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Old 05-08-2011, 02:38 AM   #9
man111
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$\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 $
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Bài 3: Tìm nguyên hàm: $\displaystyle \int{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx} $
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thay đổi nội dung bởi: man111, 05-08-2011 lúc 02:45 AM Lý do: Tự động gộp bài
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