Integrate: $\int^1_0\frac{r^3}{\sqrt{4+r^2}}dr$ $$\int_0^1\frac{r^3}{\sqrt{4+r^2}}\ \mathrm dr$$
I have attached my work. I am stuck. 
 A: Hint:
$$\tan^3(\theta) \sec (\theta)=(\sec^2(\theta)) (\sec(\theta) \tan(theta)$$
and $u=\sec(\theta)$...
P.S. You can solve the original integral faster as
$$\int^1_0\frac{r^3}{\sqrt{4+r^2}}dr=\int^1_0\frac{(r^2+4-4)r}{\sqrt{4+r^2}}dr$$
and $u=r^2+4$.
A: $$\int^1_0\frac{r^3}{\sqrt{4+r^2}}\ \mathrm dr$$
Using trigonometric substitution, we have
$$r=2\tan\phi\Rightarrow \mathrm dr=2\sec^2\phi\ \mathrm d\phi$$
Now lets find the upper and lower bounds
$$1=2\tan\phi\Rightarrow \phi=\arctan\frac12$$
$$0=2\tan\phi\Rightarrow \phi=\arctan 0=0$$
So now we have
$$\int^{\arctan\frac12}_0\frac{16\tan^3\phi\sec^2\phi}{\sqrt{4+4\tan^2\phi}}\ \mathrm d\phi$$
$$=8\int^{\arctan\frac12}_0\frac{\tan^3\phi\sec^2\phi}{\sqrt{1+\tan^2\phi}}\ \mathrm d\phi$$
$$=8\int^{\arctan\frac12}_0\frac{\tan^3\phi\sec^2\phi}{\sqrt{\sec^2\phi}}\ \mathrm d\phi$$
$$=8\int^{\arctan\frac12}_0\frac{\tan^3\phi\sec^2\phi}{\left|\sec\phi\right|}\ \mathrm d\phi$$
Since $\sec\phi\geq 0$ for $\phi\in \left[0, \arctan\frac12\right]$, we have
$$8\int^{\arctan\frac12}_0\frac{\tan^3\phi\sec^2\phi}{\sec\phi}\ \mathrm d\phi$$
$$=8\int^{\arctan\frac12}_0 \tan^3\phi\sec\phi\ \mathrm d\phi$$
$$=8\int^{\arctan\frac12}_0 \tan\phi\sec\phi\left(\sec^2\phi-1\right)\ \mathrm d\phi$$
Using $u$-substitution, we have
$$u=\sec\phi\Rightarrow\mathrm du=\tan\phi\sec\phi\ \mathrm d\phi$$
So now
$$8\int^{\frac{\sqrt 5}{2}}_1 \left(u^2-1\right)\ \mathrm du$$
$$=8\left(\int^{\frac{\sqrt 5}{2}}_1 u^2\ \mathrm du-\int_1^{\frac{\sqrt 5}{2}}\mathrm du\right)$$
$$=8\left(\frac{5\sqrt 5}{24} -\frac13-\frac{\sqrt 5}{2}+1\right)$$
$$=8\left(\frac{5\sqrt 5}{24} -\frac{\sqrt 5}{2}+\frac23\right)$$
$$=\frac83\left(\frac{5\sqrt 5}{8} -\frac{3\sqrt 5}{2}+2\right)$$
$$=\frac13\left(16-7\sqrt 5\right)$$
A: Hint: $\tan^3(x)\sec (x)=\sec^2(x)\tan(x)\sec (x)-\tan(x)\sec (x)$. Then use the substitution $u = \sec (x)$. Also change the integral limits accodring to substitutions you make.
A: HINT:
$$\int_{0}^{1}\frac{r^3}{\sqrt{r^2+4}}\space\text{d}r=$$

Substitute $u=r^2$ and $\text{d}u=2r\space\text{d}r$.
This gives a new lower bound $u=0^2=0$  and upper bound $u=1^2=1$:

$$\frac{1}{2}\int_{0}^{1}\frac{u}{\sqrt{u+4}}\space\text{d}u=$$

Substitute $s=u+4$ and $\text{d}s=\text{d}u$.
This gives a new lower bound $s=4+0=4$  and upper bound $s=4+1=5$:

$$\frac{1}{2}\int_{4}^{5}\frac{s-4}{\sqrt{s}}\space\text{d}s=$$
$$\frac{1}{2}\int_{4}^{5}\left(\sqrt{s}-\frac{4}{\sqrt{s}}\right)\space\text{d}s=$$
$$\frac{1}{2}\left(\int_{4}^{5}\sqrt{s}\space\text{d}s-\int_{4}^{5}\frac{4}{\sqrt{s}}\space\text{d}s\right)=$$
$$\frac{1}{2}\left(\int_{4}^{5}\sqrt{s}\space\text{d}s-4\int_{4}^{5}\frac{1}{\sqrt{s}}\space\text{d}s\right)=$$
$$\frac{1}{2}\left(\int_{4}^{5}s^{\frac{1}{2}}\space\text{d}s-4\int_{4}^{5}s^{-\frac{1}{2}}\space\text{d}s\right)$$
A: Try $u^2=4+x^2,$ then $2udu=2xdx$ and the extra $x^2$ on top is $u^2-4,$ making an integral without a radical.
A: It easy to compute: Take $\sqrt(4+r^{2})=u$ then $u^{2}=4+r^{2}$ and $udu=rdr$. Hence:
$$
\int_{0}^{1}\frac{r^{2}rdr}{\sqrt{4+r^{2}}}=\int_{2}^{\sqrt5}\frac{(u^{2}-4)udu}{u}.
$$
