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Question Number 117100 by bemath last updated on 09/Oct/20

$$\int \sin {}^6\left(2\mathrm{x}\right)\cos {}^6\left(2\mathrm{x}\right)\mathrm{dx}=?$$

Answered by Dwaipayan Shikari last updated on 09/Oct/20

$$\int \frac{1}{2^6}{\left(sin4x\right)}^{6}dx$$$$\frac{1}{64}.\int {\left({sin}^{2}4x\right)}^{3}dx$$$$\frac{1}{64}.\frac{1}{8}\int {(1-cos8x)}^{3}dx$$$$\frac{1}{512}\int 1-{cos}^{3}8x-3cos8x(1-cos8x)dx$$$$\frac{x}{512}-\frac{1}{512}\int {cos}^{3}8x-\frac{3}{512}\int cos8x+\frac{3}{1024}\int 1+cos16xdx$$$$\frac{x}{512}+\frac{3x}{1024}-\frac{1}{512}\int \frac{cos24x+3cos8x}{4}-\frac{3}{4096}sin8x+\frac{3}{1024.16}sin16x$$$$\frac{5x}{1024}-\frac{1}{2^{14}.3}sin24x-\frac{3}{2^{14}}sin8x-\frac{3}{2^{12}}sin8x+\frac{3}{2^{14}}sin16x+C$$$$$$

Answered by 1549442205PVT last updated on 09/Oct/20

$$\mathrm{F}=\int \sin {}^6\left(2\mathrm{x}\right)\cos {}^6\left(2\mathrm{x}\right)\mathrm{dx}=$$$$=\frac{1}{64}\int {\left[2\sin \left(2\mathrm{x}\right)\cos \left(2\mathrm{x}\right)\right]}^6\mathrm{dx}$$$$=\frac{1}{64}\int {\sin }^6\left(4\mathrm{x}\right)\mathrm{d}\underset{\mathrm{put}4\mathrm{x}=\mathrm{t}}{\mathrm{x}=}\frac{1}{256}\int {\sin }^6\mathrm{tdt}$$$$=\frac{1}{256}.\frac{1}{8}\int {(1-\cos 2\mathrm{t})}^3\mathrm{dt}$$$$=\frac{1}{2^{11}}\int (1-3\cos 2\mathrm{t}+{3\cos }^{2}2\mathrm{t}-{\cos }^{3}2\mathrm{t})\mathrm{dt}$$$$=\frac{1}{2^{11}}[\int \mathrm{dt}-\frac{3}{2}\int \cos \left(2\mathrm{t}\right)\mathrm{d}\left(2\mathrm{t}\right)+\frac{3}{2}\int (1+\cos \left(4\mathrm{t}\right))\mathrm{dt}]$$$$-\frac{1}{2^{11}}.\frac{1}{2}\int (1-{\sin }^{2}2\mathrm{t})\mathrm{d}\left(\sin 2\mathrm{t}\right))$$$$=\frac{1}{2^{11}}[\mathrm{t}-\frac{3}{2}\sin 2\mathrm{t}+\frac{3}{2}\mathrm{t}+\frac{3}{8}\sin 4\mathrm{t}]$$$$-\frac{1}{2^{12}}(\sin 2\mathrm{t}-\frac{{\sin }^{3}2\mathrm{t}}{3})$$$$=\frac{5}{2^{12}}\left(4\mathrm{x}\right)-\frac{3}{2^{12}}\sin 8\mathrm{x}+\frac{3}{2^{14}}\sin \left(16\mathrm{x}\right)$$$$-\frac{1}{2^{12}}\sin 8\mathrm{t}+\frac{1}{2^{12}}{\sin }^{3}8\mathrm{t}$$$$=\frac{5\mathrm{x}}{2^{10}}-\frac{1}{2^{10}}\sin 8\mathrm{x}+\frac{1}{2^{12}}{\sin }^{3}8\mathrm{x}+\frac{3}{2^{14}}\sin 16\mathrm{x}+\mathrm{C}$$

Answered by Bird last updated on 10/Oct/20

$$\int {sin}^6\left(2x\right){cos}^6\left(2x\right)dx$$$${=}_{2x=t}\frac{1}{2}\int {sin}^{6}t{cos}^{6}tdt$$$$\Rightarrow 2I=\int {\left(\frac{1}{2}sin\left(2t\right)\right)}^{6}dt$$$$=\frac{1}{2^6}\int {sin}^6\left(2t\right)dt$$$$=\frac{1}{2^6}\int {\left(\frac{e^{2it}-e^{-2it}}{2i}\right)}^{6}dt$$$$=\frac{1}{2^6{\left(2i\right)}^6}\int \sum _{k=0}^6{C}_6^k{\left(e^{2it}\right)}^k{(-e^{-2it})}^{6-k}$$$$=\frac{1}{{\left(4i\right)}^6}\sum _{k=0}^6{(-1)}^k{C}_6^k\int e^{2ikt}{e}^{-2(6-k)it}dt$$$$=\frac{1}{{\left(4i\right)}^6}\sum _{k=0}^6{(-1)}^k{C}_6^k\int e^{(2k+2k-12)it}dt$$$$=\frac{1}{{\left(4i\right)}^6}\sum _{k=0}^6{(-1)}^k{C}_6^k\frac{1}{(4k-12)i}{e}^{(4k-12)it}+c$$$$witht=2x$$

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