$\int\frac{x^2}{\left(x\sin x+\cos x\right)^2}dx $ Now $\left(x\sin x+\cos x\right)=\sqrt{x^2+1}\left(\frac{1}{\sqrt{x^2+1}}. \cos x+\frac{x}{\sqrt{x^2+1}}.\sin x\right) $ Now Let $\sin \theta = \frac{x}{\sqrt{x^2+1}} $ and $\cos \theta =\frac{x}{\sqrt{x^2+1}} $ so $\tan \theta =x\Leftrightarrow \theta =\tan^{-1} x $ $\left(x\sin x+\cos x\right)=\sqrt{x^2+1}\left(\cos x\cos \theta+\sin x\sin \theta\right)=\sqrt{x^2+1}.\cos(x-\theta)=\sqrt{x^2+1}.\cos\left(x-\tan^{-1} x\right) $ $\int\frac{x^2}{\left(x^2+1\right).\cos^2\left(x-\tan^{-1} x\right)}dx $ $\int\sec^2 \left(x-\tan^{-1} x\right).\frac{x^2}{1+x^2}dx $ Now $x-\tan^{-1}x=t\Leftrightarrow \left(1-\frac{1}{1+x^2}\right)dx=dt\Leftrightarrow \frac{x^2}{1+x^2}dx=dt $ $\int\sec^2 tdt =\tan \left(t\right)+C $ $=\tan\left(x-\tan^{-1} x\right)+C $ $=\frac{\tan x-\tan(\tan^{-1} x)}{1+\tan x.\tan(\tan^{-1} x)}+C $ $=\frac{\tan x- x}{1+x.\tan x}+C $ $=\frac{\sin x-x.\cos x}{\cos x+x.\sin x}+C $ ------------------------------ Bài 3: Tìm nguyên hàm: $\displaystyle \int{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx} $ [RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT] thay đổi nội dung bởi: man111, 05-08-2011 lúc 02:45 AM Lý do: Tự động gộp bài |