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Question

$\int\limits_{0}^{\Pi }cos^2x.e^xdx$

Các bạn giải chi tiết ra kết quả cho mình nhé, mình tính mãi nhg k ra. Từng phần 2 lần lận đó.

score 2 · Answer 1

$I=\dfrac{1}{2} \int e^x (1+\cos 2x)dx=\dfrac{1}{2}e^x \bigg |_0^{\pi} +\dfrac{1}{2}\int e^x \cos 2x dx=\dfrac{e^{\pi}-1}{2} +\dfrac{1}{2} I_1$

Tính

$I_1$ Đặt

$e^x = u \Rightarrow e^x dx =du$ và

$\cos 2x dx = dv \Rightarrow \dfrac{1}{2}\sin 2x =v$

$I_1= \dfrac{1}{2}e^x. \sin 2x \bigg |_0^{\pi}- \dfrac{1}{2}\int \sin 2x e^x dx=- \dfrac{1}{2}\int \sin 2x . e^x dx$

Đặt

$e^x = u;\ \sin 2x dx = dv \Rightarrow -\dfrac{1}{2}\cos 2x =v$

$I_1 = -\dfrac{1}{2} \bigg (-\dfrac{1}{2}e^x .\cos 2x \bigg |_0^{\pi} + \dfrac{1}{2}\int \cos 2x . e^x dx \bigg )$

$= -\dfrac{1}{2} \bigg (-\dfrac{1}{2}e^x .\cos 2x \bigg |_0^{\pi} + \dfrac{1}{2} I_1 \bigg )$ tự lắp lại nhé

m_internet001 · Answer 2 · 1/21/2014 6:10:09 PM

$\Leftrightarrow \frac{1}{2}\int\limits_{0}^{\pi }e^xdx+\frac{1}{2}\int\limits_{0}^{\pi }cos2x.e^xdx$
xét

$J=\int\limits_{0}^{\pi }cos2x.e^xdx=\int\limits_{0}^{\pi }cos2x.d(e^x)$

$\Leftrightarrow e^xcos2x-\int\limits_{0}^{\pi }e^xd(cos2x)$

$\Leftrightarrow e^xcos2x+2\int\limits_{0}^{\pi }e^xsin2x.dx$

$\Leftrightarrow e^xcos2x+2e^xsin2x-2\int\limits_{0}^{\pi }e^xd(sin2x)$

$\Leftrightarrow e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$
Ta có:

$J= e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$

$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$

$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=-\frac{1}{3}(e^xcos2x+2e^xsin2x)$

$\Rightarrow \int\limits_{0}^{\pi }cos^2x.e^xdx=\frac{1}{2}e^x|^{\pi }_0-\frac{1}{6}(e^xcos2x+2e^xsin2x)|^{\pi }_0$

2 Đáp án