tích phân đặc biệt 4

Question

$ \int\limits_{0}^{\pi/2}\frac{sin^{3}x}{sin^{3}x+cos^{3}x}dx $

$ \int\limits_{0}^{\pi/2}\frac{cos ^{3}x}{sin^{3}x+cos^{3}x}dx $

score 4 · Answer 1

Bình chọn tăng 4 Bình chọn giảm

Đặt $I=\int\limits_0^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx,J=\int\limits_0^{\pi/2}\frac{\cos^3x}{\sin^3x+\cos^3x}dx$
Đặt: $x=\frac{\pi}{2}-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{\pi/2}^0\frac{\sin^3(\displaystyle\frac{\pi}{2}-t)}{\sin^3(\displaystyle\frac{\pi}{2}-t)+\cos^3(\displaystyle\frac{\pi}{2}-t)}dt$
$=\int\limits_0^{\pi/2}\frac{\cos^3t}{\sin^3t+\cos^3t}dt=J$
Mà ta có:
$I+J=\int\limits_0^{\pi/2}dx=\frac{\pi}{2}$
Suy ra: $I=J=\frac{\pi}{4}$