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

Bình chọn tăng 3 Bình chọn giảm

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:
$I_n=\left\{ \begin{array}{l}\frac{(n-1)!!}{n!!}.\frac{\pi}{2}&\textrm{nếu n chẵn} \\\frac{(n-1)!!}{n!!}&\textrm{nếu n lẻ} \end{array} \right.$