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How to evaluate following integral

$$\int \frac{\mathrm dx}{\sin^4x+\cos^4x\:+\sin^2(x) \cos^2(x)}$$

Can you please also give me the steps of solving it?

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Divide the numerator & the denominator by $\cos^4x$ to find

$$I=\int\frac{\sec^4x}{\tan^4x+1+\tan^2x}dx$$

Putting $\tan x=y,$

$$I=\int\frac{1+y^2}{1+y^2+y^4}dy=\int\frac{1+\dfrac1{y^2}}{\dfrac1{y^2}+1+y^2}dy$$

$$=\int\frac{1+\dfrac1{y^2}}{\left(y-\dfrac1y\right)^2+3}dy $$

Finally set, $u=y-\dfrac1y=\dfrac{y^2-1}y=-2\cdot\dfrac1{\dfrac{2\tan x}{1-\tan^2x}}=-2\cot2x$

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  • $\begingroup$ @user282909, Yes, the rest should be easy, right? $\endgroup$ – lab bhattacharjee Nov 23 '14 at 14:02
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$$\int \frac{1}{\sin^4x+\cos^4x+\sin^2\left(x\right)\cos^2\left(x\right)}dx$$ $$=\int \frac{1}{\sin^4x+\cos^4x\:+2\sin^2\left(x\right)\cos^2\left(x\right)-\sin^2\left(x\right)\cos^2\left(x\right)}dx$$ $$=\int \frac{1}{(\sin^2x+\cos^2x)^2-\sin^2\left(x\right)\cos^2\left(x\right)}dx$$ $$=\int \frac{1}{1-\frac{1}{4}\sin^2\left(2x\right)}dx\quad\because \sin2x=2\sin x\cos x$$

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Apolozies for any typos:

My solution: $$ \int \frac{dx}{\sin^4x+\cos^4x\:+\:\sin^2\left(x\right)\cos^2\left(x\right)}\\ =\int\frac{dx}{(\sin^2x+\cos^2x)^2-\sin^2x\cos^2x}\\ =\int\frac{dx}{1-\sin^2x\cos^2x}\\ =\int\frac{(1+\tan^2x)\sec^2xdx}{1+\tan^4x+\tan^2x}\\ =\int\frac{(1+y^2)dx}{1+y^2+y^4}\quad y:=\tan x\\ =\frac12\int\frac{1}{1+y^2+y}+\frac1{1-y+y^2}\\ =\frac12\left(\frac2{\sqrt3}\left(\arctan\frac{\tan x+0.5}{\sqrt3/2}+\arctan\frac{\tan x-0.5}{\sqrt3/2}\right)\right)+c $$

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