câu 1: tích phân từ 0 đến pi/3 của ( tan X)^5
câu 2 : tích phân từ 0 đến 1 của 1/( x ^2 + 3x + 2 ) ^2 

ai rảnh sửa lại hộ t cho dễ nhìn nha ; cảm ơn –  hakunzee5897 29-12-14 08:04 PM
Câu a cách kia dài quá, ta làm cho cách này mà coi

$I=\int \bigg [ \tan^3 x (1+\tan^2 x) -\tan x (1+\tan^2 x) +\tan x \bigg ]dx= \int \tan^3 xd(\tan x) -\int \tan x d(\tan x) +\int \tan x dx$

$=\dfrac{1}{4}\tan^4 x -\dfrac{1}{2}\tan^2 x +\ln |\cos x| + C$ tự thay cận

Câu b làm như trên
cảm ơn bạn –  hakunzee5897 29-12-14 11:44 PM
+) $\int\limits_{0}^{\frac{\Pi }{3}}tan^{5}xd(x)=\int\limits_{0}^{\frac{\Pi}{3}}\frac{sin^{5}x}{cos^{5}x}d(x)=\int\limits_{0}^{\frac{\Pi }{3}}\frac{(1-cos^{2}x)^{2}sinx}{cos^{5}x}d(x)$
$=\int\limits_{0}^{\frac{\Pi }{3}}-\frac{1-2cos^2x+cos^4x}{cos^5x}d(cosx)=\int\limits_{0}^{\frac{\Pi }{3}}\left ( \frac{-1}{cos^5x}+\frac{2}{cos^3x}-\frac{1}{cosx} \right )d(cosx)$
$=\left ( \frac{1}{4cos^4x}-\frac{1}{cos^2x}-\ln \left| {cosx} \right| \right )=\frac{3}{4}-\ln \frac{1}{2}$
+) $\int\limits_{0}^{1}\frac{1}{\left ( x^2+3x+2 \right )^{2}}d(x)=\int\limits_{0}^{1}\left ( \frac{1}{(x+1)^2}+\frac{1}{(x+2)^2}+\frac{2}{x+2}-\frac{2}{x+1} \right )d(x)=\left ( \frac{-1}{x+1}-\frac{1}{x+2}+2\ln |x+2|-2\ln |x+1| \right )=\frac{2}{3}+2\ln \frac{3}{4}$
cảm ơn bạn –  hakunzee5897 29-12-14 11:44 PM

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