Question

using the method of trigonometric substitute, find the integral of 1/(x^2+4)^2

Answer 1

i am just confused about the denominator being squared and not the square root of. is it just like any other trig substitution?

Answer 2

We can use these properties because of the simple and most important identities. Now which identity can I relate to when I have \[x^{2}+4=2\]? Well using a little of creativity you may remember that \[\tan ^{2}(x)+1=\sec ^{2}(x)\]. Therefore we are going to use \[x=a \tan(u)\] which in this case \[a =2\] and you will see why: Letting \[x = 2 \tan(u)\] we know that \[dx = 2 \sec ^{2}(u) du\] Now substitute everything in the integral I you we have:\[\int\limits 1/(x ^{2}+4)^{2}dx\] transforms into\[\int\limits 2\sec ^{2}(u)du/(4\tan ^{2}(u)+4)^{2}\]. If you factor the 4 then we have\[\int\limits 2\sec ^{2}(u)du/16(\tan ^{2}(u)+1)^{2}\] which is \[\int\limits \sec ^{2}(u)du/8\sec ^{4}(u)\] by the identity. Which in turns is:\[\int\limits du/8\sec ^{2}(u)=\int\limits \cos ^{2}(u)du\]. And using the half angle formulas we have that:\[\int\limits \cos ^{2}(u)du = \int\limits 1/2(1+\cos(2u))du\] which simplifies into\[\int\limits 1/2 du + \int\limits 1/2\cos(2u)du\] which clearly is:\[(1/2)u+(1/4)\sin(2u) +C\]. But what is \[u?\] If we go back we know remember that \[x=2\tan(u)\rightarrow x/2 = \tan(u)\rightarrow u =\tan^{-1} (x/2)\]. Therefore by substituting we have \[(1/2)\tan^{-1} (x/2)+(1/4)\sin(2\tan^{-1} (x/2))\]. A little of precalculus and trigonometry would remind you that\[\sin(2u)=2\sin(u)\cos(u)\]. Therefore\[\sin(2\tan^{-1}( x/2))=2\sin(\tan^{-1} (x/2))\cos(\tan^{-1} (x/2))\]. Using the right triangles we have by the amazing SOHCAHTOA that if\[\tan(\theta) = x/2\] then \[O = x\] and \[A =2\] Therefore to find\[\sin(\theta)=O/H\] we must find H by the marvelous pithagorean theorem. \[H ^{2}=x ^{2}+4\rightarrow H=\sqrt{x ^{2}+4}\] which then we have that \[\sin(\theta)=x/\sqrt{x ^{2}+4}\] and\[\cos(\theta) = 2/\sqrt{x ^{2}+4}\]. So finally the integral will look like:\[(1/2)\int\limits dx/(x ^{2}+4)=\tan^{-1} (x/2)+x/(x^{2}+4)\]

Answer 3

Peace, Love and Happiness from Puerto Rico