Question 1121154
Find integral (dx/sqr(x^2+16))
<pre>
{{{int(dx/sqrt(x^2+16))}}}

We use trig substitution by drawing this right triangle:

{{{drawing(200,100,-.5,2,-.5,1.5,
line(0,0,1.5,0), line(0,0,1.5,1),line(1.5,0,1.5,1),
locate(.4,.3,theta), locate(1.53,.62,x),locate(.77,0.05,4),
locate(.4,1.2,sqrt(x^2+16)) )}}} 

Then {{{x/4=tan(theta)}}} and {{{sqrt(x^2+16)/4=cos(theta)}}}

So

{{{x=4tan(theta)}}} and {{{sqrt(x^2+16)=4sec(theta)}}}
                        
{{{dx=4sec^2(theta)d(theta)}}}

Substituting, {{{int(dx/sqrt(x^2+16))}}} becomes

{{{int((4sec^2(theta)d(theta))/4sec(theta))}}}

Cancel and get

{{{int(sec(theta)d(theta))}}}

We use 

{{{int(sec(u)du)}}}{{{""=""}}}{{{ln(sec(u)^""+tan(u))}}}{{{""+""}}}{{{C}}}

{{{int(sec(theta)d(theta))}}}{{{""=""}}}{{{ln(sec(theta)^""+tan(theta))}}}{{{""+""}}}{{{C[1]}}}{{{""=""}}}

Use trig ratios from above right triangle:

{{{ln(sqrt(x^2+16)/4+x/4)}}}{{{""+""}}}{{{C[1]}}}{{{""=""}}}{{{ln(expr(1/4)(sqrt(x^2+16)+x)^"")}}}{{{""+""}}}{{{C[1]}}}{{{""=""}}}{{{ln(1/4)+ln(sqrt(x^2+16)+x)}}}{{{""+""}}}{{{C[1]}}}{{{""=""}}}

We let {{{C}}}{{{""=""}}}{{{ln(1/4)+C[1]}}} to simplify and end up with:

{{{ln(sqrt(x^2+16)+x)}}}{{{""+""}}}{{{C}}}

Edwin</pre>