MHB Why does the integral of √(a² +x²) need Integration by parts?

[cbarker1]

Why this integral $\int\left\{\sqrt{{a}^{2}+{x}^{2}}\right\}dx$ needs integration by parts?

Thanks

Cbarker1

 

[MarkFL]

I have edited the thread title to remove the $\LaTeX$ code that was embedded. We have measures in place to ensure that such code will not render within thread titles. It impedes searches and we ask that text be used in thread titles. :)

The given integral does not require integration by parts, as a trig. or hyperbolic trig. substitution can be used. :D

 

[HallsofIvy]

Why this integral $\int\left\{\sqrt{{a}^{2}+{x}^{2}}\right\}dx$ needs integration by parts?

Thanks

Cbarker1

As MarkFL said, this does NOT need "integration by parts". Where did you get the idea that it did? We know that, for any angle, t, [math]sin^2(t)+ cos^2(t)= 1[/math] so, dividing both sides by [math]cos^2(t)[/math], [math]tan^2(t)+ 1= sec^2(t)[/math]. So we can write [math]a^2+ x^2= a^2(1+ \left(\frac{x}{a}\right)^2)[/math] and, with the substitution [math]\frac{x}{a}= tan(t)[/math], that becomes [math]a^2(1= tan^2(t))= a^2 sec^2(t)[/math]. Of course, if [math]\frac{x}{a}= tan(t)[/math], then [math]x= a tan(t)[/math] and [math]dx= a sec^2(t)dt[/math] so [math]\int\sqrt{a^2+ x^2}dx[/math] becomes [math]\int \sqrt{a^2sec^2(t)}(a sec^2(t)dt)= a^2\int sec^3(t)dt= a^2\int \frac{dt}{cos^3(t)}[/math].

Since that is an odd power of cosine, we can integrate by multiplying both numerator and denominator by cos(t): [math]a^2\int \frac{cos(t)dt}{cos^4(t)}= a^2\int \frac{cos(t)dt}{(1- sin^2(t))^2}[/math] and the substitution u= sin(t) gives the rational integral [math]\int \frac{du}{(1- u^2)^2}[/math] which can be done by "partial fractions".