tích phân đặc biệt 5

Question

$ \int\limits_{0}^{\pi/2}(\sqrt{sinx} - \sqrt{cosx})dx $

$ \int\limits_{0}^{\pi/2}\frac{sinx}{cosx + sinx}dx $

Bạn Duy chú ý phần b) có tận 3 bài đăng tương tự nhau nhé!

score 6 · Answer 1

Đặt $I=\int\limits_0^{\pi/2}(\sqrt{\sin x}-\sqrt{\cos x})dx$
Đặt: $x=\frac{\pi}{2}-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{\pi/2}^0\left(\sqrt{\sin(\displaystyle\frac{\pi}{2}-t)}-\sqrt{\cos(\displaystyle\frac{\pi}{2}-t)}\right)dt$
$=\int\limits_0^{\pi/2}(\sqrt{\cos t}-\sqrt{\sin t})dt=-I$
Suy ra: $2I=0\Leftrightarrow I=0$

score 6 · Answer 2

Bình chọn tăng 6 Bình chọn giảm

Đặt $I=\int\limits_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}dx,J=\int\limits_0^{\pi/2}\frac{\cos x}{\sin x+\cos x}dx$
Đặt: $x=\frac{\pi}{2}-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{\pi/2}^0\frac{\sin(\displaystyle\frac{\pi}{2}-t)}{\sin(\displaystyle\frac{\pi}{2}-t)+\cos(\displaystyle\frac{\pi}{2}-t)}dt$
$=\int\limits_0^{\pi/2}\frac{\cos t}{\sin t+\cos t}dt=J$
Mà ta có:
$I+J=\int\limits_0^{\pi/2}dx=\frac{\pi}{2}$
Suy ra: $I=J=\frac{\pi}{4}$