biet thi giup minh di

M

mastercity

minh thử giúp bạn nha

đề bài ;I1=0π4cosxsinx2+sin2xdx=0π4cosxsinx1+2cos2(xπ4)dxI_1= \int_{0}^{\frac{\pi}{4}}\frac{cosx-sinx}{\sqrt{2+sin2x}}dx=\int_{0}^{\frac{\pi}{4}} \frac{cosx-sinx}{\sqrt{1+2cos^2(x-\frac{\pi}{4})}}dx

đặt t=1+2cos2(xπ4)2cosx(xπ4)=t21 t=\sqrt{1+2cos^2(x-\frac{\pi}{4}})\Rightarrow \sqrt {2}cosx(x-\frac{\pi}{4})=\sqrt{t^2-1}dt=2sin(xπ4).cos(xπ4)tdxdt=\frac{-2sin(x-\frac{\pi}{4}).cos(x-\frac{\pi}{4})}{t}dx
với x=π4t=3;x=0t=2 x=\frac{\pi}{4}\Leftrightarrow t=\sqrt{3};x=0\Leftrightarrow t=\sqrt{2}
lại có cosxsinx=2.sin(xπ4)(sinx+cosx)dx=tdt2cos(xπ4)=tdtt21 cosx-sinx=-\sqrt{2}.sin(x-\frac{\pi}{4})\Rightarrow (sinx+cosx)dx=\frac{tdt}{\sqrt{2}cos(x-\frac{\pi}{4})}=\frac{tdt}{\sqrt{t^2-1}}
I1=231t21dt\Rightarrow I_1= \int_{\sqrt{2}}^{\sqrt{3}} \frac{1}{\sqrt{t^2-1}} dt


I1=ln/t+t21/3y2v=ln(3+22+1)I_1=ln/ t+\sqrt{t^2-1}/ \begin{vmatrix} \sqrt{3}&y \\ \sqrt{2} & v \end{vmatrix}= ln(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}+1})
 
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