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Question Number 12535 by tawa last updated on 24/Apr/17

$$\mathrm{Use}\mathrm{the}\mathrm{reduction}\mathrm{formular}.$$$${\mathrm{I}}_{\mathrm{n}}=\int {\sin }^{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}=-\frac{1}{\mathrm{n}}{\sin }^{\mathrm{n}-1}\left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)+\frac{\mathrm{n}-1}{\mathrm{n}}{\mathrm{I}}_{\mathrm{n}}-2,\mathrm{to}\mathrm{evaluate}$$$${\mathrm{I}}_{\mathrm{n}}=\int {\sin }^6\left(\mathrm{x}\right)\mathrm{dx}$$

Answered by mrW1 last updated on 25/Apr/17

$${\mathrm{I}}_{\mathrm{n}}=\int {\sin }^{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}=-\frac{1}{\mathrm{n}}{\sin }^{\mathrm{n}-1}\left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)+\frac{\mathrm{n}-1}{\mathrm{n}}{\mathrm{I}}_{\mathrm{n}-2}$$$$I_6=\int {\sin }^{6}xdx=-\frac{1}{6}{\sin }^{5}x\cos x+\frac{5}{6}\int {\sin }^{4}xdx$$$$I_4=\int {\sin }^{4}xdx=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{4}\int {\sin }^{2}xdx$$$$I_2=\int {\sin }^{2}xdx=-\frac{1}{2}\sin x\cos x+\frac{1}{2}\int dx=\frac{1}{2}x-\frac{\sin 2x}{4}$$$$$$$$I_4=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{4}(\frac{1}{2}x-\frac{\sin 2x}{4})=\frac{3}{8}x-\frac{3}{16}\sin 2x-\frac{1}{4}\sin {}^3\cos x$$$$$$$$I_6=-\frac{1}{6}{\sin }^{5}x\cos x+\frac{5}{6}[\frac{3}{8}x-\frac{3}{16}\sin 2x-\frac{1}{4}\sin {}^3\cos x]$$$$=\frac{5}{16}x-\frac{5}{32}\sin 2x-\frac{1}{24}\sin {}^3\cos x-\frac{1}{6}{\sin }^{5}x\cos x+C$$

Commented by tawa last updated on 25/Apr/17

$$\mathrm{wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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