# Evaluate the integral using a trig substitution: int x^3/(sqrt(4+x^2)) dx

In this case, this would be x = 2tan(t).

This is because the the trigonometric identity tan^2(t) + 1 = sec^2(t) can then be applied:

sqrt(4 + x^2) = sqrt(4 + 4tan^2(t)) = sqrt(4(tan^2(t) + 1))

= sqrt(4sec^2(t)) = 2sec(t)

Also, if x = 2tan(t), then dx = 2sec^2(t)dt...

In this case, this would be x = 2tan(t).

This is because the the trigonometric identity tan^2(t) + 1 = sec^2(t) can then be applied:

sqrt(4 + x^2) = sqrt(4 + 4tan^2(t)) = sqrt(4(tan^2(t) + 1))

= sqrt(4sec^2(t)) = 2sec(t)

Also, if x = 2tan(t), then dx = 2sec^2(t)dt and x^3 = 8tan^3(t)dt

Plugging all this into original integral, we get

int (8tan^3(t))/(2sec(t)) 2sec^2(t)dt

This simplifies to

int (8tan^3(t))/cos(t) dt

Rewriting tangent as tan(t) = sin(t)/cos(t) , we get

int (8sin^3(t))/(cos^4(t)) dt

Now we can rewrite sin^3(t)  as sin^2(t) * sin(t) = (1 - cos^2(t)) sin(t)

and use substitution:

u = cos(t)

du = -sin(t)dt

Then integral becomes

int (8(1 - u^2))/u^4 (-du) = int (8(u^2 - 1))/u^4 du

which can be broken up into two integrals of power functions:

int u^2/u^4 du = int u^(-2) du = - 1/u + C_1

and int 1/u^4 du= int u^(-4) du = -1/(3u^3) + C_2

Then the original integral will be, if we subtract the results and combine the constants into one:

-8/u + 8/(3u^3) + C

Now recall that u = cos(t)  and x = 2tan(t)

Use the Pythageorean identity again:

tan^2(t) + 1 = sec^2(t)

x^2/4 + 1 = 1/u^2

From here, u = 2/sqrt(x^2 + 4) . Plugging this into our result for the integral, we get

-4sqrt(x^2 + 4) + 1/3(x^2 + 4)sqrt(x^2 + 4) + C .