tích phân

[maths.dang]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
Chắc suất Đại học top - Giữ chỗ ngay!!

ĐĂNG BÀI NGAY để cùng trao đổi với các thành viên siêu nhiệt tình & dễ thương trên diễn đàn.

1)


[TEX]I=\int_{0}^{pi/2}(2x-1)cos^2x .dx[/TEX]




2)


[TEX]I=\int_{0}^{pi/3}sin^2xtanx dx[/TEX]




3)


[TEX]I=\int_{-1}^{3}\frac{X-3}{3\sqrt[]{X+1}+{X+3}}.dx[/TEX]




4)



[TEX]I=\int_{0}^{1}x^5\sqrt[]{1-x^2}.dx[/TEX]



5)



[TEX]I=\int_{0}^{pi/2}e^{3x}sin5x.dx[/TEX]




6)


[TEX]I=\int_{0}^{\sqrt[]{3}}\sqrt[]{x^3-1}x^5.dx[/TEX]]



7)



[TEX]I=\int_{0}^{pi/2}\frac{cos3x}{sinx+1}.dx[/TEX]




8)


[TEX]I=\int_{0}^{pi/2}\frac{sinx.dx}{sin^2x + 2cosx.cos^2\frac{x}{2}}[/TEX]




9)


[TEX]I=\int_{pi/4}^{pi/2}\frac{sinx-cosx}{\sqrt[]{1+sin2x}}.dx[/TEX]





10)


[TEX] I= \int_{0}^{pi/2}\frac{cos2x}{(sinx-cosx+3)^3}.dx[/TEX]


giải chi tiết nha các bạn! tk

 

[newstarinsky]

1)
2)


[TEX]I=\int_{0}^{pi/3}sin^2xtanx dx[/TEX]


giải chi tiết nha các bạn! tk

$I=\int_{0}^{\frac{\pi}{3}}\frac{sin^3x}{cosx}dx\\
=-\int_{0}^{\frac{\pi}{3}}\frac{sin^2x}{cosx}d(cosx)\\
=-\int_{0 }^{\frac{\pi}{3}}\frac{1-cos^2x}{cosx}d(cosx)\\
=-ln(cosx)+\frac{1}{2}cosx$
Thay cận vào nhé

6)
[TEX]I=\int_{0}^{\sqrt{3}}\sqrt{x^3-1}.x^5dx[/TEX]

$I=\int_{0}^{\sqrt{3}}\sqrt{x^3-1}.x^3.x^2dx\\
=\frac{1}{3}\int_{0}^{\sqrt{3}}\sqrt{x^3-1}(x^3-1+1)d(x^3)\\
=\frac{1}{3}\int_{0}^{\sqrt{3}}\sqrt{x^3-1}(x^3-1)d(x^3)+\frac{1}{3}\int_{0}^{\sqrt{3}}\sqrt{x^3-1}d(x^3)\\
=\frac{2}{15}(x^3-1)^{\frac{5}{2}}+\frac{2}{9}(x^3-1)^{\frac{3}{2}}$
Thay cận thôi

9)
[TEX]I=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{sinx-cosx}{\sqrt{1+sin2x}dx[/TEX]

Ta có $\sqrt{1+sin2x}=\sqrt{sin^2x+2sinx.cosx+cos^2x}=
\sqrt{(sinx+cosx)^2}=sinx+cosx$
Mà $d(sinx+cosx)=(cosx-sinx)dx$
Nên $I=-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{sinx+cosx}d(sinx+cosx)\\
=-ln(sinx+cosx)$
Ok

 

[newstarinsky]

4)



[TEX]I=\int_{0}^{1}x^5\sqrt[]{1-x^2}.dx[/TEX]


giải chi tiết nha các bạn! tk

Đặt $x^2=sint\Rightarrow 2xdx=costdt$
Đổi cận $x=0\Rightarrow t=0\\x=1\Rightarrow t=\frac{\pi}{2}$

$I=\int_{0}^{\frac{\pi}{2}}\frac{1}{2}cost.sint
\sqrt{1-sint}dt\\
=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(1-1+sint)\sqrt{1-sint}d(sint)\\
=-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(1-sint)\sqrt{1-sint}d(sint)+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}
\sqrt{1-sint}d(sint)\\
=\frac{1}{5}(1-sint)^{\frac{5}{2}}-\frac{1}{3}(1-sint)^{\frac{3}{2}}$
Thay cận

7)
[TEX]I=\int_{0}^{\frac{\pi}{2}}\frac{cos3x}{sinx+1}[/TEX]

$I=\int_{0}^{\frac{\pi}{2}}\frac{cosx(4cos^2x-3)}{sinx+1}dx\\
=\int_{0}^{\frac{\pi}{2}}\frac{-3+4(1-sin^2x)}{sinx+1}d(sinx)\\
=\int_{0}^{\frac{\pi}{2}}\frac{-3}{sinx+1}+\int_{0}^{\frac{\pi}{2}}4(1-sinx)d(sinx)\\
=-3.ln(sinx+1)-4(sinx-1)^2$

 

[maths.dang]

còn mấy bài mấy bạn giải lun nhaz'..................................................

 

[nguyenbahiep1]

1)
[TEX]I=\int_{0}^{\frac{\pi}{2}}(2x-1)cos^2x .dx\\ \int_{0}^{\frac{\pi}{2}}(x -\frac{1}{2}).(1 +cos 2x).dx\\ \int_{0}^{\frac{\pi}{2}}(x -\frac{1}{2}).dx + \int_{0}^{\frac{\pi}{2}}(x -\frac{1}{2}).cos 2x.dx\\\frac{x^2 -x}{2} + \int_{0}^{\frac{\pi}{2}}(x -\frac{1}{2}).cos 2x.dx = \frac{(\pi)^2 -2.\pi}{8} + \frac{1}{2}.sin2x.(x -\frac{1}{2}) - \int_{0}^{\frac{\pi}{2}}\frac{1}{2}.sin2x.dx\\ \frac{(\pi)^2 -2.\pi}{8} - \frac{co2x}{4} = \frac{(\pi)^2 -2.\pi +4}{8}[/TEX]

3)
[TEX]I=\int_{-1}^{3}\frac{x-3}{3\sqrt{x+1}+{x+3}}.dx\\ \sqrt{x+1} = u \Rightarrow x+1 = u^2 \Rightarrow 2u.du =dx\\I=\int_{0}^{2}\frac{(u^2 -4).2udu}{u^2 +3u +2} = \int_{0}^{2}\frac{(u-2).(u+2).2udu}{(u+1).(u+2)} = \int_{0}^{2}\frac{(2.u^2-4u)du}{u+1} \\ \int_{0}^{2}( 2u -6 + \frac{6.du}{u+1}) = u^2 - 6u + 6ln|u+1| = 6ln3 - 8[/TEX]
5)
[TEX]I=\int_{0}^{\frac{\pi}{2}}e^{3x}sin5x.dx\\ I= \frac{e^{3x}}{3} .sin5x - \int_{0}^{\frac{\pi}{2}}\frac{e^{3x}}{3}.5.cos5x.dx\\ \frac{e^{\frac{3.\pi}{2}}}{3} - \frac{5.e^{3x}.cos5x}{9} - \frac{25}{9}.I\\ \frac{34}{9}.I = \frac{e^{\frac{3.\pi}{2}}}{3} + \frac{5}{9}[/TEX]
8)
[TEX]I=\int_{0}^{\frac{\pi}{2}}\frac{sinx.dx}{sin^2x + 2cosx.cos^2\frac{x}{2}}=\int_{0}^{\frac{\pi}{2}} \frac{sinx.dx}{sin^2x + cosx.(1 +cosx)} \\ I=\int_{0}^{\frac{\pi}{2}}\frac{sinx.dx}{sin^2x + cos^2x +cosx} = \int_{0}^{\frac{\pi}{2}}\frac{sinx.dx}{sin^2x + cos^2x +cosx} = \int_{0}^{\frac{\pi}{2}}\frac{sinx.dx}{1+cosx}\\ -ln|1+cosx|[/TEX]

 

[nguyenbahiep1]

10)
[TEX]I= \int_{0}^{\frac{\pi}{2}} \frac{cos2x}{(sinx-cosx+3)^3}.dx \\ I= \int_{0}^{\frac{\pi}{2}} \frac{(cosx -sinx)(sinx+cosx)}{(sinx-cosx+3)^3}.dx [/TEX]

đặt
[TEX]sinx - cosx +3 = u \Rightarrow du = sinx +cosx[/TEX]

[TEX]I= \int_{2}^{4} \frac{(3-u)}{u^3}.du \\ I= \int_{2}^{4} (\frac{3}{u^3} - \frac{1}{u^2})du = \frac{1}{u} - \frac{3}{2.u^2} = \frac{1}{32}[/TEX]

 

[maths.dang]

cám ơn mấy bạn nhiều lắm !.......................................