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

Textbook Question

Chapter 7, 7.3 - Problem 2 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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ishpiro | College Teacher | (Level 1) Educator

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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` .

This is the final answer.

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