Once upon a time, @dragon_dodo asked me to help with:

$\int_{- \piby 2}^{\piby 2} \frac{1}{2007^x+1} \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x)} \dx$.

Heaven - or the other place - only knows where she got that thing from.

### Where do you even begin?

My usual approach when I don't know where to start is to start *somewhere*. My first thought was to see what happens if you reflect the function in the $y$-axis, reasoning that the second fraction is even and I don't have a clue how to integrate the first one.

A little bit of book-keeping: I'm going to call the whole integral $I$ and the second fraction $S$ , just to save on typing.

Replacing $x$ with $-x$, I get:

$I = -\int_{\piby 2}^{-\piby 2} \frac{1}{2007^{-x}+1} S \dx$

However, we can flip the limits and the sign, and multiply top and bottom by $2007^x$ to make it:

$I = \int_{-\piby 2}^{\piby 2} \frac{2007^x}{2007^{x}+1} S \dx$

That looks vaguely familiar - in fact, the only thing that differs from the original $I$ is the numerator. If we added the two expressions together, we'd get:

$2I = \int_{-\piby 2}^{\piby 2} \frac{1+2007^x}{1+2007^x} S \dx$

And that all cancels out nicely, giving:

$I = \frac{1}{2} \int_{-\piby 2}^{\piby 2} \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x)} \dx$

### A second sneaky trick

You know what? This is an even function, and we're integrating over a symmetric domain! If we restrict it to $0 \le x \le \piby 2$, it also gets rid of the horrible half out front.

$I = \int_{0}^{\piby 2} \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x)} \dx$

### Trick the third

It's already looking much nicer, but I'm not sure this particular identity is in the formula book. However, we do know that $\sin\br{ \piby 2 - x} = \cos(x)$ and we can probably use that.

Substituting $\piby 2 - x$ for $x$, I get:

$I = -\int_{\piby 2}^{0} \frac{\cos^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x)} \dx$

Again, the limits are upside down and we can change those and lose the minus sign at once:

$I = \int_{0}^{\piby 2} \frac{\cos^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x)} \dx$

Now, *that* is very similar to what we had after the second trick. Adding those two versions together gives:

$2I = \int_{0}^{\piby 2} \frac{\sin^{2008}(x) + \cos^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x)} \dx$

Oh, lookit! *That* cancels out, too.

$2I = \int_{0}^{\piby 2} \dx = \piby 2$

So $I = \piby 4$ - without doing anything much by way of integration!

## 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.