trigonometric substitution integrals

[CellCoree]

what would be the appropriate trigonometic substition for the following:

1.) \int(7x^2-7)^{3/2}) dx

2.) \int \frac{x^2}{\sqrt{(7x^2 + 8)}}

ok i have a book with a list of trigonometric functions in front of me, but i dont get it. i dont see anything that i can sub in for.

ok i kinda get the first one...

7(x^2-1)^{3/2}

sin^2(\theta) + cos^2(\theta) = 1/ so...
divided by cos^2(\theta) and get tan^2(\theta)

7(x^2 - tan^2(\theta))^{3/2}

7(x^3 - tan^3(\theta))

7x^3 - 7tan^3(\theta)
7x^3=7tan^3(\theta)

divide by 7.... (lol, dont think i can do this)

x^3 = tan^3(\theta)

cube root...

x= tan(\theta) is that right?

[Pyrrhus]

Why don't you check this thread?

http://www.physicsforums.com/showthread.php?t=42657

[CellCoree]

lol, thanks. i just made a dumb mistake

[CellCoree]

\int \frac{x^2}{\sqrt{(7x^2 + 8)}}

ok this one is a bit hard....i been thinking about this one for awhile.

can i do \int \frac{x^2}{\sqrt{(7x^2 + 7 + 1)}}
?

sin^2(\theta) + cos^2(\theta) = 1 so...

well i know it's sec^2(\theta) from solving the previous problem

\int \frac{x^2}{\sqrt{(7x^2 + 7 + sec^2(\theta))}}

\int \frac{x^2}{\sqrt{7(x^2 + 1 + sec^2(\theta))}}

agian, there's another 1! so i use sec^2(\theta)
(just working with the bottom part now)

\sqrt{7(x^2 + sec^2(\theta) + sec^2(\theta)

if i squared 7, it would be a mess, can someone help?

[Parth Dave]

You might want to review trigonometric substitution in your textbook.

[Pyrrhus]

Check what are you doing again, it seems a bit... mixed up...

Here's a hint

\sqrt{7}x = \sqrt{8}tan\theta

so you should have

\sqrt{7x^2 + 8} = \sqrt{8}sec\theta

[CellCoree]

i get it now! thanks.

the trig. sub. should be:

\frac{\sqrt(8)tan(t)}{\sqrt(7)}