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

1 Answer
Sep 5, 2016

#(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#