# Ask Uncle Colin: a nasty integral

Dear Uncle Colin,

How would you integrate $e^x \sin(x)$ (with respect to $x$, obviously)?

- Difficult Integral, Just Kan't See The Right Answer

Hi, DIJKSTRA, and thanks for your message! As seems to be the way recently, there are several ways to approach this.

### My favourite way

One of the best bits about further maths for me were things like proof by induction, reduction formulas, and setting a value on an integral to help you solve it. They all feel like cheating, somehow, and I get a small frisson of excitement every time I get to use them.

Here, we can let $I = \int e^x \sin(x) \dx$ and proceed by parts (it really doesn't matter which part we pick to differentiate.) Let's let $u = e^x$ and $\diff vx = \sin(x)$, making $\diff ux = e^x$ and $v = -\cos(x)$.

This gives us $I = -\cos(x)e^x + \int e^x \cos(x) \dx$

Repeating the trick, with $U = e^x$ and $\diff Vx = \cos(x)$, gives us $\diff Ux = e^x$ and $V = \sin(x)$. This means:

$I = -\cos(x) e^x + \sin(x) e^x - \int e^x \sin(x)\dx$.

However, that last term is just $I$! With a bit of rearrangement and tidying up:

$2I = e^x \br{\sin(x) - \cos(x)}$, so $I = \frac{1}{2} e^x \br{\sin(x) - \cos(x)}$. Plus a constant.

### The Ansatz method

I don't have a good translation for Ansatz, a word introduced to me by a German lecturer in my final year; the best I have is 'template': you make a guess about what the answer ought to look like and work backwards from there. It's very much a Mathematical Pirate trick, but it does drop out nicely.

Here, we might assume an answer of the form $e^x \br{a \sin(x) + b\cos(x)}$. The derivative of this (which ought to match the thing we're trying to integrate) is $e^x \br{(a-b)\sin(x) + (a+b)\cos(x) }$.

Matching coefficients with the integrand gives $a-b=1$ and $a+b=0$, so $a=\frac{1}{2}$ and $b=-\frac{1}{2}$ - which leads to the same answer as before.

### Complex variables!

A third option is to use the representation $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$ to make the integral $\frac{1}{2i} \br{e^{(1+i)x}-e^{(1-i)x}}$.

This integrates to $\frac{1}{2i} \br { \frac{1}{1+i}e^{(1+i)x} - \frac{1}{1-i}e^{(1-i)x}}$, which is a bit of a mess.

Let's note that $\frac{1}{1+i} = \frac{1-i}{2}$ if you realise the denominator; similarly, $\frac{1}{1-i} = \frac{1+i}{2}$. There's also a factor of $e^x$ all the way through, so I'll bring that out. This leaves us with:

$\frac{e^x}{4i} \br { (1-i)e^{ix} - (1+i)e^{-ix}}$

We could try to pattern-match $\sin(x)$ and $\cos(x)$ back onto that, but it's probably simpler to say $e^{\pm ix} = \cos(x) \pm i\sin(x)$.

We've then got $\frac{e^x}{4i} \br{ (1-i)(\cos(x)+i\sin(x))- (1+i)(\cos(x)-i\sin(x))}$.

Multiplying out: $\frac{e^x}{4i} \br{ (\cos(x)+\sin(x)-\cos(x)-\sin(x)) + i(\sin(x)-\cos(x)+\sin(x)-\cos(x))}$. There's a lot of cancelling available there - the "real" part of the bracket vanishes, and the $i$ alongside the complex part and the $i$ on the bottom of the fraction simplify out. This gives us: $\frac{e^x}{4} \br{2\sin(x)+2\cos(x)}$, which simplifies down as before.

Plus a constant.

There are probably several other ways to approach it - I'd love to hear of them - but I hope that helps.

- Uncle Colin ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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