Question

How do you integrate `int sqrt(13+25x^2)` using trig substitutions?

Answer 1

To integrate an irrational function with a sum of squares under the root, you use the substitution with the tangent.

Explanation:

First reduce the radical to a sum of squares:

`int(sqrt(13+25x^2))dx = 1/5int(sqrt((sqrt(13)/5)^2+x^2))dx`

Pose `x=sqrt(13)/5tan(t)` and as `(d tan(t))/dt = sec^2(t)`

`dx=sqrt(13)/5sec^2(t)dt`

`int(sqrt(13+25x^2))dx = 1/5int(sqrt(13/25+13/25tan^2(t))sqrt(13)/5sec^2(t)dt = 13/125int(sqrt(1+tan^2(t))sec^2(t)dt`

Use the trigonometric identity:

`1+tan^2t=sec^2t`

`int(sqrt(13+25x^2))dx = 13/125int(sqrt(sec^2(t))sec^2(t)dt =`
`= 13/125int|sec(t)|^3dt`.

Now, to integrate this last integral let's limit ourselves in the interval `t in [-pi/2,pi/2]` where `sec(t)` is positive and use integration by parts:

`intsec^3(t)dt = int (sect*sec^2t)dt = intsect * d(tant) = sect tant -int tan t* d(sec t) = sect tant -int tan ^2t*sec t dt =`

but `tan^2t = sec^2t-1`, so

`intsec^3(t)dt = sect tant + int (sect -sec^3t)dt= sect tant +int sect dt -int sec^3(t)tdt`

or

`2intsec^3(t)dt = sect tant +int sect dt`

Finally:

`intsec^3(t)dt = 1/2sect tant +1/2 ln(|sec t + tan t|)`

Now reverse the substitution:

`tan t = 5/sqrt(13)x`

and `sect = sqrt(1+tan^2(t)) = sqrt(1+25/13x^2)`

So:

`int(sqrt(13+25x^2))dx = 5/(2sqrt(13))xsqrt(1+25/13x^2)+1/2ln(|5/sqrt(13)x+sqrt(1+25/13x^2)|)`