Answer to Question #189880 in Calculus for Moel Tariburu

Question

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

Answer 1

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

put x=4tan(t)

So dx=$4sec^2tdt$

Now integration becomes $\int((4\sqrt(1+tan^2t))*4sec^2t)dt/(4^4tan^4(t))$

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

=$\int(1/cos^3(t))*(16(cos^4(t)/sin^4(t)))$

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

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

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

= -1/(48z^3)

=-1/(48sin^3(t)) As $\int(x^ndx)=(x^(n+1))/(n+1)$ 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=$\sqrt(x^2+16)$ ,So $sin(t)=p/h=x/(\sqrt (x^2+16))$ $=-((\sqrt(x^2+16))^3)/(48x^3) + C$ (ans)

Here C is constant of integration.