Question #189880

Use an appropriate trigonometry substitution to evaluate ∫▒√(x^2+16)/x^4  dx


1
Expert's answer
2021-05-07T14:14:27-0400

We will calculate ((x2+16))dx/(x4)\int(\sqrt(x^2+16))dx/(x^4)

put x=4tan(t)

So dx=4sec2tdt4sec^2tdt

Now integration becomes ((4(1+tan2t))4sec2t)dt/(44tan4(t))\int((4\sqrt(1+tan^2t))*4sec^2t)dt/(4^4tan^4(t))

=(sec3(t)dt)/(16tan4(t))\int(sec^3(t) dt)/(16tan^4(t))

=(1/cos3(t))(16(cos4(t)/sin4(t)))\int(1/cos^3(t))*(16(cos^4(t)/sin^4(t)))

=(cos(t))/(16sin4(t))\int(cos(t))/(16sin^4(t))

Now let sin(t)=z So cos(t)dt=dz

So integration becomes (dz/16z4)=((z(4)dz))/16\int(dz/16z^4) =\int((z^(-4)dz))/16

= -1/(48z^3)

=-1/(48sin^3(t)) As (xndx)=(x(n+1))/(n+1)\int(x^ndx)=(x^(n+1))/(n+1) We had taken x=4tan(t) ,so tan(t)=x/4 ,in a right angle triangle tan(t)=p/b , so p=x & b=4

So h=(x2+16)\sqrt(x^2+16) ,So sin(t)=p/h=x/((x2+16))sin(t)=p/h=x/(\sqrt (x^2+16)) =(((x2+16))3)/(48x3)+C=-((\sqrt(x^2+16))^3)/(48x^3) + C (ans)

Here C is constant of integration.


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