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# Trig Identity Proof

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Prove that  $$\int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx = 2^{-m} \int_0^{\frac{\pi}{2}} \cos^m x \, dx$$

Dec 21, 2021

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\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 \\
=2^{-m}\int_0^{\pi } (sing)^m \, \frac{dx}{dg}\;dg \\
=2^{-m}\int_0^{\pi } (sing)^m \, \frac{1}{2}\;d

Dec 23, 2021