$$\int_0^1\frac{r^3}{\sqrt{4+r^2}}\ \mathrm dr$$
I have attached my work. I am stuck.
6 Answers
$\begingroup$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 $u=r^2+4$.
$\endgroup$ $\begingroup$$$\int^1_0\frac{r^3}{\sqrt{4+r^2}}\ \mathrm dr$$ 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 $$8\int^{\frac{\sqrt 5}{2}}_1 \left(u^2-1\right)\ \mathrm du$$ $$=8\left(\int^{\frac{\sqrt 5}{2}}_1 u^2\ \mathrm du-\int_1^{\frac{\sqrt 5}{2}}\mathrm du\right)$$ $$=8\left(\frac{5\sqrt 5}{24} -\frac13-\frac{\sqrt 5}{2}+1\right)$$ $$=8\left(\frac{5\sqrt 5}{24} -\frac{\sqrt 5}{2}+\frac23\right)$$ $$=\frac83\left(\frac{5\sqrt 5}{8} -\frac{3\sqrt 5}{2}+2\right)$$ $$=\frac13\left(16-7\sqrt 5\right)$$
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ $\begingroup$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 $s=4+1=5$:
$$\frac{1}{2}\int_{4}^{5}\frac{s-4}{\sqrt{s}}\space\text{d}s=$$ $$\frac{1}{2}\int_{4}^{5}\left(\sqrt{s}-\frac{4}{\sqrt{s}}\right)\space\text{d}s=$$ $$\frac{1}{2}\left(\int_{4}^{5}\sqrt{s}\space\text{d}s-\int_{4}^{5}\frac{4}{\sqrt{s}}\space\text{d}s\right)=$$ $$\frac{1}{2}\left(\int_{4}^{5}\sqrt{s}\space\text{d}s-4\int_{4}^{5}\frac{1}{\sqrt{s}}\space\text{d}s\right)=$$ $$\frac{1}{2}\left(\int_{4}^{5}s^{\frac{1}{2}}\space\text{d}s-4\int_{4}^{5}s^{-\frac{1}{2}}\space\text{d}s\right)$$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$It easy to compute: Take $\sqrt(4+r^{2})=u$ then $u^{2}=4+r^{2}$ and $udu=rdr$. Hence: $$ \int_{0}^{1}\frac{r^{2}rdr}{\sqrt{4+r^{2}}}=\int_{2}^{\sqrt5}\frac{(u^{2}-4)udu}{u}. $$
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