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Question Number 89318 by M±th+et£s last updated on 16/Apr/20

$$\int \frac{dx}{{sin}^3\left(2x\right)+{cos}^3\left(2x\right)}$$

Answered by MJS last updated on 16/Apr/20

$$\int \frac{dx}{{\sin }^{3}2x+{\cos }^{3}2x}=$$$$[t=\tan x\to dx={\cos }^{2}xdt]$$$$=-\int \frac{{(t^2+1)}^2}{(t^2-2t-1)(t^2+(1-\sqrt{3})t+2-\sqrt{3})(t^2+(1+\sqrt{3})t+2+\sqrt{3})}dt=$$$$=-\frac{2}{3}\int \frac{dt}{t^2-2t-1}+$$$$-\frac{1-\sqrt{3}}{6}\int \frac{dt}{t^2+(1-\sqrt{3})t+2-\sqrt{3}}-$$$$-\frac{1+\sqrt{3}}{6}\int \frac{dt}{t^2+(1+\sqrt{3})t+2+\sqrt{3}}=$$$$\left[\mathrm{now}\mathrm{use}\mathrm{formulas}\right]$$$$=\frac{\sqrt{2}}{6}\ln \frac{t-1+\sqrt{2}}{t-1-\sqrt{2}}+\frac{1}{3}(\arctan ((1+\sqrt{3})t-1)+\arctan ((1-\sqrt{3})t-1))$$$$\mathrm{with}t=\tan x$$

Commented by M±th+et£s last updated on 16/Apr/20

$$thankyousir$$

Answered by M±th+et£s last updated on 17/Apr/20

$$I=\int \frac{dx}{(sin\left(2x\right)+cos\left(2x\right))(1-cos2xsin2x}$$$$=2\int \frac{dx}{(sin2x+cos2x)(2-sin4x)}$$$$I=2\int \frac{cos\left(2x\right)+sin\left(2x\right)}{(1+sin\left(4x\right))(2-sin4x)}dx$$$$y=cos\left(2x\right)-sin\left(2x\right)\Rightarrow \Rightarrow dy=(-2sin\left(2x\right)-2cos\left(2x\right))\Rightarrow -2(sin\left(2x\right)+cos\left(2x\right))$$$$y^2=1-sin\left(4x\right)$$$$sin\left(4x\right)=1-y^2$$$$=2\int \frac{\frac{-1}{2}dy}{(1+1-y^2)(2-1+y^2)}$$$$=-\int \frac{dy}{(2-y^2)(1+y^2)}$$$$=-\int \frac{dy}{(\sqrt{2}-y)(\sqrt{2}+y)(1+y^2)}$$$$=-\frac{2}{3}\int (\frac{1}{({\left(\sqrt{2}\right)}^2-y^2)}+\frac{1}{1+y^2})dy$$$$=\frac{-2}{3}\frac{1}{2\sqrt{2}}ln\mid \frac{\sqrt{2}+y}{\sqrt{2}-y}\mid -\frac{2}{3}{tan}^{-1}\left(y\right)+c$$$$=\frac{-2}{3}\frac{1}{2\sqrt{2}}ln\mid \frac{\sqrt{2}+cos2x-sin\left(2x\right)}{\sqrt{2}-cos\left(2x\right)+sin\left(2x\right)}\mid -\frac{2}{3}{tan}^{-1}(cos\left(2x\right)-sin\left(2x\right))+c$$$$$$

Commented by MJS last updated on 17/Apr/20

$$\mathrm{the}\mathrm{path}\mathrm{is}\mathrm{ok}\mathrm{but}\mathrm{the}\mathrm{constants}\mathrm{are}\mathrm{wrong}$$$$\mathrm{I}\mathrm{get}\mathrm{in}\mathrm{your}\mathrm{line}10$$$$\frac{1}{3}\int \frac{dy}{y^2-2}-\frac{1}{3}\int \frac{dy}{y^2+1}=$$$$=-\frac{\sqrt{2}}{12}\ln \frac{y+\sqrt{2}}{y-\sqrt{2}}-\frac{1}{3}\arctan y$$

Commented by M±th+et£s last updated on 17/Apr/20

$$thankyousirbutcanyoushowyour$$$$workhowdidyouget\frac{1}{3}\int \frac{dy}{y^2-2}-\frac{1}{3}\int \frac{dy}{y^2+1}$$$$andiamgreatfultoyouforyournotice$$

Commented by M±th+et£s last updated on 17/Apr/20

$$andithinkits-\frac{1}{3}\int \frac{1}{x^2-2}dx+\frac{1}{3}\int \frac{1}{x^2+1}dx$$

Commented by M±th+et£s last updated on 17/Apr/20

$$oksirthankyouigetit$$

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