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Question Number 118436 by bramlexs22 last updated on 17/Oct/20

$$\int \cos {}^4\left(x\right)\cos {}^4\left(2x\right)dx$$

Answered by benjo_mathlover last updated on 17/Oct/20

$$\left(1\right)\cos {}^4\left(x\right)={(\frac{1}{2}+\frac{1}{2}\cos \left(2x\right))}^2$$$$=\frac{1}{4}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{4}(\frac{1}{2}+\frac{1}{2}\cos \left(4x\right))$$$$=\frac{3}{8}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{8}\cos \left(4x\right)$$$$\left(2\right)\cos {}^4\left(2x\right)={(\frac{1}{2}+\frac{1}{2}\cos \left(4x\right))}^2$$$$=\frac{1}{4}+\frac{1}{2}\cos \left(4x\right)+\frac{1}{4}(\frac{1}{2}+\frac{1}{2}\cos \left(8x\right))$$$$=\frac{3}{8}+\frac{1}{2}\cos \left(4x\right)+\frac{1}{8}\cos \left(8x\right)$$$$Then\left(1\right)\times \left(2\right):$$$$=\frac{9}{64}+\frac{3}{16}\cos \left(4x\right)+\frac{3}{64}\cos \left(8x\right)+\frac{3}{16}\cos \left(2x\right)$$$$+\frac{1}{8}\cos \left(6x\right)+\frac{1}{8}\cos \left(2x\right)+\frac{1}{32}\cos \left(10x\right)$$$$+\frac{1}{32}\cos \left(6x\right)+\frac{3}{64}\cos \left(4x\right)+\frac{1}{32}+\frac{1}{32}\cos \left(8x\right)$$$$+\frac{1}{128}\cos \left(12x\right)+\frac{1}{128}\cos \left(4x\right)$$

Answered by mathmax by abdo last updated on 17/Oct/20

$$\mathrm{I}=\int {\cos }^4\mathrm{x}{\cos }^4\left(2\mathrm{x}\right)\mathrm{dx}\Rightarrow \mathrm{I}=\int {\left(\frac{1+\cos \left(2\mathrm{x}\right)}{2}\right)}^2{\left(\frac{1+\cos \left(4\mathrm{x}\right)}{2}\right)}^2\mathrm{dx}$$$$=\frac{1}{16}\int {(1+\cos \left(2\mathrm{x}\right))}^2{(1+\cos \left(4\mathrm{x}\right))}^2\mathrm{dx}$$$$=\frac{1}{16}\int (1+2\cos \left(2\mathrm{x}\right)+{\cos }^2\left(2\mathrm{x}\right))(1+2\cos \left(4\mathrm{x}\right)+{\cos }^2\left(4\mathrm{x}\right))\mathrm{dx}$$$$16\mathrm{I}=\int (1+2\cos \left(4\mathrm{x}\right)+{\cos }^2\left(4\mathrm{x}\right)+2\cos \left(2\mathrm{x}\right)+4\cos \left(2\mathrm{x}\right)\cos \left(4\mathrm{x}\right)$$$$+2\cos \left(2\mathrm{x}\right){\cos }^2\left(4\mathrm{x}\right)+{\cos }^2\left(2\mathrm{x}\right)+{2\cos }^2\left(2\mathrm{x}\right)\cos \left(4\mathrm{x}\right)+{\cos }^2\left(2\mathrm{x}\right){\cos }^2\left(4\mathrm{x}\right))\mathrm{dx}$$$$=\mathrm{x}+\frac{1}{2}\sin \left(4\mathrm{x}\right)+\int {\cos }^2\left(4\mathrm{x}\right)\mathrm{dx}+\sin \left(2\mathrm{x}\right)+4\int \cos \left(2\mathrm{x}\right)\cos \left(4\mathrm{x}\right)\mathrm{dx}$$$$+2\int \cos \left(2\mathrm{x}\right){\cos }^2\left(4\mathrm{x}\right)\mathrm{dx}+\int {\cos }^2\left(2\mathrm{x}\right)\mathrm{dx}+2\int {\cos }^2\left(2\mathrm{x}\right)\cos \left(4\mathrm{x}\right)\mathrm{dx}$$$$+\int {\cos }^2\left(2\mathrm{x}\right){\cos }^2\left(4\mathrm{x}\right)\mathrm{dx}\mathrm{we}\mathrm{have}$$$$\int {\cos }^2\left(4\mathrm{x}\right)\mathrm{dx}=\int \frac{1+\cos \left(8\mathrm{x}\right)}{2}\mathrm{dx}=\frac{\mathrm{x}}{2}+\frac{1}{16}\sin \left(8\mathrm{x}\right)+{\mathrm{c}}_0$$$$\int \cos \left(2\mathrm{x}\right)\cos \left(4\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}\int (\cos \left(6\mathrm{x}\right)+\cos \left(2\mathrm{x}\right))\mathrm{dx}$$$$=\frac{1}{12}\sin \left(6\mathrm{x}\right)+\frac{1}{4}\sin \left(2\mathrm{x}\right)+{\mathrm{c}}_1$$$$\int \cos \left(2\mathrm{x}\right){\cos }^2\left(4\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}\int \cos \left(2\mathrm{x}\right)(1+\cos \left(8\mathrm{x}\right))\mathrm{dx}$$$$=\frac{1}{4}\sin \left(2\mathrm{x}\right)+\frac{1}{2}\int \cos \left(2\mathrm{x}\right)\cos \left(8\mathrm{x}\right)\mathrm{dx}$$$$=\frac{1}{4}\sin \left(2\mathrm{x}\right)+\frac{1}{4}\int (\cos \left(10\mathrm{x}\right)+\cos \left(6\mathrm{x}\right))\mathrm{dx}$$$$=\frac{1}{4}\sin \left(2\mathrm{x}\right)+\frac{1}{40}\sin \left(10\mathrm{x}\right)+\frac{1}{24}\sin \left(6\mathrm{x}\right)+{\mathrm{c}}_2$$$$\int {\cos }^2\left(2\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}\int (1+\cos \left(4\mathrm{x}\right))\mathrm{dx}=\frac{\mathrm{x}}{2}+\frac{1}{8}\sin \left(4\mathrm{x}\right)+{\mathrm{c}}_3$$$$\int {\cos }^2\left(2\mathrm{x}\right)\cos \left(4\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}\int (1+\cos \left(4\mathrm{x}\right))\cos 4\mathrm{xdx}$$$$=\frac{1}{2}\int \cos \left(4\mathrm{x}\right)\mathrm{dx}+\frac{1}{4}\int (1+\cos \left(8\mathrm{x}\right))\mathrm{dx}$$$$=\frac{1}{8}\sin \left(4\mathrm{x}\right)+\frac{\mathrm{x}}{4}+\frac{1}{32}\sin \left(8\mathrm{x}\right)+{\mathrm{c}}_4\mathrm{also}$$$$\int {\cos }^2\left(2\mathrm{x}\right){\cos }^2\left(4\mathrm{x}\right)\mathrm{dx}=\frac{1}{4}\int (1+\cos \left(4\mathrm{x}\right))(1+\cos \left(8\mathrm{x}\right))\mathrm{dx}$$$$=\frac{1}{4}\int (1+\cos \left(8\mathrm{x}\right)+\cos \left(4\mathrm{x}\right)+\cos \left(4\mathrm{x}\right)\cos \left(8\mathrm{x}\right))\mathrm{dx}$$$$=\frac{\mathrm{x}}{4}+\frac{1}{32}\sin \left(8\mathrm{x}\right)+\frac{1}{16}\sin \left(4\mathrm{x}\right)+\frac{1}{8}\int \cos \left(12\mathrm{x}\right)+\cos \left(4\mathrm{x}\right))\mathrm{dx}$$$$=\frac{\mathrm{x}}{4}+\frac{1}{32}\sin \left(8\mathrm{x}\right)+\frac{1}{16}\sin \left(4\mathrm{x}\right)+\frac{1}{8.12}\sin \left(12\mathrm{x}\right)+\frac{1}{32}\sin \left(4\mathrm{x}\right)$$$$\mathrm{rest}\mathrm{to}\mathrm{collect}\mathrm{the}\mathrm{integrals}..$$$$$$

Answered by MJS_new last updated on 17/Oct/20

$$\int {\cos }^{4}x{\cos }^{4}2xdx=$$$$[t=\tan x\to dx={\cos }^{2}xdt]$$$$=\int \frac{{(t-1)}^4{(t+1)}^4}{{(t^2+1)}^7}dt=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{t(165t^{10}+935t^8+1986t^6+3006t^4+1305t^2+795)}{960{(t^2+1)}^6}+$$$$+\frac{11}{64}\int \frac{dt}{t^2+1}[\mathrm{which}=\frac{11}{64}\arctan t]$$$$\mathrm{now}\mathrm{put}t=\tan x$$

Answered by 1549442205PVT last updated on 17/Oct/20

$${\cos }^4\mathrm{x}={\left(\frac{1+\cos 2\mathrm{x}}{2}\right)}^2=\frac{1+2\cos 2\mathrm{x}+{\cos }^{2}2\mathrm{x}}{4}$$$$\mathrm{Put}\mathrm{F}=\int \cos {}^4\left(x\right)\cos {}^4\left(2x\right)\mathrm{dx}\Rightarrow 4\mathrm{F}=$$$$\int {\cos }^{4}2\mathrm{xdx}+2\int {\cos }^{4}2\mathrm{xcos}2\mathrm{xdx}+\int {\cos }^{6}2\mathrm{xdx}$$$${\cos }^{4}2\mathrm{x}={\left(\frac{1+\cos 4\mathrm{x}}{2}\right)}^2=\frac{1+2\cos 4\mathrm{x}+{\cos }^{2}4\mathrm{x}}{4}$$$$=\frac{1}{4}(1+2\cos 4\mathrm{x}+\frac{1+\cos 8\mathrm{x}}{2})\left(1\right)$$$$={(1-{\sin }^{2}2\mathrm{x})}^2=1-{2\sin }^{2}2\mathrm{x}+{\sin }^{4}2\mathrm{x}\left(2\right)$$$${{\cos }^{6}2\mathrm{x}=\frac{1+\cos 4\mathrm{x}}{2})}^3=\frac{1}{8}(1+3\cos 4\mathrm{x}$$$$+{3\cos }^{2}4\mathrm{x}+{\cos }^{3}4\mathrm{x})$$$$=\frac{1}{8}+\frac{3}{8}\cos 4\mathrm{x}+\frac{3}{16}(1+\cos 8\mathrm{x})+(1-{\sin }^{2}4\mathrm{x})\cos 4\mathrm{x}$$$$=\frac{5}{16}+\frac{3}{8}\cos 4\mathrm{x}+\frac{3}{16}\cos 8\mathrm{x}+(1-{2\sin }^{2}4\mathrm{x}+{\sin }^{4}4\mathrm{x})\cos 4\mathrm{x}\left(3\right)$$$$\mathrm{From}\left(1\right)\left(2\right)\left(3\right)\mathrm{we}\mathrm{get}$$$$4\mathrm{F}=\int \frac{11}{16}\mathrm{dx}+\frac{7}{8}\int \cos 4\mathrm{xdx}+\frac{5}{16}\int \cos 8\mathrm{xdx}$$$$+\frac{1}{4}\int (1-{\sin }^{2}4\mathrm{x})\mathrm{d}\left(\sin 4\mathrm{x}\right)$$$$+\int (1-{2\sin }^{2}2\mathrm{x}+{\sin }^{4}2\mathrm{x})\mathrm{d}\left(\sin 2\mathrm{x}\right)$$$$=\frac{11}{16}\mathrm{x}+\frac{7}{32}\sin 4\mathrm{x}+\frac{5}{128}\sin 8\mathrm{x}+\frac{1}{4}\sin 4\mathrm{x}$$$$-\frac{1}{12}{\sin }^{3}4\mathrm{x}+\sin 2\mathrm{x}-\frac{2}{3}{\sin }^{3}2\mathrm{x}+\frac{1}{5}{\sin }^{5}2\mathrm{x}$$$$\mathrm{F}=\frac{1}{4}(\frac{11}{16}\mathrm{x}+\frac{7}{32}\sin 4\mathrm{x}+\frac{5}{128}\sin 8\mathrm{x}$$$$-\frac{1}{12}{\sin }^{3}4\mathrm{x}+\sin 2\mathrm{x}-\frac{2}{3}{\sin }^{3}2\mathrm{x}$$$$+\frac{1}{5}{\sin }^{5}2\mathrm{x})+\mathrm{C}$$

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