We went through an example in class today which was

\[\int tan^6\theta \sec^4\theta d\theta\]

In this case we took out two powers of sec and then converted all the other \(\sec\) into $latex\ tan$, which left a function of tan times \(sec^2\theta d\theta\). We wanted to do this because the derivative of \(\tan\) is \(\sec^2\) and so we can do a simple substitution. If we have an odd power of \(\tan\), we can employ a different trick. Let’s look at:

\(I=\int \tan^5\theta\sec^7\theta d\theta\).

Here, sec is an odd power and so we can’t employ the same trick as before. Now we want to convert everything to a function of \(\sec\) and have only a factor which is the derivative of \(\sec\) left over. The derivative of \(\sec\) is \(\sec\tan\), so let’s try and take this out:

\(I=\int \tan^5\theta\sec^7\theta d\theta=\int \tan^4\theta\sec^6\theta (\sec\theta\tan\theta)d\theta\).

Now convert the \(\tan\) into \(\sec\) by \(\tan^2\theta=\sec^2\theta-1\):

\[I=\int (\sec^2\theta-1)^2\sec^6\theta (\sec\theta\tan\theta)d\theta=\int (\sec^{10}\theta-2\sec^8\theta+\sec^6\theta) (\sec\theta\tan\theta)d\theta\]

where here we have just expanded out the bracket and multiplied everything out.

Now let’s use the substitution \(\sec\theta=u\implies\tan\theta\sec\theta d\theta=du\). Plugging this in, we get:

\[I=\int (u^{10}-2u^8+u^6) du=\frac{u^{11}}{11}-2\frac{u^9}{9}+\frac{u^7}{7}+C=\frac{\sec^11\theta}{11}-2\frac{\sec^9\theta}{9}+\frac{\sec^7\theta}{7}+C\]

And in general this will always be the pattern we use when come across an integral like this with an odd power of \(\tan\). Look at the following example and see if you can get this:

\(\int\frac{\tan^3\theta}{\sqrt{\sec\theta}}d\theta\).