(1-x) \to \displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=\displaystyle\int\limits_0^1 f(1-x)\mathrm{\,d}x$.$(2)$
Từ $(1)$ và $(2)$,
suy ra $\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=\displaystyle\int\limits_0^1 f(1-x)\mathrm{\,d}x=21$.
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Lời Giải:
Đặt $u=f(x)$, ta thu được $u^3+u=x$. Suy ra $(3u^2+1)\mathrm{\,d}u=\mathrm{d}x$.\\
Từ $u^3+u=x$,
đổi cận $\begin{cases}&x=0 \to u=0\\&x=2 \to u=1\end{cases}$.
Khi đó $I=\displaystyle\int\limits_0^1 u(3u^2+1)\mathrm{\,d}u=\dfrac{5}{4}$.\\
Cách khác. Nếu bài toán cho $f(x)$ có đạo hàm liên tục thì ta làm như sau:\\
Từ giả thiết $f^3(x)+f(x)=x \to \begin{cases}&f^3(0)+f(0)=0\\&f^3(2)+f(2)=2\end{cases} \to \begin{cases}&f(0)=0\\&f(2)=1\end{cases}$.$(*)$\\
Cũng từ giả thiết $f^3(x)+f(x)=x$, ta có $f'(x) \cdot f^3(x)+f'(x) \cdot f(x)=x \cdot f'(x)$.\\
Lấy tích phân hai vế $\displaystyle\int\limits_0^2 \left[f'(x) \cdot f^3(x)+f'(x) \cdot f(x)\right]\mathrm{d}x=\displaystyle\int\limits_0^2 x \cdot f'(x)\mathrm{\,d}x$\\
$ \to \left(\dfrac{\left[f(x)\right]^4}{4}+\dfrac{\left[f(x)\right]^2}{2}\right)\bigg|_0^2 =xf(x)\bigg|_0^2 -\displaystyle\int\limits_0^2 f(x)\mathrm{\,d}x\xrightarrow{(*)}\displaystyle\int\limits_0^2 f(x)\mathrm{\,d}x=\dfrac{5}{4}$.\\