∫sin^4(x)cos^2(x)dx…（x：0→π/2）の積分 のバックアップの現在との差分(No.1)

\[ \int^{\frac{\pi}{2}}_{0} \sin^4 x \cos^2 x dx \]
\[ = \int^{\frac{\pi}{2}}_{0} \sin^2 x ( \sin x \cos x )^2 dx \]
\[ = \int^{\frac{\pi}{2}}_{0} \frac{1-\cos 2x}{2} \frac{\sin^2 2x}{4} dx \]
\[ = \int^{\frac{\pi}{2}}_{0} \frac{1-\cos 2x}{2} \frac{\sin^2 2x}{4} dx \]
\[ = \frac{1}{8} \int^{\frac{\pi}{2}}_{0} (1-\cos 2x) \sin^2 2x dx \]
\[ = \frac{1}{8} \int^{\frac{\pi}{2}}_{0} \sin^2 2x dx - \frac{1}{8} \int^{\frac{\pi}{2}}_{0} \cos 2x \sin^2 2x dx \]
\[ = \frac{1}{8} \int^{\frac{\pi}{2}}_{0} \frac{1-\cos 4x}{2} dx - \frac{1}{8} \int^{\frac{\pi}{2}}_{0} \cos 2x \sin^2 2x dx \]
\[ = \frac{1}{16} \int^{\frac{\pi}{2}}_{0} (1-\cos 4x) dx - \frac{1}{8} \int^{\frac{\pi}{2}}_{0} \cos 2x \sin^2 2x dx \]
\[ = \frac{1}{16} \left[ x-\frac{\sin 4x}{4} \right]^{\frac{\pi}{2}}_{0} - \frac{1}{8} \left[ \frac{\sin^3 2x}{6} \right]^{\frac{\pi}{2}}_{0} \]
\[ = \frac{1}{16} \left( \frac{\pi}{2} - 0 \right) \]
\[ = \frac{\pi}{32} \]

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=(1/8) ∫ ( 1-cos(2x) ) sin^2(2x) dx
=(1/8) ∫ sin^2(2x) dx
　　　+ (1/8) ∫ sin^2(2x) cos(2x) dx
=(1/8) ∫ ( 1-cos(4x) )/2 dx
　　　+ (1/8) ∫ sin^2(2x) cos(2x) dx
=(1/16) ∫ ( 1-cos(4x) ) dx
　　　+ (1/8) ∫ sin^2(2x) cos(2x) dx
=(1/16) [ x-(1/4)sin(2x) ]
　　　+(1/8) [ (1/6) sin^3(2x) ]
=(1/16) × (π/2)
=π/32