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Question Number 140447 by 676597498 last updated on 07/May/21

Answered by EDWIN88 last updated on 07/May/21

$$\mathrm{L}+\mathrm{M}=\underset{0}{\stackrel{\pi /2}{\int }}\cos {}^2\mathrm{x}\sin {}^2\mathrm{x}(\cos {}^2\mathrm{x}+\sin {}^2\mathrm{x})\mathrm{dx}$$$$\mathrm{L}+\mathrm{M}=\underset{0}{\stackrel{\pi /2}{\int }}\cos {}^2\mathrm{x}\sin {}^2\mathrm{x}\mathrm{dx}$$$$=\underset{0}{\stackrel{\pi /2}{\int }}\cos {}^2\mathrm{x}-\cos {}^4\mathrm{x}\mathrm{dx}$$$$=\underset{0}{\stackrel{\pi /2}{\int }}\left(\frac{1+\cos 2\mathrm{x}}{2}\right)-{\left(\frac{1+\cos 2\mathrm{x}}{2}\right)}^2\mathrm{dx}$$$$={\left[\frac{\mathrm{x}+\frac{1}{2}\sin 2\mathrm{x}}{2}\right]}_0^{\pi /2}-\underset{0}{\stackrel{\pi /2}{\int }}\left(\frac{1+2\cos 2\mathrm{x}+\frac{1+\cos 4\mathrm{x}}{2}}{4}\right)\mathrm{dx}$$$$=\left(\frac{\pi }{4}\right)-\underset{0}{\stackrel{\pi /2}{\int }}\frac{3+4\cos 2\mathrm{x}+\cos 4\mathrm{x}}{8}\mathrm{dx}$$$$=\frac{\pi }{4}-{\left[\frac{3\mathrm{x}+2\sin 2\mathrm{x}+\frac{1}{4}\sin 4\mathrm{x}}{8}\right]}_0^{\pi /2}$$$$=\frac{\pi }{4}-\frac{3\pi }{16}=\frac{\pi }{16}.$$

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