[toán 12] tích phân

T

trantien.hocmai

$$\int_0^{\dfrac{\pi}{2}} \dfrac{3\cos x-2\sin x}{(\sin x+\cos x)^3}dx=\int_0^{\dfrac{\pi}{2}} \dfrac{2(\cos x-\sin x)}{(\sin x+\cos x)^3}dx+\int_0^{\dfrac{\pi}{2}} \dfrac{\cos x}{(\sin x+\cos x)^3}dx\\
=I_1+I_2 \\$$
$$I_1=\int_0^{\dfrac{\pi}{2}} \dfrac{2(\cos x-\sin x)}{(\sin x+\cos x)^3}dx$$
$\text{đặt t=}\sin x+\cos x \rightarrow dt=(\cos x-\sin x)dx \\$
$$I_2=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos x}{(\sin x+\cos x)^3}dx \\$$
$\text{đặt }x=\dfrac{\pi}{2}-t \rightarrow dx=-dt \\
\text{đổi cận} x=0 \rightarrow t=\dfrac{\pi}{2} \\
x=\dfrac{\pi}{2} \rightarrow t=0 \\
\text{ta có} \\$
$$I_2=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos (\dfrac{\pi}{2}-t)}{(\sin (\dfrac{\pi}{2}-t)+\cos (\dfrac{\pi}{2}-t))^3}dt \\
=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin t}{(\sin t+\cos t)^3}dt=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin x}{(\sin x+\cos x)^3}dx \\$$
$\text{tự làm tiếp nhá}$
 
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