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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.


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.