tích phân

Question

1, $\int\limits_{\frac{1}{2}}^{2}\sqrt{x^2+\frac{1}{x^2}-2} dx$

2, $\int\limits_{0}^{\pi}\sqrt{1+\cos2x} dx$

3, $\int\limits_{0}^{2\pi}\sqrt{1+\sin x} dx$

score 3 · Answer 1

Bình chọn tăng 3 Bình chọn giảm

3. Ta có:

$\int\limits_0^{2\pi}\sqrt{1+\sin x}dx$

$=\int\limits_0^{2\pi}\sqrt{(\sin\dfrac{x}{2}+\cos\dfrac{x}{2})^2}dx$

$=\int\limits_0^{2\pi}\left|\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right|dx$

$=\int\limits_0^{\frac{3\pi}{2}}\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)dx-\int\limits_{\frac{3\pi}{2}}^{2\pi}\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)dx$

$=2\left(\sin\dfrac{x}{2}-\cos\dfrac{x}{2}\right)\left|\begin{array}{l}\dfrac{3\pi}{2}\\0\end{array}\right.-2\left(\sin\dfrac{x}{2}-\cos\dfrac{x}{2}\right)\left|\begin{array}{l}2\pi\\\dfrac{3\pi}{2}\end{array}\right.$

$=4\sqrt2$

ừ nhỉ.... – hanhphucnhe989 16-11-13 09:54 PM

score 3 · Answer 2

Bình chọn tăng 3 Bình chọn giảm

2. Ta có:

$\int\limits_0^{\pi}\sqrt{1+\cos2x}dx$

$=\int\limits_0^{\pi}\sqrt{2\cos^2x}dx$

$=\int\limits_0^{\pi}\sqrt2|\cos x|dx$

$=\int\limits_0^{\frac{\pi}{2}}\sqrt2\cos xdx-\int\limits_{\frac{\pi}{2}}^{\pi}\sqrt2\cos xdx$

$=\sqrt2\sin x\left|\begin{array}{l}\dfrac{\pi}{2}\\0\end{array}\right.-\sqrt2\sin x\left|\begin{array}{l}\pi\\\dfrac{\pi}{2}\end{array}\right.=2\sqrt2$

score 3 · Answer 3

Bình chọn tăng 3 Bình chọn giảm

1. Ta có:

$\int\limits_{\frac{1}{2}}^2\sqrt{x^2+\dfrac{1}{x^2}-2}dx$

$=\int\limits_{\frac{1}{2}}^2\left|x-\dfrac{1}{x}\right|dx$

$=\int\limits_{\frac{1}{2}}^1\left(\dfrac{1}{x}-x\right)dx+\int\limits_1^2\left(x-\dfrac{1}{x}\right)dx$

$=\left(\ln x-\dfrac{x^2}{2}\right)\left|\begin{array}{l}1\\\dfrac{1}{2}\end{array}\right.+\left(\dfrac{x^2}{2}-ln x\right)\left|\begin{array}{l}2\\1\end{array}\right.=\dfrac{9}{8}$