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