Can You Integrate This Tricky Expression?

[karush]

$\int \frac{t^3} {(2-t^2)^{5/2} } dt$
Was going to expand
$$\frac{2t}{\left(2-{t}^{2} \right)^{5 /2 } }
-\frac{t}{\left(2-t^2 \right)^{3/2 } }$$

 

[Opalg]

$\int \frac{t^3} {(2-t^2)^{5/2} } dt$
Was going to expand
$$\frac{2t}{\left(2-{t}^{2} \right)^{5 /2 } }
-\frac{t}{\left(2-t^2 \right)^{3/2 } }$$

Substitution: $t = \sqrt2\sin\theta$.

 

[Prove It]

$\int \frac{t^3} {(2-t^2)^{5/2} } dt$
Was going to expand
$$\frac{2t}{\left(2-{t}^{2} \right)^{5 /2 } }
-\frac{t}{\left(2-t^2 \right)^{3/2 } }$$

$\displaystyle \begin{align*} \int{ \frac{t^3}{ \left( 2 - t^2 \right) ^{\frac{5}{2}}}\,\mathrm{d}t} &= \frac{1}{2} \int{ t^2\, \left( 2 - t^2 \right) ^{-\frac{5}{2}}\,2t\,\mathrm{d}t } \end{align*}$

Substitute $\displaystyle \begin{align*} u = 2 - t^2 \implies \mathrm{d}u = 2t\,\mathrm{d}t \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{1}{2}\int{ t^2\,\left( 2 - t^2 \right) ^{-\frac{5}{2}}\,2t\,\mathrm{d}t} &= \frac{1}{2} \int{ \left( 2 - u \right) \, u^{-\frac{5}{2}}\,\mathrm{d}u} \\ &= \frac{1}{2} \int{ 2u^{-\frac{5}{2}} - u^{-\frac{3}{2}}\,\mathrm{d}u } \end{align*}$

Finish it.

 

[Opalg]

$\displaystyle \begin{align*} \int{ \frac{t^3}{ \left( 2 - t^2 \right) ^{\frac{5}{2}}}\,\mathrm{d}t} &= \frac{1}{2} \int{ t^2\, \left( 2 - t^2 \right) ^{-\frac{5}{2}}\,2t\,\mathrm{d}t } \end{align*}$

Substitute $\displaystyle \begin{align*} u = 2 - t^2 \implies \mathrm{d}u = 2t\,\mathrm{d}t \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{1}{2}\int{ t^2\,\left( 2 - t^2 \right) ^{-\frac{5}{2}}\,2t\,\mathrm{d}t} &= \frac{1}{2} \int{ \left( 2 - u \right) \, u^{-\frac{5}{2}}\,\mathrm{d}u} \\ &= \frac{1}{2} \int{ 2u^{-\frac{5}{2}} - u^{-\frac{3}{2}}\,\mathrm{d}u } \end{align*}$

That suggestion is much better than mine (but notice that $du = -2t\,dt$, with a minus sign).

 

[karush]

$$\frac{1}{2}\left[\frac{4}{3 u^\left\{3/2\right\}}-\frac{2}{{u}^{1/2}}\right]+C$$

$$2-{t}^{2} = u$$

$$\frac{2}{3 \left(2-t^2 \right)^{3/2} }
-\frac{1}{3\left(2-{t}^{2}\right)^{1/2} } +C$$

I hope?

 

[karush]

One thing I really like about MHB is the creative ways to solve problems that are not in textbook examples.