Kỹ thuật tích phân từng phần trong hàm ẩn (VDC)

Vấn đề 3. Kỹ thuật tích phân từng phần.
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Lời Giải:
Đặt $\begin{cases}&u=x\\&\mathrm{d}v=f'(x) \cdot \mathrm{e}^{f(x)}\mathrm{\,d}x\end{cases} \Rightarrow \begin{cases}&\mathrm{d}u=\mathrm{d}x\\&v=\mathrm{e}^{f(x)}\end{cases}$. Khi đó $\displaystyle\int\limits_0^3 x \cdot f'(x) \cdot \mathrm{e}^{f(x)}\mathrm{\,d}x=x \cdot \mathrm{e}^{f(x)}\bigg|_0^3 -\displaystyle\int\limits_0^3 \mathrm{e}^{f(x)}\mathrm{\,d}x$.
Suy ra $8=3 \cdot \mathrm{e}^{f(3)}-\displaystyle\int\limits_0^3 \mathrm{e}^{f(x)}\mathrm{\,d}x \to \displaystyle\int\limits_0^3 \mathrm{e}^{f(x)}\mathrm{\,d}x=9-8=1$.
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Lời Giải:
Xét $\displaystyle\int\limits_0^{\tfrac{\pi}{2}} f'(x)\cos^2x\mathrm{\,d}x=10$,
đặt $\begin{cases}u=\cos^2x\\\mathrm{d}v=f'(x)\cos^2x\mathrm{\,d}x\end{cases} \Rightarrow \begin{cases}&\mathrm{d}u=-\sin 2x\mathrm{\,d}x\\&v=f(x)\end{cases}.$
Khi đó $10=\displaystyle\int\limits_0^{\tfrac{\pi}{2}} f'(x)\cos^2x\mathrm{\,d}x=\cos^2xf(x)\bigg|_0^{\tfrac{\pi}{2}} +\displaystyle\int\limits_0^{\tfrac{\pi}{2}} f(x)\sin 2x\mathrm{\,d}x$
$\Leftrightarrow 10=-f(0)+\displaystyle\int\limits_0^{\tfrac{\pi}{2}} f(x)\sin 2x\mathrm{\,d}x \to \displaystyle\int\limits_0^{\tfrac{\pi}{2}} f(x)\sin 2x\mathrm{\,d}x=10+f(0)=13$.
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Lời Giải:
Ta có $\displaystyle\int\limits_1^2 f(x-1)\mathrm{\,d}x=3\xrightarrow{t=x-1}\displaystyle\int\limits_0^1 f(t)\mathrm{\,d}t=3$ hay $\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=3$.
Xét $\displaystyle\int\limits_0^1 x^3f'(x^2)\mathrm{\,d}x\xrightarrow{t=x^2}\dfrac{1}{2}\displaystyle\int\limits_0^1 tf'(t)\mathrm{\,d}t=\dfrac{1}{2}\displaystyle\int\limits_0^1 xf'(x)\mathrm{\,d}x$. Đặt $\begin{cases}&u=x\\&\mathrm{d}v=f'(x)\mathrm{\,d}x\end{cases} \Rightarrow \begin{cases}&\mathrm{d}u=\mathrm{d}x\\&v=f(x)\end{cases}.$
Khi đó $\displaystyle\int\limits_0^1 x^3f'(x^2)\mathrm{\,d}x\xrightarrow{t=x^2}\dfrac{1}{2}\displaystyle\int\limits_0^1 tf'(t)\mathrm{\,d}t=\dfrac{1}{2}\left[xf(x)\bigg|_0^1-\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x \right]=\dfrac{1}{2}[4-3]=\dfrac{1}{2}$.
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Lời Giải:
Từ giả thiết $f(x)f(2-x)=\mathrm{e}^{2x^2-4x}\xrightarrow{x=2}f(2)=1$.\\
Ta có $I=\displaystyle\int\limits_0^2 \dfrac{(x^3-3x^2)f'(x)}{f(x)}\mathrm{\,d}x$.
Đặt $\begin{cases}&u=x^3-3x^2\\&\mathrm{d}v=\dfrac{f'(x)}{f(x)}\mathrm{\,d}x\end{cases} \Rightarrow \begin{cases}&\mathrm{d}u=(3x^2-6x)\mathrm{\,d}x\\&v=\ln |f(x)|\end{cases}.$
Khi đó $I=(x^3-3x^2)\ln |f(x)|\bigg|_0^2 -\displaystyle\int\limits_0^2 (3x^2-6x)\ln |f(x)|\mathrm{\,d}x\overset{f(2)=1}=-3\displaystyle\int\limits_0^2 (x^2-2x)\ln |f(x)|\mathrm{\,d}x=-3J$.
Ta có $J=\displaystyle\int\limits_0^2 (x^2-2x)\ln |f(x)|\mathrm{\,d}x \overset{x=2-t}= \displaystyle\int\limits_2^0 \left[(2-t)^2-2(2-t)\right]\ln |f(2-t)|\mathrm{d}(2-t)$
$=\displaystyle\int\limits_2^0 \left[(2-x)^2-2(2-x)\right]\ln |f(2-x)|\mathrm{d}(2-x)=\displaystyle\int\limits_0^2 (x^2-2x)\ln |f(2-x)|\mathrm{\,d}x$.
Suy ra $2J=\displaystyle\int\limits_0^2 (x^2-2x)\ln |f(x)|\mathrm{\,d}x+\displaystyle\int\limits_0^2 (x^2-2x)\ln |f(2-x)|\mathrm{\,d}x=\displaystyle\int\limits_0^2 (x^2-2x)\ln |f(x)f(2-x)|\mathrm{\,d}x$
$=\displaystyle\int\limits_0^2 (x^2-2x)\ln \mathrm{e}^{2x^2-4x}\mathrm{\,d}x=\displaystyle\int\limits_0^2 (x^2-2x)(2x^2-4x)\mathrm{\,d}x=\dfrac{32}{15} \to J=\dfrac{16}{15}$.
Vậy $I=-3J=-\dfrac{16}{5}$.
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Lời Giải:
Ta có $\displaystyle\int\limits_{\tfrac{\pi}{4+m^2}}^{\tfrac{\pi}{2}} \left(2-\sin 2x\right)\mathrm{e}^{2\cot x}\mathrm{\,d}x=2\displaystyle\int\limits_{\tfrac{\pi}{4+m^2}}^{\tfrac{\pi}{2}} \mathrm{e}^{2\cot x}\mathrm{\,d}x-\displaystyle\int\limits_{\tfrac{\pi}{4+m^2}}^{\tfrac{\pi}{2}} \sin 2x\mathrm{e}^{2\cot x}\mathrm{\,d}x$.$(1)$
Xét $\displaystyle\int\limits_{\tfrac{\pi}{4+m^2}}^{\tfrac{\pi}{2}} \sin 2x\mathrm{e}^{2\cot x}\mathrm{\,d}x=\displaystyle\int\limits_{\tfrac{\pi}{4+m^2}}^{\tfrac{\pi}{2}} \mathrm{e}^{2\cot x}\mathrm{d}\left(\sin^2x\right)=\sin^2x \cdot \mathrm{e}^{2\cot x}\bigg|_{\tfrac{\pi}{4+m^2}}^{\tfrac{\pi}{2}} -\displaystyle\int\limits_{\tfrac{\pi}{4+m^2}}^{\tfrac{\pi}{2}} \sin^2x\left(-\dfrac{2}{\sin^2x}\right)\mathrm{e}^{2\cot x}\mathrm{\,d}x$
$=\sin^2x \cdot \mathrm{e}^{2\cot x}\bigg|_{\tfrac{\pi}{4+m^2}}^{\tfrac{\pi}{2}} +2\displaystyle\int\limits_{\tfrac{\pi}{4+m^2}}^{\tfrac{\pi}{2}} \mathrm{e}^{2\cot x}\mathrm{\,d}x$.$(2)$\\
Từ $(1)$ và $(2)$, suy ra $I=\sin^2x \cdot \mathrm{e}^{2\cot x}\bigg|_{\tfrac{\pi}{4+m^2}}^{\tfrac{\pi}{2}} =-1+\sin^2\dfrac{\pi}{4+m^2} \cdot \mathrm{e}^{2\cot \dfrac{\pi}{4+m^2}}$.
$ \to S=\ln \left(\sin^2\dfrac{\pi}{4+m^2} \cdot \mathrm{e}^{2\cot \dfrac{\pi}{4+m^2}}\right)=2\cot \left(\dfrac{\pi}{4+m^2}\right)+2\ln \left(\sin \dfrac{\pi}{4+m^2}\right)$.