Bài 107095

Question

Tìm nguyên hàm của hàm số:

$f(x)=\frac{x^3}{\sqrt{1-x^2} }$

score 0 · Answer 1

Đặt

$x=\sin t$ ,

$\frac{\pi}{2}<t<\frac{\pi}{2}$ , suy ra:

$dx=\cos t.dt ; \frac{x^3dx}{\sqrt{1-x^2} }=\frac{\sin^3t.\cos tdt}{\cos t}=\sin^3t.dt=\frac{1}{4}(3\sin t-\sin3t)dt$

Khi đó:

$\int\limits f(x)dx=\frac{1}{4}\int\limits (3\sin t-\sin 3t)dt=-\frac{3}{4}\cos t+\frac{1}{12}\cos 3t+C$

$=\int\limits -\frac{3}{4}\cos t+\frac{1}{12}(4\cos^3t-3\cos t)+C=\frac{1}{3}\cos ^3t-\cos t+C$

$=(\frac{1}{3}\cos^2t-1)\cos t+C=[\frac{1}{3}(1-\sin^2t)-1]\cos t+C$

$=[\frac{1}{3}(1-x^2)-1]\sqrt{1-x^2}+C=-\frac{1}{3}(x^2+2)\sqrt{1-x^2}+C$

1 Đáp án