I am emailing a link of this to everyone on the class list every Friday afternoon. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.ie and I will add you to the mailing list.

## Test 2 Results

Hopefully this week.

## Schedule — Recently updated

Thursday 6 December 20:00 – 22:00: Finish double integrals and start Triple integrals

Tuesday 11 December 18:00 – 22:00: Finish off multiple integration; Exam Format Review Lecture; Tutorial; etc. Will make a plan on Friday.

## Integral

In class we found the anti-derivative $\int\cos^2x\sin x\,dx$ by using the substitution $u=\cos x$. Here we present an alternative method that manipulates the integrand instead.

Firstly we have that

$\cos^2 x=\frac12 (1+\cos 2x)$.

Hence we have

$\cos^2x\sin x=\frac{1}{2}(1+\cos 2x)\sin x=\frac{1}{2}(\sin x+\cos 2x\sin x)$.

We also have the product-to-sum formula

$\cos A\sin B=\frac12 (\sin(A+B)-\sin(A-B))$.

Hence we have

$\cos^2x\sin x=\frac{1}{2}\left(\sin x+\frac{1}{2}(\sin 3x-\sin x)\right)=\frac12 \sin x+\frac14 \sin 3x -\frac14 \sin x$

$=\frac14 \sin x+\frac{1}{4}\sin 3x$.

Now we integrate directly:

$\int \cos^2 x\sin x\,dx=\frac{1}{4}\int \sin x\,dx+\frac{1}{4}\int\sin 3x\,dx$

$\int \cos^2 x\sin x\,dx=\frac{1}{4}(-\cos x)+\frac{1}{4}\left(-\frac{\cos 3x}{3}\right)$

$\int \cos^2 x\sin x\,dx=-\frac{1}{4}\cos x-\frac{1}{12}\cos 3x+C$.

This does not look the same as the anti-derivative in the notes but there are identities relating $\cos x$, $\cos 3x$ and $\cos^3x$… these anti-derivatives are indeed equal.