# Browsing category integration

## Ask Uncle Colin: A trigonometric integral

Dear Uncle Colin, I'm trying to find a definite integral: $\int_0^\pi \sin(kx) \sin(mx) \dx$, where $m$ and $k$ are positive integers and the answer needs to be simplified as far as possible. I've wound up with $\left[\frac{ (k+m) \sin((k-m) \pi) - (k-m)\sin((k+m)\pi) }{2(k-m)(k+m)}\right]$, but it's been marked wrong. -- Flat

## Ask Uncle Colin: Integration rules

Dear Uncle Colin, Why can't I work out $\int \left( \ln(x) \right)^2 \dx$ using the reverse chain rule? -- Previously Acceptable, Reasonable Technique Stumbles Hello, PARTS, There are two answers to this: the first is, you can't use the reverse chain rule -- which I learned as 'function-derivative' when I

## Integrating $\sec^4(x)$

A student asks: How do you integrate $\int_{0}^{\frac{\pi}{4}} \sec^4(x) \d x$? Yuk. Let me say that again for good measure: yuk. That's going to need a trigonometric identity and, I think, a substitution. But that's ok: we can do that. Let's roll up our sleeves. Step 1: get rid of

## “A little biter of a question”

This problem came via the lovely @realityminus3 and caused me no end of problems - although I got there in the end. I thought it'd be useful to look at not just the answer, but the mistakes I made on the way. Maths is usually presented as 'here's what you

## That pesky constant

A student asks: I've got to work out: $\int \cosec^2(x) \cot(x) \d x$. I did it letting $u = \cosec(x)$ and got an answer -- but when I did it with $u = \cot(x)$, I got something else. What gives? Ah! A substitution question! My favourite -- and it sounds

## A student asks: Why is there a $+c$ when you integrate?

A student asks: We've just started integration and I don't understand why there's always a $+c$ - I understand it's a constant, I just don't understand why it's there! Great question! The simple answer is, because constants vanish when you differentiate, they have to appear when you integrate - it's