How do I find the integral using the Tables of Integrals??? Please help?

Question

integral (sqrt(7-2x^2)) / x^2

Using the tables I found that a=sqrt 7 , u=x (sqrt2). But when I plug it into the required integral its not the right answer. WHY?

Answer 1

This is of the form sqrt(a^2 - x^2)/x^2 so you would have a^2 = 7 and x-> 2x

So make the subsitution for x --> y =sqrt(2)x ---> dy = sqrt(2)dx

SO now you have sqrt(a^2-y^2)/y^2 * 4/sqrt(2) dy

Then 4/sqrt(2)*(-sqrt(7 -2x^2)/(4x^2) -sin^-1(sqrt(2/7)x))

-sqrt([7-2x^2]/2)/x^2 - 4/sqrt(2)*sin^-1(sqrt(2/7)x)

Answer 2

properly purely look at the back of your e book to work out what integrals of those varieties provide you. there's no longer too lots extra to declare actual :). i visit declare that a million/(5+y^2) will combine to an inverse tangent function. it would be something like (a million/sqrt(5))tan^-a million(y/sqrt(5)) + C. For the 2d, you're making the substitution cos^2(6t) = a million-sin^2(6t) and that provide you with: Int cos(6t)[sin^2(6t) - sin^4(6t)] dt Make a substitution u=6t, du=6dt so the quintessential turns into: (a million/6) Int cos(u)[sin^2(u) - sin^4(u)] du Now make yet another substitution w=sin(u), dw=cos(u)du. Then the quintessential turns into: (a million/6) Int w^2 - w^4 dw = (a million/6)[(a million/3)w^3 - (a million/5)w^5] + C Subbing back supplies us: (a million/18) sin^3(6t) - (a million/30)sin^5(6t) + C it fairly is the way you actual locate those integrals, yet each e book has a table of integrals the two in the front of back of the e book. reliable success!

Answer 3

To use a table of integrals you need to find an example that is as close you your specific finction as possible.
I don't know which integral you used in the table, so I can't help much, but I think you may just have some minor calculation or conversion error.