How can I use Trigonometric substitution to evaluate this integral?

[Vadim]

The question I'm working on is "use the appropriate Trigonometric substitution to evaluate the following integral: \int \frac{x^2}{(4x^2-49)^\frac{3}{2}}dx " Now the first thing i did was make it so that it had a root in it by making it \int \frac{x^2}{(4x^2-49)(4x^2-49)^\frac{1}{2}}dx
this then allows me to make the substitution of x=\frac{7sec\theta}{2} and dx=\frac{7sec\theta tan\theta}{2}d\theta.

making the substitution i get \int\frac{(49)(sec^2\theta) * (7)(tan\theta) (sec\theta)}{(4)(2)(49)(tan^2\theta)(7)(tan\theta)}d\theta which simplifies to \int\frac{(sec^2\theta)(tan\theta)(sec\theta)}{(8)(tan^3\theta)}d\theta = \frac{1}{8}\int\frac{sec^3\theta}{tan^2\theta}d\theta from there I've tired all sorts of different things, messed around in all the ways i know to get it so that i can integrate, but i just don't know where to go from there.

 

[OlderDan]

\frac{1}{8}\int\frac{sec^3\theta}{tan^2\theta}d\theta

I think this might help, but I think I still see some partial fractions in your future

\frac{sec^3\theta}{tan^2\theta} = \frac{sec \theta}{sin^2\theta} = \frac{cos \theta}{cos^2\theta sin^2\theta} = \frac{cos \theta}{(1 - sin^2\theta) sin^2\theta}

 

[dextercioby]

I=\int \frac{x^{2}}{\left(4x^{2}-49\right)^{3/2}} \ dx (1)

Write it like that

I=\frac{1}{49^{3/2}} \int\frac{x^{2}}{\left[\left(\frac{2x}{7}\right)^{2}-1\right]^{3/2}} \ dx (2)

and make the sub

\frac{2x}{7}=\cosh t (3)

under which the integration element behaves

2\frac{dx}{7}=\sinh t \ dt (4)

The transformed integral is

I=\frac{1}{7^{3}}\frac{7^{3}}{2^{3}}\int \coth^{2}t \ dt=\frac{1}{8}\left(\int \frac{\sinh^{2}t +1}{\sinh^{2}t} \ dt \right)=\frac{1}{8}\left(t-\coth t\right) +\mathcal{C} (5)

Reverse the substitution #3.

Daniel.