1.

Starting with we note that this isn’t in the tables so perhaps we need to write it as a sum of things that are in the table and this means do a partial fraction expansion.

The first step here is to factorise the denominator:

It appears that we can’t factorise so we look to see does it have any roots using the formula: no real roots means no real factors. We examine

hence we have complex roots and hence no (real) factors.

This is not ideal but we can rescue the situation if we note that there is a similarity

.

In particular if we could complete the square

,

then we might be able to write our transform as the sum of a shifted sine and a shifted cosine.

So we do what we want to do

.

Hence we want and . Hence we have

.

Now this is a nice situation because this is exactly a shifted

,

so comes back with a shift of :

.

2.

Everything is the same down until we get

.

Now at this point we should say that perhaps we can write this as a sum of shifted cosines and sines:

.

First we make the shifted sine happen by getting an in the numerator:

.

Now we can split this

.

The first term is just the shifted cosine that we saw above while the second can be made into a shifted sine by taking out the four:

.

Regards,

J.P.

Using all the steps as per some of your examples I worked out Question 2 (iv). The answer I got was:

from

Can you let me know if this is close?

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