$I=\int_{0}^{\frac{\pi }{3}}cos^{4}xsin^{2}xdx$
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Đã gửi 13-04-2012 - 18:40
duongtoi
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Thiếu úy
$I=\int_0^{\frac{\pi}{3}}\cos^4x\sin^2 x{\rm d}x=\int_0^{\frac{\pi}{3}}\cos^2x\left ( \sin^2x\cos^2x \right ){\rm d}x$
$=\int_0^{\frac{\pi}{3}}\frac{1}{2}(1+\cos2x)\frac{1}{4}\sin^22x{\rm d}x=\frac{1}{8}\int_0^{\frac{\pi}{3}}\sin^22x{\rm d}x+\frac{1}{16}\int_0^{\frac{\pi}{3}}\sin^22x{\rm d}\sin2x$
$=\frac{1}{16}\int_0^{\frac{\pi}{3}}\left (1-\cos4x \right ){\rm d}x+\frac{1}{16}.\frac{1}{3}\sin^32x\Bigg|_0^{\frac{\pi}{3}}$
$=\frac{1}{16}\int_0^{\frac{\pi}{3}}\left (1-\cos4x \right ){\rm d}x+\frac{1}{16}.\frac{1}{3}\sin^32x\Bigg|_0^{\frac{\pi}{3}} =\frac{\pi}{48}+\frac{1}{64}.\frac{\sqrt3}{2}+\frac{1}{48}.\frac{3\sqrt3}{8}$
$=\frac{\pi}{48}+\frac{\sqrt3}{64}$
$=\int_0^{\frac{\pi}{3}}\frac{1}{2}(1+\cos2x)\frac{1}{4}\sin^22x{\rm d}x=\frac{1}{8}\int_0^{\frac{\pi}{3}}\sin^22x{\rm d}x+\frac{1}{16}\int_0^{\frac{\pi}{3}}\sin^22x{\rm d}\sin2x$
$=\frac{1}{16}\int_0^{\frac{\pi}{3}}\left (1-\cos4x \right ){\rm d}x+\frac{1}{16}.\frac{1}{3}\sin^32x\Bigg|_0^{\frac{\pi}{3}}$
$=\frac{1}{16}\int_0^{\frac{\pi}{3}}\left (1-\cos4x \right ){\rm d}x+\frac{1}{16}.\frac{1}{3}\sin^32x\Bigg|_0^{\frac{\pi}{3}} =\frac{\pi}{48}+\frac{1}{64}.\frac{\sqrt3}{2}+\frac{1}{48}.\frac{3\sqrt3}{8}$
$=\frac{\pi}{48}+\frac{\sqrt3}{64}$
Bài viết đã được chỉnh sửa nội dung bởi duongtoi: 13-04-2012 - 18:42
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