How do you integrate #int sqrt(x^2-25)# by trigonometric substitution?
1 Answer
Explanation:
We have:
#I=intsqrt(x^2-25)dx#
Let
#I=5intsec(u)tan(u)sqrt(25sec^2(u)-25)du#
Factoring
#I=25intsec(u)tan(u)sqrt(sec^2(u)-1)du#
Note that since
#I=25intsec(u)tan^2(u)du#
Now, rewrite this using
#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
#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
#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
#I_1=sec(u)tan(u)-I_1+ln(abs(tan(u)+sec(u)))#
Adding
#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
#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
#I=25/2sec(u)sqrt(sec^2(u)-1)-25/2ln(abs(sqrt(sec^2(u)-1)+sec(u)))#
Note that
#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
#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
#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#