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:

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

 How clear is this post?