Bài 111796

Question

Tính các tích phân sau :
a) $I_n = \int\limits_{0}^{1} x^n \sqrt{1-x}dx (0 \leq n \in Z)$
b) $J_n = \int\limits_{0}^{\pi} \sin ^ {n-1} x \cos (n+1)xdx (1 \leq n \in Z).$

score 0 · Answer 1

a) Đặt $\begin{cases}u = x^n \\ dv = \sqrt{1-x}dx \end{cases} \Rightarrow \begin{cases}du= nx^{n-1}dx \\ v = \frac{2}{3}(1-x)\sqrt{1-x} \end{cases} $
Khi đó : $I_n = \frac{2}{3}x^n (1-x)\sqrt{1-x}\left| \begin{array}{l}
1\\
0
\end{array} \right.+\frac{2}{3}n \int\limits_{0}^{1} x^{n-1} (1-x) \sqrt{1-x}dx$
$= \frac{2}{3}n [ \int\limits_{0}^{1}x^{n-1} \sqrt{1-x}dx - \int\limits_{0}^{1}x^n \sqrt{1-x}dx ] =\frac{2}{3}n(I_{n-1} - I_n)$
$\Rightarrow I_n =\frac{2n}{2n+3}.I_{n-1}$
$\Rightarrow \frac{I_n}{I_{n-1}}.\frac{I_{n-1}}{I_{n-2}}...\frac{I_2}{I_1}.\frac{I_1}{I_0} =\frac{2n}{2n+3}.\frac{2(n-1)}{2n+1}...\frac{2.2}{7}.\frac{2}{5}$
$\Rightarrow \frac{I_n}{I_0} = \frac{2^n.n!}{1.3.5...(2n+3)}.3$
Mặt khác: $I_0 = \int\limits_{0}^{1}\sqrt{1-x}dx = -\frac{2}{3} (1-x)^{\frac{3}{2}}\left| \begin{array}{l}
1\\
0
\end{array} \right. = \frac{2}{3}$
$\Rightarrow I_n = \frac{2^{n+1}.n!}{1.3.5...(2n+8)}.$

b) Đặt $ \begin{cases}u = \cos nx \\ dv = \sin ^{n-1}x \cos xdx \end{cases} \Rightarrow \begin{cases}du = - n \sin xdx \\ v = \frac{\sin ^n x}{n} \end{cases}$
$\Rightarrow \int\limits_{0}^{\pi} \sin ^{n-1} x . \cos nx.\cos dx$
$= \left ( \cos nx . \frac{\sin ^nx}{n} \right )\left| \begin{array}{l}
\pi \\
0
\end{array} \right.+ \int\limits_{0}^{\pi} \sin ^nx.\sin xdx = \int\limits_{0}^{\pi} \sin ^n x.\sin xdx$
$\Rightarrow \int\limits_{0}^{\pi} \sin ^{n-1} x.\cos nx.\cos xdx = \int\limits_{0}^{\pi}\sin ^nx.\sin nxdx$
$\Leftrightarrow \int\limits_{0}^{\pi} (\sin ^{n-1} x.\cos nx.có x- \sin ^nx.\sin nx)dx =0$
$\Leftrightarrow \int\limits_{0}^{\pi} \sin ^{n-1}x ( \cos nx.\cos x - \sin nx .\sin x)dx=0$
$\Leftrightarrow I_n = \int\limits_{0}^{\pi} \sin ^{n-1} x.\cos (n+1)xdx=0.$

Bài 111796

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Bài 111796

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