Question

To help find the velocity of particles requires the evaluation of the indenite integral of the acceleration function, a(t), i.e. v = Z
∫a(t)dt
Evaluate the following indenite integrals: (a)
∫??lntt4dt
(b)
∫??(t2+1)sin9tdt

Answer 1

To be absolutely clear, is that
\[\int t^4 \ln t~dt~~\text{and}~~\int (t^2+1)\sin9t~dt~~?\]

Answer 2

a) \[\int\limits_{?}^{?} \frac{ lnt }{ t^4 } dt\]

Answer 3

b) yes

Answer 4

I think integrating by parts will work for both:
\[\begin{matrix}u=\ln t&&&dv=\frac{dt}{t^4}\\
du=\frac{dt}{t}&&&v=-\frac{1}{3t^3}\end{matrix}\]
Then
\[\int \frac{\ln t}{t^4}~dt=-\frac{\ln t}{3t^3}+\int\frac{dt}{t^4}\]
Oof, the second one is significantly more tedious...

Answer 5

could you show me the full working out at all please?

Answer 6

That would be doing your work for you, so no, but I'm willing to start you off. I think I've given you enough to tackle the first.

Speaking of which, there's a slight mistake, it should be
\[\int \frac{\ln t}{t^4}~dt=-\frac{\ln t}{3t^3}+\color{red}{\frac{1}{3}}\int\frac{dt}{t^4}\]

Answer 7

For the second one, you integrate by parts twice:
\[\begin{matrix}f_1=t^2+1&&&dg_1=\sin9t\\
df_1=2t~dt&&&g_1=-\frac{1}{9}\cos9t
\end{matrix}\]
So
\[\begin{align*}\int(t^2+1)\sin9t~dt&=-\frac{1}{9}(t^2+1)\cos9t+\frac{2}{9}\int t\cos9t~dt
\end{align*}\]
By parts again, letting
\[\begin{matrix}f_2=t&&&dg_2=\cos9t\\
df_2=dt&&&g_2=\frac{1}{9}\sin9t
\end{matrix}\]
Carry on...

Answer 8

Perfect :) @SithsAndGiggles

Answer 9

Thanks :3

Answer 10

what was your final answer? sorry thank you for your help, just checking I have done it right.

Answer 11

@SithsAndGiggles