This is just an indefinite integral and by this point we should be comfortable doing them so here is the answer to this part.

\[\int{{\cos \left( x \right) - \frac{3}{{{x^5}}}\,dx}} = \int{{\cos \left( x \right) - 3{x^{ - 5}}\,dx}} = \sin \left( x \right) + \frac{3}{4}{x^{ - 4}} + c = \require{bbox} \bbox[2pt,border:1px solid black]{{\sin \left( x \right) + \frac{3}{{4{x^4}}} + c}}\]

Don’t forget to add on the “+c” since we are doing an indefinite integral!

Recall that in order to do a definite integral the integrand (i.e. the function we are integrating) must be continuous on the interval over which we are integrating, \(\left[ { - 3,4} \right]\) in this case.

We can clearly see that the second term will have division by zero at \(x = 0\) and \(x = 0\) is in the interval over which we are integrating and so this function is not continuous on the interval over which we are integrating.

Therefore, this integral cannot be done.

Now, the function still has a division by zero problem in the second term at \(x = 0\). However, unlike the previous part \(x = 0\) does not fall in the interval over which we are integrating, \(\left[ {1,4} \right]\) in this case.

This integral can therefore be done. Here is the work for this integral.

\[\begin{align*}\int_{1}^{4}{{\cos \left( x \right) - \frac{3}{{{x^5}}}\,dx}} & = \int_{1}^{4}{{\cos \left( x \right) - 3{x^{ - 5}}\,dx}} = \left. {\left( {\sin \left( x \right) + \frac{3}{{4{x^4}}}} \right)} \right|_1^4\\ & = \sin \left( 4 \right) + \frac{3}{{4\left( {{4^4}} \right)}} - \left( {\sin \left( 1 \right) + \frac{3}{{4\left( {{1^4}} \right)}}} \right)\\ & = \sin \left( 4 \right) + \frac{3}{{1024}} - \left( {\sin \left( 1 \right) + \frac{3}{4}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\sin \left( 4 \right) - \sin \left( 1 \right) - \frac{{765}}{{1024}}}}\end{align*}\]