minh thử giúp bạn nha
đề bài ;
I 1 = ∫ 0 π 4 c o s x − s i n x 2 + s i n 2 x d x = ∫ 0 π 4 c o s x − s i n x 1 + 2 c o s 2 ( x − π 4 ) d x I_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 I 1 = ∫ 0 4 π 2 + s i n 2 x c o s x − s i n x d x = ∫ 0 4 π 1 + 2 c o s 2 ( x − 4 π ) c o s x − s i n x d x
đặt
t = 1 + 2 c o s 2 ( x − π 4 ) ⇒ 2 c o s x ( x − π 4 ) = t 2 − 1 t=\sqrt{1+2cos^2(x-\frac{\pi}{4}})\Rightarrow \sqrt {2}cosx(x-\frac{\pi}{4})=\sqrt{t^2-1} t = 1 + 2 c o s 2 ( x − 4 π ) ⇒ 2 c o s x ( x − 4 π ) = t 2 − 1 có
d t = − 2 s i n ( x − π 4 ) . c o s ( x − π 4 ) t d x dt=\frac{-2sin(x-\frac{\pi}{4}).cos(x-\frac{\pi}{4})}{t}dx d t = t − 2 s i n ( x − 4 π ) . c o s ( x − 4 π ) d x
với
x = π 4 ⇔ t = 3 ; x = 0 ⇔ t = 2 x=\frac{\pi}{4}\Leftrightarrow t=\sqrt{3};x=0\Leftrightarrow t=\sqrt{2} x = 4 π ⇔ t = 3 ; x = 0 ⇔ t = 2
lại có
c o s x − s i n x = − 2 . s i n ( x − π 4 ) ⇒ ( s i n x + c o s x ) d x = t d t 2 c o s ( x − π 4 ) = t d t t 2 − 1 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}} c o s x − s i n x = − 2 . s i n ( x − 4 π ) ⇒ ( s i n x + c o s x ) d x = 2 c o s ( x − 4 π ) t d t = t 2 − 1 t d t ⇒ I 1 = ∫ 2 3 1 t 2 − 1 d t \Rightarrow I_1= \int_{\sqrt{2}}^{\sqrt{3}} \frac{1}{\sqrt{t^2-1}} dt ⇒ I 1 = ∫ 2 3 t 2 − 1 1 d t I 1 = l n / t + t 2 − 1 / ∣ 3 y 2 v ∣ = l n ( 3 + 2 2 + 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}) I 1 = l n / t + t 2 − 1 / ∣ ∣ ∣ ∣ ∣ 3 2 y v ∣ ∣ ∣ ∣ ∣ = l n ( 2 + 1 3 + 2 )
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