How do you integrate #int (25+x^2)/sqrt(x^2-4)dx# using trigonometric substitution?

Question

Elijah Abram · Accepted Answer

#int (25+x^2)/sqrt(x^2-4)dx = 27 ln abs (x+sqrt(4-x^2)) + (x+sqrt(4-x^2)) /2 + C#

Substitute #x = 2 sect# #dx =2sect tant dt#, and consider first #x in (2,+oo)# so that #t in (0,pi/2)#

#int (25+x^2)/sqrt(x^2-4)dx = 2int ((25+4sec^2t)sect tant)/(sqrt(4sec^2t-4) dt#

Use now the trigonometric identity:

#sec^2t -1 = 1/cos^2t-1 = (1-cos^2t)/cos^2t =sin^2t/cos^2t = tan^2t#

and note that for #t in (0,pi/2)# the tangent is positive so: #sqrt(sec^2-1) = tant#

and:

#int (25+x^2)/sqrt(x^2-4)dx = 1/2 int ((25+4sec^2t)sect tant)/sqrt(sec^2t-1) dt = int (25+4sec^2t)sect dt#

using the linearity of the integral:

#int (25+x^2)/sqrt(x^2-4)dx = 25 int sect dt +4 int sec^3t dt#

This are fairly known integrals, but we can go through the solution:

#int sect dt = int sect (sect + tant)/(sect+tant)dt#

#int sect dt = int (sec^2 t + sect tant)/(sect+tant)dt#

#int sect dt = int (d(sec t + tant))/(sect+tant)#

#int sect dt = ln abs (sect+tant) + C#

and:

#int sec^3 t dt = int sect sec^2t dt#

#int sec^3 t dt = int sect d(tan t) #

#int sec^3 t dt = sect tant - int sect tan^2 t dt #

#int sec^3 t dt = sect tant - int sect (sec^2 t -1) dt #

#int sec^3 t dt = sect tant - int sec^3tdt +int sectdt #

#2int sec^3 t dt = sect tant + ln abs (sect+tant) + C#

#int sec^3 t dt = (sect tant)/2 +1/2 ln abs (sect+tant) + C#

Putting it together:

#int (25+x^2)/sqrt(x^2-4)dx = 25 ln abs (sect+tant) +2sect tant + 2ln abs (sect+tant) + C#

#int (25+x^2)/sqrt(x^2-4)dx = 27 ln abs (sect+tant) + 2sect tant + C#

and undoing the substitution, considering that:

#sect = x/2#

#tan t = sqrt(1-sec^2t) = sqrt(1-x^2/4)#

#int (25+x^2)/sqrt(x^2-4)dx = 27 ln abs (x/2+sqrt(1-x^2/4)) +2( x/2sqrt(1-x^2/4)) + C#

#int (25+x^2)/sqrt(x^2-4)dx = 27 ln abs (x+sqrt(4-x^2)) + (x+sqrt(4-x^2)) /2 + C#

William Abdalla · Answer

undefined To integrate $\int \frac{{25 + x^2}}{{\sqrt{{x^2 - 4}}}} dx$ using trigonometric substitution, follow these steps:

1. Recognize the form of the integral and the need for trigonometric substitution.
2. Substitute $x = 2\sec(\theta)$, which implies $dx = 2\sec(\theta) \tan(\theta) d\theta$, and simplify the expression.
3. Express $x^2$ in terms of $\theta$ using the substitution.
4. Rewrite the integral using trigonometric identities involving secant and tangent.
5. Simplify the expression by factoring and canceling common terms.
6. Integrate the resulting expression in terms of $\theta$.
7. Express the result in terms of $x$ by substituting back the expression for $\theta$.
8. Simplify the final expression if necessary.

Following these steps, the integral can be evaluated using trigonometric substitution.

Adam Adkins · Answer

undefined To integrate $\int \frac{25 + x^2}{\sqrt{x^2 - 4}} \, dx$ using trigonometric substitution, let $x = 2\sec(\theta)$. Then, $dx = 2\sec(\theta)\tan(\theta) \, d\theta$. After substitution and simplification, the integral becomes $\int \frac{25 + 4\sec^2(\theta)}{2\tan(\theta)} \, (2\sec(\theta)\tan(\theta)) \, d\theta$. Simplify further to obtain $\int (25\sin(\theta) + 2\sec(\theta)\tan^2(\theta)) \, d\theta$. Finally, integrate term by term to get the result.