Lời giải (Toán)Bài tập tích phân đặc biệt : KSTN K66 (Solved by Sơn)

Được viết bởi: Hust IT1



Question 1:$\int\frac{(x+3)dx}{x^2(x-1)(x-2)(x^2+1)}=? $

Solution:

$$\int\frac{(x+3)dx}{x^2(x-1)(x-2)(x^2+1)} =\int{\frac{\frac{3}{2}}{x^2}+\frac{\frac{11}{4}}{x}+\frac{-2}{x-1}+\frac{\frac{1}{4}}{x-2}+\frac{-x}{x^2+1}dx} =-\frac{3}{2x}+\frac{11}{4}.ln{|x|}-2.ln{|x-1|}+\frac{1}{4}.ln{|x-2|}-\frac{1}{2}.ln{|x^2+1|}+C$$


Question 2: $\int_{-1}^{1}\frac{\sqrt{x^2+1}+x-1}{\sqrt{x^2+1}+x+1}dx=?$

Solution:

$$ I=\int_{-1}^{1}\frac{\sqrt{x^2+1}+x-1}{\sqrt{x^2+1}+x+1}dx=\int_{-1}^{1}{1\ -\ \frac{2}{\sqrt{x^2+1}+x+1}}dx=2-\int_{-1}^{1}\frac{2}{\sqrt{x^2+1}+x+1}dx=2-\int_{-1}^{1}{\frac{-1}{x}.\left(\sqrt{x^2+1}-x+1\right)}dx=2-\int_{-1}^{1}{\frac{-\sqrt{x^2+1}}{x}+1-\frac{-1}{x}}dx $$


Xét:

  • $f(x)=\frac{-\sqrt{x^2+1}}{x}$ ta có $f(-x)=\frac{-\sqrt{(-x)^2+1}}{-x}=\frac{\sqrt{x^2+1}}{x} $ mà $f(x)=-f(-x)\ => $f(x) hàm lẻ
  • $f(x)=1$ là hàm chẵn
  • $f(x)=\frac{-1}{x}$ ta có $f(x)=\frac{1}{x}$ => $f(x)$ là hàm lẻ

$$\Rightarrow I=2-\int_{-1}^{1}{\frac{-\sqrt{x^2+1}}{x}+1-\frac{-1}{x}}dx=2-\int_{-1}^{1}1dx=0$$


Question 3: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{{cos}^3{x}}{1-x+\sqrt{1+x^2}}dx=?$

Solution:

$$I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{{cos}^3{x}}{1-x+\sqrt{1+x^2}}dx=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{{cos}^3{x}.(1-x-\sqrt{1+x^2})}{2x}dx=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}{\frac{{cos}^3{x}}{2x}-\frac{{cos}^3{x}}{2}-\frac{{cos}^3{x}.\sqrt{1+x^2}}{2x}}dx$$


Xét:

  • $f(x)=\frac{{cos}^3{x}}{2x}$ ta có $f(-x)=-\frac{{cos}^3{x}}{2x}$ => f(x) là hàm lẻ
  • $f(x)=\frac{{cos}^3{x}}{2}$ ta có $f(-x)=\frac{{cos}^3{x}}{2}$ => f(x) là hàm chẵn
  • $f(x)=\frac{{cos}^3{x}.\sqrt{1+x^2}}{2x}$ ta có $f(-x)=-\frac{{cos}^3{-x}.\sqrt{1+(-x)^2}}{2x}=-\frac{{cos}^3{x}.\sqrt{1+x^2}}{2x}$ => f(x) là hàm lẻ

$$=> I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{{cos}^3{x}}{2}dx=\frac{1}{2}.\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\cos^3{x}dx=\frac{1}{8}.\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}{cos{3x}+3.cos{x}}dx=\frac{1}{8}.(\frac{-2}{3}+3.2)=\frac{2}{3}$$


Question 4: $I=\int_{-0.5}^{0.5}\frac{x.ln{\frac{1+x}{1-x}}}{e^x+1}dx=?$

Solution:

  • Áp dụng $$ \int_{-a}^{a} \frac{f(x)}{b^x+1}dx = \int_{0}^{a} f(x)dx $$
  • $x.ln{\frac{1+x}{1-x}}$ Là hàm chẵn
  • Gọi f(x) là $x.ln{\frac{1+x}{1-x}}$ Và là một hàm chẵn
  • $e>0$
  • $$I=\int_{-0.5}^{0.5}\frac{x.ln{\frac{1+x}{1-x}}}{e^x+1}dx=\int_{0}^{0.5}{x.ln{\frac{1+x}{1-x}}}dx$$
  • Đặt $$\begin{matrix}u=ln{\frac{1+x}{1-x}}&dv=xdx\\du=\frac{2}{1-x^2}dx&v=\frac{x^2}{2}\end{matrix}$$

$$I=\frac{x^2}{2}ln{\frac{1+x}{1-x}}|_0^{0.5}-\int_{0}^{0.5}\frac{x^2}{2}.\frac{2}{1-x^2}dx= \frac{1}{8}ln{3}-\int_{0}^{0.5}\frac{x^2}{2}.\frac{2}{1-x^2}dx=\frac{1}{8}ln{3}-\int_{0}^{0.5}{-1+\frac{1}{1-x^2}}dx=\frac{1}{8}ln{3}+\frac{1}{2}-(\frac{1}{2}\ln{\frac{1+x}{1-x}})|_0^{0.5}=\frac{1}{8}ln{3}+\frac{1}{2}-\frac{1}{2}\ln{3=\frac{1}{2}-\frac{3}{8}\ln{3}} I=\frac{1}{2}-\frac{3}{8}\ln{3}$$


Question 5: $\int_{-1}^{1}{\sin{x}.\cos{2x}.\sin{3x}.\tan{\frac{x}{4}}}dx=?$

Solution:

Xét:

  • $f(x)=sin{x}$ ta có $f(-x)=sin{-x}=-sin{x}$ => f(x) là hàm lẻ
  • $f(x)=cos{2x}$ ta có $f(-x)=cos{-2x}=cos{2x}$ => f(x) là hàm chẵn
  • $f(x)=sin{3x}$ ta có $f(-x)=sin{-3x}=-sin{3x}$ => f(x) là hàm lẻ
  • $f(x)=tan{\frac{x}{4}}$ ta có $f(-x)=tan{-\frac{x}{4}}=-tan{\frac{x}{4}}$ => f(x) là hàm lẻ => $f(x)=\sin{x}.\cos{2x}.\sin{3x}.\tan{\frac{x}{4}}$ là hàm lẻ

=> $\int_{-1}^{1}{\sin{x}.\cos{2x}.\sin{3x}.\tan{\frac{x}{4}}}dx=0$


Question 6: $I=\int_{0}^{\frac{\pi}{2}}\frac{{sin}^{2021}{x}\ .\ cos{x}}{{sin}^{2020}{x}+{cos}^{2020}{x}}dx=?$

Solution:

  • Đặt $x=\frac{\pi}{2}-t => dx=-dt$ ta có:

$$I=-\int_{\frac{\pi}{2}}^{\frac{-\pi}{2}}\frac{{cos}^{2021}{t}\ .\ sin{t}}{{cos}^{2020}{t}+{sin}^{2020}{t}}dt=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{{cos}^{2021}{x}\ .\ sin{x}}{{cos}^{2020}{x}+{sin}^{2020}{x}}dx => 2I=\int_{0}^{\frac{\pi}{2}}\frac{{sin}^{2021}{x}\ .\ cos{x}}{{sin}^{2020}{x}+{cos}^{2020}{x}}dx+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{{cos}^{2021}{x}\ .\ sin{x}}{{cos}^{2020}{x}+{sin}^{2020}{x}}dx=\int_{0}^{\frac{\pi}{2}}\frac{{sin}^{2021}{x}\ .\ cos{x}+{cos}^{2021}{x}\ .\ sin{x}}{{sin}^{2020}{x}+{cos}^{2020}{x}}dx=\int_{0}^{\frac{\pi}{2}}\frac{sin{x}cos{x}({sin}^{2020}{x}+{cos}^{2020}{x})}{{sin}^{2020}{x}+{cos}^{2020}{x}}dx=\int_{0}^{\frac{\pi}{2}}{sin{x}cos{x}}dx=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}sin{2x}dx=\frac{1}{2} => I=\frac{1}{4}$$


Question 7: $I=\int_{3\pi}^{7\pi}{\sin{x}\cos^2{2x}\sin{3x}\sin{4x}}dx=?$

Solution:

  • Đặt $x=5\pi-t => dx=-dt$ ta có: $$I=-\int_{2\pi}^{-2\pi}{\sin{(5\pi-t)}\cos^2{2(5\pi-t)}\sin{3(5\pi-t)}\sin{4(5\pi-t)}}dt=\int_{-2\pi}^{2\pi}{\sin{t}\cos^2{2t}\sin{3t}\sin{(-4t)}}dt=-\int_{-2\pi}^{2\pi}{\sin{t}\cos^2{2t}\sin{3t}\sin{4t}}dt=-\int_{-2\pi}^{2\pi}{\sin{x}\cos^2{2x}\sin{3x}\sin{4x}}dx$$

Xét:

  • $f(x)=sin{x}$ ta có $f(-x)=sin{-x}=-sin{x}$ => f(x) là hàm lẻ
  • $f(x)={cos}^2{2x}$ ta có $f(-x)=\cos^2{(-2x)}={cos}^2{2x}$ => f(x) là hàm chẵn
  • $f(x)=sin{3x}$ ta có $f(-x)=sin{-3x}=-sin{3x}$ => f(x) là hàm lẻ
  • $f(x)=sin{4x}$ ta có $f(-x)=sin{-4x}=-sin{4x}$ => f(x) là hàm lẻ

$=> f(x)=\sin{x}\cos^2{2x}\sin{3x}\sin{4x}$ là hàm lẻ $=> I=0$


Question 8: $I=\int_{0}^{\pi}{x\sin{x}\cos{2x}}dx=?$

Solution:

  • Đặt $x=\pi-t => dx=-dt$ ta có $$I=-\int_{\pi}^{0}{(\pi-t)\sin{(\pi-t)}\cos{2(\pi-t)}}dt=\int_{0}^{\pi}{(\pi-t)\sin{t}\cos{2t}}dt => I=\frac{\pi}{2}\int_{0}^{\pi}{\sin{t}\cos{2t}}dt=\frac{\pi}{2}\int_{0}^{\pi}{\sin{t}\cos{2t}}dt=\frac{\pi}{2}\int_{0}^{\pi}{2\sin{t}cos^2{t}}-\sin{t}dt==\pi\int_{0}^{\pi}{\sin{t}cos^2{t}}dt-\frac{\pi}{2}\int_{0}^{\pi}\sin{t}dt=\pi\int_{0}^{\pi}{\sin{t}cos^2{t}}dt-\pi$$
  • Đặt $u=cos{t} => dx=-\frac{1}{sin{t}}dt$ ta có $I=-\pi\int_{1}^{-1}u^2du-\pi=\frac{-\pi}{3}$

Question 9: $I=\int_{1}^{4}\frac{(x+1)^{2021}}{(x+1)^{2021}+(6-x)^{2021}}dx=?$

Solution:

  • Đặt $x=5-t => dx=-dt$ ta có $$I=-\int_{4}^{1}\frac{(5-t\ +1)^{2021}}{(5-t\ +1)^{2021}+(6-5+t)^{2021}}dt=\int_{1}^{4}\frac{(6-t)^{2021}}{(6-t)^{2021}+(1+t)^{2021}}dt=\int_{1}^{4}\frac{(6-x)^{2021}}{(6-x)^{2021}+(1+x)^{2021}}dx$$

$$=> 2I=\int_{1}^{4}\frac{(6-x)^{2021}}{(6-x)^{2021}+(1+x)^{2021}}dx+\int_{1}^{4}\frac{(x+1)^{2021}}{(x+1)^{2021}+(6-x)^{2021}}dx=\int_{1}^{4}\frac{(6-x)^{2021}+(x+1)^{2021}}{(6-x)^{2021}+(1+x)^{2021}}dx=\int_{1}^{4}1dx=3 => I=\frac{3}{2}$$


Question 10: $I=\int_{0}^{1}\frac{x^3}{1+x+\frac{x^2}{2}+\frac{x^3}{6}}dx=?$

Solution: $$I=\int_{0}^{1}\frac{{6x}^3}{6+6x+3x^2+x^3}dx=\int_{0}^{1}{6-\frac{6.(6+6x+3x^2)}{6+6x+3x^2+x^3}}dx=\int_{0}^{1}6dx-\int_{0}^{1}\frac{6.(6+6x+3x^2)}{6+6x+3x^2+x^3}dx=6.(1-\int_{0}^{1}{\frac{(6+6x+3x^2)}{6+6x+3x^2+x^3})}dx=6.(1-\int_{0}^{1}{\frac{(6+6x+3x^2)}{6+6x+3x^2+x^3})}dx Ta đặt t=6+6x+3x^2+x^3 => dx=\frac{1}{6+6x+{3x}^2}dt I=6.(1-\int_{6}^{16}\frac{1}{t}dt)=6.(1-\ln{\frac{8}{3}})$$

Posted on July 21, 2021 12:37:00 AM


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