$\int_{0}^{\frac{\pi }{2}}\sqrt{1-sin^{4}x}.dx$
$\int_{0}^{\frac{\pi }{2}}\sqrt{1-sin^{4}x}.dx$
Bắt đầu bởi thuanhoang1712, 27-03-2012 - 14:45
#1
Đã gửi 27-03-2012 - 14:45
#2
Đã gửi 27-03-2012 - 18:25
Biến đổi, $1-\sin^4 x=(1-\sin^2x)(1+\sin^2x)=\cos^2x(1+\sin^2x)$
$\Rightarrow \sqrt{1-\sin^4 x}=\cos x\sqrt{(1+\sin^2x)}$ (vì $0\le x\le \frac{\pi}{2}$)
Do vậy,
$\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin^4x}{\rm d}x=\int_{0}^{\frac{\pi}{2}}\sqrt{1+\sin^2x}\cos x{\rm d}x.$
Đặt $t=\sin t$, ta được
$\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin^4x}{\rm d}x=\int_{0}^{1}\sqrt{1+t^2}{\rm d}t$$\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin^4x}{\rm d}x=\int_{0}^{1}\sqrt{1+t^2}{\rm d}t=\arctan t \Big|_0^1=\frac{\pi}{4}.$
$\Rightarrow \sqrt{1-\sin^4 x}=\cos x\sqrt{(1+\sin^2x)}$ (vì $0\le x\le \frac{\pi}{2}$)
Do vậy,
$\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin^4x}{\rm d}x=\int_{0}^{\frac{\pi}{2}}\sqrt{1+\sin^2x}\cos x{\rm d}x.$
Đặt $t=\sin t$, ta được
$\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin^4x}{\rm d}x=\int_{0}^{1}\sqrt{1+t^2}{\rm d}t$$\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin^4x}{\rm d}x=\int_{0}^{1}\sqrt{1+t^2}{\rm d}t=\arctan t \Big|_0^1=\frac{\pi}{4}.$
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