stuck on this
use a substitution u=t^2 to solve the integration t^5/sqrt(3-t^4
So you know du = 2t dt and u^2 = t^4. Im completely stuck on where to process from that as this was given in our very first calc class and our lecturer refuses to teach us or help us with the step by step method.
Applying integration by parts first would put this into a form where this substitution is more useful...
\(\displaystyle \displaystyle \begin{align*} \int{\frac{t^5}{\sqrt{3 - t^4}}\,\mathrm{d}t} = -\frac{1}{4} \int{ t^2 \, \left( -\frac{4\,t^3}{\sqrt{3 - t^4}} \right) \,\mathrm{d}t } \end{align*}\)
So apply integration by parts with \(\displaystyle \displaystyle \begin{align*} u = t^2 \implies \mathrm{d}u = 2\,t\,\mathrm{d}t \end{align*}\) and \(\displaystyle \displaystyle \begin{align*} \mathrm{d}v = -\frac{4\,t^3}{\sqrt{3 - t^4}}\,\mathrm{d}t \implies v = 2\,\sqrt{ 3 - t^4 } \end{align*}\) and the integral becomes
\(\displaystyle \displaystyle \begin{align*} -\frac{1}{4}\int{ t^2\,\left( -\frac{4\,t^3}{\sqrt{3 - t^4}} \right) \,\mathrm{d}t } &= -\frac{1}{4} \,\left( 2\,t^2\,\sqrt{3 - t^4} - \int{ 4\,t\,\sqrt{ 3 - t^4 } \,\mathrm{d}t } \right) \\ &= -\frac{1}{2}\,t^2\,\sqrt{ 3 - t^4 } + \frac{1}{2} \int{ 2\,t\,\sqrt{ 3 - \left( t^2 \right) ^2 } \,\mathrm{d}t } \end{align*}\)
Now your original choice of substitution \(\displaystyle \displaystyle \begin{align*} x = t^2 \implies \mathrm{d}x = 2\,t\,\mathrm{d}t \end{align*}\) is more appropriate.
