Đề luyện thi

C

conga2222

tịch phan tu 0 den pi/4 cua can bac 2 tan x


:Mfoyourinfo::Mfoyourinfo::Mfoyourinfo::Mfoyourinfo::Mfoyourinfo::Mfoyourinfo:

\[\begin{array}{l}
t = \sqrt {\tan x} \\
dt = \frac{1}{{2{{\cos }^2}x\sqrt {\tan x} }}dx = \frac{{1 + {{\tan }^2}x}}{{2\sqrt {\tan x} }}dx\\
dx = \frac{{2tdt}}{{1 + {t^4}}}\\
I = \int\limits_0^1 {\frac{{2{t^2}}}{{1 + {t^4}}}dt} \\
t = \tan u\\
dt = \frac{1}{{{{\cos }^2}u}}du\\
I = \int\limits_0^{\frac{\pi }{4}} {\frac{{2{{\tan }^2}u}}{{(1 + {{\tan }^4}u){{\cos }^2}u}}du} \\
= \int\limits_0^{\frac{\pi }{4}} {\frac{{2{{\sin }^2}u}}{{{{\cos }^4}u + {{\sin }^4}u}}du} \\
= \int\limits_0^{\frac{\pi }{4}} {\frac{{1 - \cos 2u}}{{1 - {{\sin }^2}u{{\cos }^2}u}}du} \\
= 4\int\limits_0^{\frac{\pi }{4}} {\frac{{1 - \cos 2u}}{{4 - {{\sin }^2}2u}}du} \\
= 4(\int\limits_0^{\frac{\pi }{4}} {\frac{1}{{4 - {{\sin }^2}2u}}du} - \int\limits_0^{\frac{\pi }{4}} {\frac{{\cos 2u}}{{4 - {{\sin }^2}2u}}du} )\\

\end{array}\]
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