+876 Prove that \( \int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx = 2^{-m} \int_0^{\frac{\pi}{2}} \cos^m x \, dx \)
+118667 \int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx = 2^{-m} \int_0^{\frac{\pi}{2}} \cos^m x \, dx
\(\int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx = 2^{-m} \int_0^{\frac{\pi}{2}} \cos^m x \, dx \)
\(LHS\\=\int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx\\ =\int_0^{\frac{\pi}{2}} (\cos x \sin x)^m \, dx \\ =2^{-m}\int_0^{\frac{\pi}{2}} (2\cos x \sin x)^m \, dx \\ =2^{-m}\int_0^{\frac{\pi}{2}} (sin(2x))^m \, dx \\ \qquad let \;\;g=2x\\ \qquad when\;\;x=0,\;g=0\qquad when\;\;x=\frac{\pi}{2}\;\;g=\pi\\ =2^{-m}\int_0^{\pi } (sing)^m \, \frac{dx}{dg}\;dg \\ =2^{-m}\int_0^{\pi } (sing)^m \, \frac{1}{2}\;dg \\ \)
Below is perhaps less formal.
It can be deduced by knowing what the cos and sin graphs look like
and knowing they are the same except there is a phase shift of pi/2 radians
\(=2^{-m}\int_{\frac{\pi}{2}} ^{\pi } (sing)^m \;dg\\ =2^{-m}\int_{0} ^{\frac{\pi}{2}} (cosg)^m \;dg\\ =2^{-m}\int_{0} ^{\frac{\pi}{2}} (cosx)^m \;dx\\ =RHS\)
QED
LaTex:
LHS\\=\int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx\\
=\int_0^{\frac{\pi}{2}} (\cos x \sin x)^m \, dx \\
=2^{-m}\int_0^{\frac{\pi}{2}} (2\cos x \sin x)^m \, dx \\
=2^{-m}\int_0^{\frac{\pi}{2}} (sin(2x))^m \, dx \\
\qquad let \;\;g=2x\\
\qquad when\;\;x=0,\;g=0\qquad when\;\;x=\frac{\pi}{2}\;\;g=\pi\\
=2^{-m}\int_0^{\pi } (sing)^m \, \frac{dx}{dg}\;dg \\
=2^{-m}\int_0^{\pi } (sing)^m \, \frac{1}{2}\;d
