giải tích phân hàm lượng giác

Question

Tính tích phân
a . $I=\int\limits_{0}^{1}\frac{dx}{2x^{2}+5x+2}$
b. $I=\int\limits_{0}^{\pi} \sin x.\sin 3x.\sin 5xdx$

score 2 · Answer 1

b) $\int\limits_{0}^{\pi}sinxsin3xsin5xdx=\frac{1}{2}(\int\limits_{0}^{\pi}sin5xcos2xdx-\int\limits_{0}^{\pi}sin5xcos4xdx)$
=$\frac{1}{4}(\int\limits_{0}^{\pi}sin7xdx+\int\limits_{0}^{\pi}sin3xdx-\int\limits_{0}^{\pi}sin9xdx-\int\limits_{0}^{\pi}sinxdx)$
Den day cung la dang tich phan co ban.

em cảm ơn nhiều nha! hj – linhdajka95 14-01-13 08:17 PM

score 2 · Answer 2

Bình chọn tăng 2 Bình chọn giảm

a) $I=\int\limits_{0}^{1}\frac{1}{2x^{2}+5x+2}dx=\frac{1}{2}\int\limits_{0}^{1}\frac{1}{(x+\frac{5}{4})^{2}-\frac{9}{16}}dx$
Dat: $x+\frac{5}{4}=t\Rightarrow dx=dt$. Doi can: $x=0\Rightarrow t=\frac{5}{4}$ ; $x=1\Rightarrow t=\frac{9}{4}$.
Ta thu duoc tich phan moi:
$\frac{1}{2}\int\limits_{\frac{5}{4}}^{\frac{9}{4}}\frac{1}{t^{2}-\frac{9}{16}}dt=\frac{1}{3}\int\limits_{\frac{5}{4}}^{\frac{9}{4}}\frac{1}{t-\frac{3}{4}}dt-\int\limits_{\frac{5}{4}}^{\frac{9}{4}}\frac{1}{t+\frac{3}{4}}dt$. Den day la tich phan co ban.