- #1
Vadim
- 15
- 0
The question I'm working on is "use the appropriate Trigonometric substitution to evaluate the following integral: [tex]\int \frac{x^2}{(4x^2-49)^\frac{3}{2}}dx[/tex] " Now the first thing i did was make it so that it had a root in it by making it [tex]\int \frac{x^2}{(4x^2-49)(4x^2-49)^\frac{1}{2}}dx[/tex]
this then allows me to make the substitution of [tex]x=\frac{7sec\theta}{2}[/tex] and [tex]dx=\frac{7sec\theta tan\theta}{2}d\theta[/tex].
making the substitution i get [tex]\int\frac{(49)(sec^2\theta) * (7)(tan\theta) (sec\theta)}{(4)(2)(49)(tan^2\theta)(7)(tan\theta)}d\theta[/tex] which simplifies to [tex]\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[/tex] 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.
this then allows me to make the substitution of [tex]x=\frac{7sec\theta}{2}[/tex] and [tex]dx=\frac{7sec\theta tan\theta}{2}d\theta[/tex].
making the substitution i get [tex]\int\frac{(49)(sec^2\theta) * (7)(tan\theta) (sec\theta)}{(4)(2)(49)(tan^2\theta)(7)(tan\theta)}d\theta[/tex] which simplifies to [tex]\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[/tex] 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.