Này thì tích phân,đố các ads

Question

Tính các tích phân sau theo

$n \in N$ :
a)

$I_n = \int\limits_{0}^{\frac{\pi}{2} } \cos ^n xdx$ b)

$J_n = \int\limits_{0}^{1} (1-x^2)^n dx.$

score 3 · Answer 1

Bình chọn tăng 3 Bình chọn giảm

b)
Đặt

$x=\sin t\Rightarrow dx=\cos tdt$
Đổi cận:

$x=0\Rightarrow t=0$

$x=1\Rightarrow t=\frac{\pi}{2}$
Suy ra:

$J_n=\int\limits_0^{\frac{\pi}{2}}\cos^{2n+1}tdt$

$=I_{2n+1}=\frac{(2n)!!}{(2n+1)!!}$

lời giải hay – congiola297 31-10-12 10:30 PM

score 3 · Answer 2

a.
Đặt

$u=\cos^{n-1}x,dv=\cos xdx$

$\Rightarrow du=-(n-1)\cos^{n-2}x\sin xdx, v=\sin x$

$\Rightarrow I_n=\cos^{n-1}x\sin x\left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.+(n-1)\int\limits_0^{\frac{\pi}{2}}\cos^{n-2}x(1-\cos^2x)dx$

$=(n-1)I_{n-1}-(n-1)I_n\Rightarrow I_n=\frac{n-1}{n}I_{n-2}$
Ta có:

$I_0=\int\limits_0^{\frac{\pi}{2}}dx=\frac{\pi}{2}\Rightarrow I_{2m}=\frac{2m-1}{2m}.\frac{2m-3}{2m-2}\ldots\frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}=\frac{(2m-1)!!}{(2m)!!}.\frac{\pi}{2}$

$I_1=\int\limits_0^{\frac{\pi}{2}}\cos xdx=1\Rightarrow I_{2m+1}=\frac{2m}{2m+1}.\frac{2m-2}{2m-1}\ldots\frac{2}{3}.1=\frac{(2m)!!}{(2m+1)!!}$
Từ đó: