Question

How do you integrate `int sqrt(x^2-25)` by trigonometric substitution?

Answer 1

`(xsqrt(x^2-25)-25ln(abs(sqrt(x^2-25)+x)))/2+C`

Explanation:

We have:

`I=intsqrt(x^2-25)dx`

Let `x=5sec(u)`. Note that this implies `dx=5sec(u)tan(u)du`. Thus:

`I=5intsec(u)tan(u)sqrt(25sec^2(u)-25)du`

Factoring `sqrt25=5`:

`I=25intsec(u)tan(u)sqrt(sec^2(u)-1)du`

Note that since `tan^2(u)+1=sec^2(u)`, we can say that `sqrt(sec^2(u)-1)=tan(u)`, which is why we chose the original subsitution of `x=5sec(u)`. This yields:

`I=25intsec(u)tan^2(u)du`

Now, rewrite this using `tan^2(u)=sec^2(u)-1`:

`I=25intsec(u)(sec^2(u)-1)du`

`I=25intsec^3(u)du-25intsec(u)du`

The second integral is a common integral:

`I=25intsec^3(u)du-25ln(abs(tan(u)+sec(u)))" "color(red)barulabsstar`

Letting `I_1=intsec^3(u)du`, we will solve for it using integration by parts, in the form:

`intsdt=st-inttds`

So, let:

`{(s=sec(u)" "=>" "ds=sec(u)tan(u)du),(dt=sec^2(u)du" "=>" "t=tan(u)):}`

Thus:

`I_1=sec(u)tan(u)-intsec(u)tan^2(u)`

Letting `tan(u)=sec^2(u)-1`:

`I_1=sec(u)tan(u)-intsec(u)(sec^2(u)-1)du`

`I_1=sec(u)tan(u)-intsec^3(u)+intsec(u)`

Here, we see that `intsec^3(u)=I_1`, the integral we're currently solving for. Additionally, we've already integrated `sec(u)` once:

`I_1=sec(u)tan(u)-I_1+ln(abs(tan(u)+sec(u)))`

Adding `I_1` to both sides:

`2I_1=sec(u)tan(u)+ln(abs(tan(u)+sec(u)))`

`I_1=1/2sec(u)tan(u)+1/2(abs(tan(u)+sec(u)))`

Returning to `color(red)barulabsstar`, we see that:

`I=25(1/2sec(u)tan(u)+1/2(abs(tan(u)+sec(u))))-25ln(abs(tan(u)+sec(u)))`

`I=25/2sec(u)tan(u)-25/2ln(abs(tan(u)+sec(u)))`

Since we have `x=5sec(u)`, write this all in terms of `sec(u)`:

`I=25/2sec(u)sqrt(sec^2(u)-1)-25/2ln(abs(sqrt(sec^2(u)-1)+sec(u)))`

Note that `x/5=sec(u)` and `sec^2(u)=x^2/25`:

`I=25/2(x/5)sqrt(x^2/25-1)-25/2ln(abs(sqrt(x^2/25-1)+x/5))`

`I=5/2xsqrt((x^2-25)/25)-25/2ln(abs(sqrt((x^2-25)/25)+x/5))`

Factoring out `sqrt(1/25)=1/5`:

`I=1/2xsqrt(x^2-25)-25/2ln(abs(1/5sqrt(x^2-25)+x/5))`

`I=1/2xsqrt(x^2-25)-25/2ln(abs(1/5(sqrt(x^2-25)+x)))`

`I=1/2xsqrt(x^2-25)-25/2ln(abs(sqrt(x^2-25)+x))-25/2ln(1/5)+C`

Note that `-25/2ln(1/5)` is absorbed into `C`:

`I=1/2xsqrt(x^2-25)-25/2ln(abs(sqrt(x^2-25)+x))+C`

And, finally:

`I=(xsqrt(x^2-25)-25ln(abs(sqrt(x^2-25)+x)))/2+C`