Đáp án: $I=0$
Giải thích các bước giải:
Đặt
$f(x)=\dfrac{\sin^{2020}x}{\sin^{2020}x+\cos^{2020}x}$
$\to f(-x)=\dfrac{\sin^{2020}(-x)}{\sin^{2020}(-x)+\cos^{2020}(-x)}$
$\to f(-x)=\dfrac{\sin^{2020}x}{\sin^{2020}x+\cos^{2020}x}$
$\to f(-x)=f(x)$
Lại có:
$ I=\displaystyle\int^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}\dfrac{\sin^{2020}x}{\sin^{2020}x+\cos^{2020}x}dx$
$\to I=\displaystyle\int^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}f(x)dx$
$\to I=\displaystyle\int^{\dfrac{\pi}{2}}_0f(x)dx+\displaystyle\int^0_{-\dfrac{\pi}{2}}f(x)dx$
$\to I=\displaystyle\int^{\dfrac{\pi}{2}}_0f(x)dx+\displaystyle\int^0_{\dfrac{\pi}{2}}f(-x)d(-x)$
$\to I=\displaystyle\int^{\dfrac{\pi}{2}}_0f(x)dx-\displaystyle\int^0_{\dfrac{\pi}{2}}f(x)dx$
$\to I=\displaystyle\int^{\dfrac{\pi}{2}}_0f(x)dx-\displaystyle\int^{\dfrac{\pi}{2}}_0f(x)dx$
$\to I=0$
