what is the definite integral from 0 to 4 of x to the 3 dx over the square root of (x squared + 3)?

Answer 1

You can use a trig-substitution, but I always use u-substitutions and a bit of algebra when possible:

Let u equal the radicand, find du, then "pay the consequences" and manipulate the rest (the numerator) to make it fit. If it doesn't fit, then try something else, but in this case it does:

Let u=(x^2 + 3), du=(2x*dx)

Now you have the (1/2) * integral( (u-3) / sqrt(u) ) du, which simplifies to the difference of two simple integrals:

(1/2) * integral( sqrt(u) ) du - (3/2) * integral( u^(-1/2) ) du

For reference, I got (1/3) * 19^(3/2) - (1/3) * 3^(3/2) - 3sqrt(19) + 3sqrt(3), which simplifies to (10/3)sqrt(19) + 2sqrt(3), or approximately 17.99.

Answer 2

I do not have a textbook or calculater available to me, but i do remember something about the square root of (a^2 + b^2)
in this case a = x and b = the square root of three

If i remember right i think it was the arctan formula, but I am not sure. I hope this gets you on the right track

Answer 3

Use the trig substitution x = sqrt(3) tan(t)
dx = sqrt(3) sec^2(t) dt

You will need to integrate sec^3(t). You can find this in tables or solve it by parts.

Answer 4

I'm pretty sure you'd have to use the methods of fractions or something like that from Calc II. I did it on my calculator (use the fnInt button under Calc if you have a TI-83, 85 or 86). I got 17.99...although I'm not sure about how to exactly get the antiderivative.