tinh tich phan

N

newstarinsky

ta có
$sin^2x-cos^2x+2sinx.cosx=sin2x-cos2x=\sqrt{2}sin(2x-\dfrac{\pi}{4})=2\sqrt{2}sin(x-\dfrac{\pi}{8}).cos(x-\dfrac{\pi}{8})$
Nên $I=\dfrac{1}{2\sqrt{2}}\int \dfrac{dx}{sin(x-\dfrac{\pi}{8}).cos(x-\dfrac{\pi}{8})}$
$=\dfrac{1}{2\sqrt{2}}\int \dfrac{d(sin(x-\dfrac{\pi}{8}))}{1-sin^2(x-\dfrac{\pi}{8})}$
$=\dfrac{1}{4\sqrt{2}}(\int\dfrac{d(sin(x-\dfrac{\pi}{8})}{1-sin(x-\dfrac{\pi}{8})}+\int\dfrac{d(sin(x-\dfrac{\pi}{8}))}{1+sin(x-\dfrac{\pi}{8})})$
$=\dfrac{1}{4\sqrt{2}}(ln|1+sin(x-\dfrac{\pi}{8})|-ln|1-sin(x-\dfrac{\pi}{8})|)$

Bạn thay cận nhé
 
Last edited by a moderator:
L

luffy_95

tớ nguyên hàm thui nhé!

$$ \int_{0}^{\frac{\pi}}{4}\frac{dx}{sin2x-cos2x} $$

\Leftrightarrow $$ -\int_{0}^{\frac{\pi}}{4}\frac{dx}{\sqrt{2}sin(frac{\pi}{4})-2x) $$

ok!
 
N

nguyenbahiep1

[TEX]\int\limits_{0}^{pi/4}\frac{dx}{sin^2x+2sinxcosx-cosx^2}[/TEX]
giai toi dua ket qua cuoi cung bai nay lunk nhe''thanks

cách mình thấy là hay hơn cả

[TEX]\int_{0}^{\frac{\pi}{4}} \frac{dx}{cos^2x.(tan^2x + 2tanx-1)} \\ u = tanx \\ \int_{0}^{1}\frac{dx}{u^2+2u-1} \\ \int_{0}^{1}\frac{dx}{(u+1)^2-2} = \int_{0}^{1}\frac{1}{2.\sqrt{2}}.(\frac{1}{u+1-\sqrt{2}} - \frac{1}{u+1+\sqrt{2}}) \\ \frac{1}{2.\sqrt{2}}.ln (| \frac{u+1-\sqrt{2}}{u+1+\sqrt{2}}|) = \frac{1}{2.\sqrt{2}}.ln \sqrt{2}[/TEX]
 
H

hoan1793



cách mình thấy là hay hơn cả

[TEX]\int_{0}^{\frac{\pi}{4}} \frac{dx}{cos^2x.(tan^2x + 2tanx-1)} \\ u = tanx \\ \int_{0}^{1}\frac{dx}{u^2+2u-1} \\ \int_{0}^{1}\frac{dx}{(u+1)^2-2} = \int_{0}^{1}\frac{1}{2.\sqrt{2}}.(\frac{1}{u+1-\sqrt{2}} - \frac{1}{u+1+\sqrt{2}}) \\ \frac{1}{2.\sqrt{2}}.ln (| \frac{u+1-\sqrt{2}}{u+1+\sqrt{2}}|) = \frac{1}{2.\sqrt{2}}.ln \sqrt{2}[/TEX]

hehe cảm ơn anh hiệp nhá


..........................................................................................
 
Top Bottom