Integral of sinx cosx/sin^4x cos^4x - Weebly

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Integral of sinx cosx/sin^4x cos^4x

Last updated at Dec. 23, 2019 by Teachoo Transcript Misc 26 Evaluate the definite integral _0^( /4)(sin cos )/(cos^4 + sin^4 ) _0^( /4)(sin cos )/(cos^4 + sin^4 ) Adding and subtracting 2 ^2 ^2 from denominator = _0^( /4)(sin cos )/(cos^4 + sin^4 + 2 ^2 ^2 - 2 ^2 ^2 ) = _0^( /4)(sin cos )/((cos^2 + sin^2 )^2 - 2 ^2 ^2 ) = _0^( /4)(sin cos )/(1 - (4 ^2 ^2 )/2) = _0^( /4)(sin cos )/(1 - 1/2 (2 sin cos )^2 )

= _0^( /4)(sin cos )/(1 - ( ^2 2 )/2) = _0^( /4)(2 sin cos )/(2 - ^(2 ) 2 ) = _0^( /4)sin2 /(1 + (1 - ^(2 ) 2 )) = _0^( /4)sin2 /(1 + ^2 2 ) ( ^2 + ^2 =1" " ) ("2 sin " cos =sin2 " " ) ("2 sin " cos =sin2 " " ) Let t = cos 2 / =-2 sin2 (- )/2=sin2 So, our equation becomes _0^( /4)sin2 /(1 + ^2 2 ) = (-1)/2 1_1^0 /(1 + ^2 ) = (-1)/2 [ ^(-1) ( )]_1^0 = (-1)/2 [ ^(-1) (0)- ^(-1) (1)] = (-1)/2 (0- /4) = / Hence put the value of t in the above expression to get the final answer. Hope you have got it. Please log in or register to add a comment. Please log in or register to answer this question. Please log in or register to add a comment. First, some preliminary manipulation. $$\frac{\sin x \cos x}{\sin^4x + \cos^4x} = \frac{\sin x \cos x}{(1\cos^2x)^2 + \cos^4x}\\=\frac{\sin x \cos x}{2\cos^4x -2\cos^2x+1}=\frac{4\sin x \cos x}{8\cos^4x -8\cos^2x+1 + 3}\\=\frac{4\sin x \cos x}{\cos4x+ 3}$$ The last step uses the quadruple angle formula for cosine. Now $\sin 2x = 2\sin x \cos x$, using this twice yields:$$\frac{4\sin x \cos x}{\cos4x+ 3}=\frac{2\sin 2x }{\cos4x+ 3} =\frac{2\sin 2x \cos 2x}{(\cos4x+ 3)\cos 2x} = \frac{\sin 4x }{(\cos4x+ 3)\cos 2x} $$ We can now use the half-angle formula for cosine, which is $\cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}$. $$\frac{\sin 4x }{(\cos4x+ 3)\cos 2x}=\frac{\sin 4x }{(\cos4x+ 3)\sqrt{\frac{1+\cos 4x}{2}}} \\=\frac{\sqrt{2} \times \sin 4x }{(\cos4x+ 3)\sqrt{\cos 4x + 3 - 2}}$$ The time is ripe to substitute with $u=\cos 4x +3$. Then $du = -4\sin 4x \ dx$ and $$\int \frac{\sqrt{2} \times \sin 4x }{(\cos4x+ 3)\sqrt{\cos 4x + 3 - 2}} \ dx\\ = \frac{\sqrt{2}}{-4}\int \frac{1}{u\sqrt{u - 2}} \ du$$ To finish this relatively simple integral, I did another substitution, this time with $t=\sqrt{u-2}$ and $du =2t \ dt$. $$\frac{\sqrt{2}}{-4}\int \frac{1}{u\sqrt{u - 2}} \ du = \frac{\sqrt{2}}{-4}\int \frac{2t}{ut} \ dt\\=\frac{\sqrt{2}}{-2}\int \frac{1}{t^2 +2} \ dt$$ This is an integral that involves $\arctan$:$$\frac{\sqrt{2}}{-2}\int \frac{1}{t^2 +2} \ dt=-\frac{1} {2}\arctan(\frac{t}{\sqrt{2}})=-\frac{1}{2}\arctan(\frac{\sqrt{u-2}}{\sqrt{2}})\\=-\frac{1}{2}\arctan(\sqrt{\frac{\cos 4x +1}{2}})=-\frac{1}{2} \arctan(\cos 2x)$$ Checking with wolframalpha, this differentiates to the correct result.

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