Giải thích các bước giải:
Ta có :
$I=\int^2_{-2}(x^3\cos\dfrac x2+\dfrac 12)\sqrt{4-x^2}dx$
$=\int^2_{-2}x^3\cos\dfrac x2.\sqrt{4-x^2}+\dfrac 12\sqrt{4-x^2}dx$
$=\int^2_{-2}x^3\cos\dfrac x2.\sqrt{4-x^2}dx+\int^2_{-2}\dfrac 12\sqrt{4-x^2}dx$
Lại có :
$f(x)=x^3\cos\dfrac x2.\sqrt{4-x^2}$
$\to f(-x)=(-x)^3\cos\dfrac{-x}2\sqrt{4-(-x)^2}=-x^3\cos\dfrac x2\sqrt{4-x^2}=-f(x)$
$\to f(x)$ là hàm lẻ
$\to \int^2_{-2}x^3\cos\dfrac x2.\sqrt{4-x^2}dx=0$
$\to I=\int^2_{-2}\dfrac 12\sqrt{4-x^2}dx$
$\to I=\int^2_{-2}\sqrt{1-(\dfrac{x}{2})^2}dx$
$\to I=\int^2_{-2}2\sqrt{1-(\dfrac{x}{2})^2}d(\dfrac{x}{2})$
$\to I=\arcsin \left(\dfrac{1}{2}x\right)+\dfrac{1}{2}\sin \left(2\arcsin \left(\dfrac{1}{2}x\right)\right)|^2_{-2}$
$\to I=\pi$
