### Video Transcript

Find the definite integral between
zero and ๐ by four of negative nine tan ๐ง times sec squared ๐ง with respect to ๐ง.

Firstly, letโs not worry that this
function is in terms of ๐ง. Weโre integrating with respect to
๐ง, so we perform the process as normal. Then, we notice that sec squared ๐ง
is the derivative of tan of ๐ง. This tells us we can use
integration by substitution to evaluate it. Weโre going to let ๐ข be equal to
tan of ๐ง. And we know that the first
derivative of tan of ๐ง is sec squared ๐ง. And whilst d๐ข by d๐ง is not a
fraction, we treat it a little like one, and we say that d๐ข is equal to sec squared
๐ง d๐ง.

And we now see that we can replace
tan of ๐ง with ๐ข and we can replace sec squared ๐ง d๐ง with d๐ข. Before we go any further though,
weโre going to need to work out what our new limits are. So, we use our substitution. We said ๐ข is equal to tan of ๐ง
and our lower limit is when ๐ง is equal to zero, which is when ๐ข is equal to tan of
zero, which is zero. Our upper limit is ๐ง is equal to
๐ by four. So, ๐ข is equal to tan of ๐ by
four, which is one. And thatโs great because we can now
rewrite our definite integral as the definite integral between zero and one of
negative nine ๐ข with respect to ๐ข.

Now, we could, if we wanted, take
negative nine out as a constant factor. Of course, we donโt need to. And if we donโt, we find that the
integral of negative nine ๐ข is negative nine ๐ข squared divided by two. Weโre going to evaluate this
between zero and one. Thatโs negative nine times one
squared over two minus negative nine times zero squared over two, which is just
simply negative nine over two. Our definite integral is equal to
negative nine over two.