[TEX][/TEX]
1. [tex]\int\limits_{0}^{x}xsinxcos^2xdx[/tex]
2. [tex]\int\limits_{0}^{ln3}\frac{dx}{e^x + 2}[/tex]
3. [tex]\int\limits_{0}^{pi/2}e^(sin^2x)sinxcos^3xdx[/tex] ghi chú [tex]e^sin^2x[/tex]
4. [tex]\int\limits_{-1}^{1}\frac{\sqrt{1 - x^2}}{1 + 2^x}dx[/tex]
5. [tex]\int\limits_{0}^{pi/4}\frac{sin4x}{1 + cos^2x}dx[/tex]
6. [tex]\int\limits_{}^{}\frac{sinx}{1 + sin2x}dx[/tex] (cái này là nguyên hàm)
7. [tex]\int\limits_{0}^{pi/4}\frac{cos2xdx}{(sinx + cosx + 2)^3}[/tex]
8. [tex]\int\limits_{pi/6}^{pi/3}\frac{dx}{sin^4xcosx}[/tex]
1) dùng từng phần
2) thêm bớt 2,-2 ở tử
3) từng phần
[TEX]\left{\begin{u=cos^2x}\\{dv=e^{sin^2x}.sinx.cosxdx} \Rightarrow \left{\begin{du=-2sinx.cosxdx}\\{v=\frac{e^{sin^2x}}{2}}[/TEX]
\Rightarrow [TEX]I=(\frac{e^{sin^2x}}{2}.cos^2x)|_0^{\frac{\pi}{2}}[/TEX][TEX]+\int_{0}^{\frac{\pi}{2}}e^{sin^2x}.sinx.cosxdx[/TEX]
[TEX]=\frac{e^{sin^2x}}{2}.cos^2x|_0^{\frac{\pi}{2}}+[/TEX][TEX]\int_{0}^{\frac{\pi}{2}}e^{sin^2x}d(sin^2x)[/TEX]
4)
[TEX]\int\limits_{-1}^{1}\frac{\sqrt{1 - x^2}}{1 + 2^x}dx=\int_{0}^{1}\sqrt[]{1-x^2}dx[/TEX] đặt [TEX]x=sint[/TEX]
5)
[tex]\int\limits_{0}^{pi/4}\frac{sin4x}{1 + cos^2x}dx[/tex][TEX]=\int_{0}^{\frac{\pi}{4}}\frac{4cosx(2cos^2x-1)sinxdx}{1+cos^2x}[/TEX]
đặt [TEX]t=cosx \Rightarrow I=\int_{\frac{1}{\sqrt[]{2}}}^{1}\frac{4t(2t^2-1)dt}{1+t^2}=8\int_{\frac{1}{\sqrt[]{2}}}^{1}\frac{t^2.tdt}{1+t^2}-4\int_{\frac{1}{\sqrt[]{2}}}^{1}\frac{tdt}{1+t^2}[/TEX]
6)
[tex]\int\limits_{}^{}\frac{sinx}{1 + sin2x}dx[/tex] [TEX]=\int_{}^{}\frac{sinx}{(sinx+cosx)^2}dx[/TEX] cái này liên kết với cái này [TEX]\int_{}^{}\frac{cosx}{(sinx+cosx)^2}dx[/TEX]
7)
[TEX]\int_{0}^{\frac{\pi}{4}}\frac{cos2xdx}{(sinx+cosx+2)^3}=\int_{0}^{\frac{\pi}{4}}\frac{(cosx-sinx)(cosx+sinx)dx}{(sinx+cosx+2)^3}[/TEX]
đặt [TEX]t=sinx+cosx+2 \Rightarrow dt=(cosx-sinx)dx[/TEX]
[TEX]\Rightarrow I=\int_{3}^{2+\sqrt[]{2}}\frac{(t-2)dt}{t^3}[/TEX]
8)
[TEX]\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{sin^4x.cosx}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{cosxdx}{sin^4x(1-sin^2x)}[/TEX]
[TEX]=\int_{\frac{1}{2}}^{\frac{\sqrt[]{3}}{2}}\frac{dt}{t^4(t^2-1)}[/TEX] (đặt [TEX]t=sinx[/TEX])
[TEX]=\int_{\frac{1}{2}}^{\frac{\sqrt[]{3}}{2}}(-\frac{1}{t^2}-\frac{1}{t^4}+\frac{1}{t^2-1})dt[/TEX]
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