Tính: 1, $\int\limits_{0}^{1}\frac{4x+11}{x^2+5x+6}dx$

2, $\int\limits_{0}^{\frac{\pi}{6}}(\sin^6x+\cos^6x)dx$

3, $\int\limits_{0}^{\frac{\pi}{2}}\frac{4\sin^3x}{1+ \cos x}dx$

4, $\int\limits_{0}^{\frac{\pi}{4}}\frac{1+\sin 2x}{\cos^2x}dx$
$I=\int \dfrac{1}{\cos^2 x}dx +\int \dfrac{2\sin x \cos x}{\cos^2 x}dx=\tan x+2\int \dfrac{\sin x}{\cos x}dx$

$=\tan x \bigg |_0^{\frac{\pi}{4}} -2\int \dfrac{d(\cos x)}{\cos x} = 1 -2\ln |\cos x| \bigg |_0^{\frac{\pi}{4}} = ...$ làm nốt đi
$I=-4\int \dfrac{\sin^2 x d(\cos x)}{1+\cos x}=-4\int \dfrac{(1-\cos^2 x)d(\cos x)}{1+\cos x}=-4\int \dfrac{1-t^2}{1+t}dt$

$=-4\int(1-t)dt =-4t+2t^2 +C=( -4\cos x +2\cos^2 x)\bigg |_0^{\frac{\pi}{2}}=4-2=2$ 
$\sin^6 x +\cos^6 x= (\sin^2 x +\cos^2 x)(\cos^4 x -\sin^2 x \cos^2 x+\cos^4 x) = (\cos^2 x +\sin^2 x)^2 -3\sin^2 x \cos^2 x$

$=1-\dfrac{3}{4}\sin^2 2x =1-\dfrac{3}{8}(1-\cos 4x) = \dfrac{5}{8}+\dfrac{3}{8}\cos 4x$

Vậy $I = \int (\dfrac{5}{8}+\dfrac{3}{8}\cos 4x) dx = \dfrac{5}{8}\int dx + \dfrac{3}{8} \int \cos 4x dx = \dfrac{5}{8}x +\dfrac{3}{32} \int \cos 4x d(4x)$

$= \dfrac{5}{8}x +\dfrac{3}{32} \sin 4x + C$ tự thế cận
uh nhầm đó e sửa đi –  Dép Lê Con Nhà Quê 09-11-13 08:35 PM
$(\cos^2 x \sin^2 x)^2 -3\sin^2 x \cos^2 x$$=1-\dfrac{3}{2}\sin^2 2x$ chỗ này nhầm thì phải anh ạ, phải là $=1-\dfrac{3}{4}\sin^2 2x$ chứ –  hanhphucnhe989 09-11-13 08:31 PM
Dùng hệ số bất định ta tính được

$\int_0^1 \dfrac{4x+1}{x^2 +5x +6} dx = \int_0^1 \bigg (\dfrac{11}{x+3} -\dfrac{7}{x+2}\bigg )dx$

$=\bigg ( 11\ln |x+3| -7\ln|x+2| \bigg ) \bigg |_0^1$ tự thế cận nha
hướng dẫn em phương pháp dùng hệ số bất định đc k –  hanhphucnhe989 09-11-13 12:37 PM

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