# Integration of $\int_0^4 2t^3\sqrt{4t^2+4}\,\,dt$

I tried integrating $\int_0^4 2t^3\sqrt{4t^2+4}\,\,dt$, by substituting $u=t^2+1$. Substitution would seem to result in $8\int_1^{17} (u^2-2u+1)u^{1/2}\,\, du$. Then I integrated and realized that this does not match with the value (I used the first integral form) I computed using calculator and wolframalpha. And I also computed my own "should-be-equivalent-but-with-substitution-of-t-with-u integral" using calculator and wolframalpha, and the results did not match up.

So what did I do wrong here?

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$t^2\sqrt{4(t^2+1)}(2t\,dt) = (u-1)\sqrt{4u}\,du$ –  Daniel Fischer Nov 8 '13 at 23:11

With $u = t^2 + 1 \implies du = 2t\,dt$

$t^3 = t^2\cdot t$, and $t^2 = (t^2 + 1) - 1 = u - 1$.

$$\int_0^4 2t^3\sqrt{4t^2+4}\,\,dt\quad = \quad\int_0^4 t^2\sqrt{4(t^2+1)}(2t\,dt)\quad = \quad 2\int_1^5 (u-1)\sqrt{u}\,du$$

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Indeed nice. +1 –  DonAntonio Nov 8 '13 at 23:25

$$u=t^2\;\;,\;\;u'=2t\\v'=2t\sqrt{4t^2+4}=\frac14\frac23(4t^2+4)^{3/2}=\frac16(4t^2+4)^{3/2}$$

so

$$\int\limits_0^42t^3\sqrt{4t^2+4}dt=\left.\frac{t^2(4t^2+4)^{3/2}}6\right|_0^4-\frac13\int\limits_0^4t(4t^2+4)^{3/2}dt=$$

$$=\frac{64}3(17)^{3/2}-\left.\frac1{24}\frac25(4t^2+4)^{5/2}\right|_0^4=\ldots$$

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+1 it nice too dear Don. –  Babak S. Nov 9 '13 at 14:42

Another method: $$\int_0^42t^3\sqrt{4t^2+4}dt = 4\int_0^4t^3\sqrt{t^2+1}$$

Substitute$$u=t^2; \,t=\sqrt{u};\, dt = \frac{1}{2\sqrt{u}}du;\, t^3 = u^{3/2}$$

$$=4\int_0^{16}u^{3/2}\sqrt{u+1}\frac{1}{2\sqrt{u}}du = 2\int_0^{16}u\sqrt{u+1}\\=2\int_0^{16}(u+1)\sqrt{u+1}-\sqrt{u+1}\,\,\,du$$

from which we find that an antiderivative for $2t^3\sqrt{4t^2+4}$ is $$\frac{4}{5}(t^2+1)^{5/2}-\frac{4}{3}(t^2+1)^{3/2}.$$

Note that $u=t^2$ is a differentiable bijection (diffeomorphism) on the relevant interval.

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